This page contains the solutions to the puzzles in the blog post. They are semi randomized, so that while reading the solution of one puzzle you will not automatically read the solution of the next puzzle.
1) Prisoners dilemma
1) Label the prisoners A, B and C. Suppose that A would have had a white cap. Then B would have seen a white cap and a black cap. He can then reason: if I had had a white cap, C would immediately have noticed that there are 2 whitecaps and claimed that he has a black cap. But C does not immediately claim this. So I can not have a white cap, so I must have a black cap.
So if A would have had a white cap, B could have concluded that he has a black cap.
As B does not draw that conclusion – it remains quiet for a pretty period – A obviously can not have a white cap, and must have a black cap.
2) The most common wrong answer seems to be 5. There are just 3 socks necessary, as the first two socks can have both colors, the third must have a color already present.
|Left shore||Right shore|
|Farmer + Fox + Chicken + Lettuce|
|1||Farmer + Chicken ?||Fox + Lettuce||Farmer + Chicken|
|2||Farmer ?||Fox + Lettuce + Farmer||Chicken|
|3||Farmer + Fox ?||Lettuce||Farmer + Fox + Chicken|
|4||Farmer + Chicken ?||Lettuce||Fox|
|5||Farmer + Lettuce ?||Chicken||Farmer + Lettuce + Fox|
|6||Farmer ?||Chicken||Fox + Lettuce|
|7||Farmer + chicken ?||Farmer + Fox + Chicken + Lettuce|
Let’s abbreviate the man with “M”, his wife by “W” and the two children by “c”.
|Nr||Movement||Left shore||Right shore|
5) Camel inheritance
The mullah lends one of his own camels out, bringing the total up to 18. The oldest son receives 9, his second son 6 and his youngest son 2. The mullah takes his own camel back. This solution is possible because the sheik did not divide his entire herd.
7) Next number
The next number is 17. We have 2 series, one remaining constant, the other adds 3 to the previous number.
8) Dollar auction
Jeremy Stangroom reports that this problem was first posed by economist Martin Shubik in 1971. It has been tried and bids around 3$ are not uncommon.
The situation has been extensively analyzed, by Barry O’Neill, “International escalation and the dollar auction”, Journal of Conflict Resolution, vol. 30, pp. 33–50, 1986, by Wolfgang Leininger, “Escalation and cooperation in conflict situations: The dollar auction game revisted”, Journal of Conflict Resolution, vol. 33, pp. 231–254, 198, and also by Sandeep Pathak. The latter proofs:
For a dollar auction game with stake v units and equal wealth w units for the players, the rational course of action is for player 1 to bid (w -1) mod (v – 1) + 1 units, and for player 2 to drop out.
In more lay terms, I think there is a simple strategy for bidders to make money: If you can do the first bid, do so by bidding 99 cents. You make a 1 cent profit, and for no one it is worthwhile to overbid you.
Strange enough I did not notice this strategy mentioned in Jeremy’s book or in one of the mathematical articles.
(4 + 4/4)^4=81.
SEND MORE -----+ MONEY O = 0, M = 1, Y = 2, E = 5, N = 6, D = 7, R = 8, and S = 9. 9567 1085 -----+ 10652
11) Person B will reason: IF I had a triangular hat, A would see two triangular hats, and would know that he had a square hat. But A dos not yell this, so I must have a square hat myself.
|Left shore||Right shore|
|Farmer Fox Chicken Ctp Lettuce|
|1||Farmer Chicken Lettuce ->||Fox Ctp||Chicken Lettuce Farmer|
|2||? Farmer||Fox Ctp Farmer||Chicken Lettuce|
|3||Farmer – Fox – Ctp ?||Farmer Fox Chicken Ctp Lettuce|
14) Next number
The next number is 660.The sequence alternates subtracting 10 and doubling.
15) How many camels?
There were 23 camels to be divided. Add 1 from the mullah, and you have 24. Each of the 5 sons got 1/6=4 camels, the daughter got 24/8=3 camels, for a total of 23.
16) The Sudoku toilet paper?
Suppose Bill speaks the truth. Then Charles stole it, and lies. But both Albert and Bill would be speaking the truth, which contradicts the statement that only one person speaks the truth.
So Bill must be lying, and Charles must be innocent.
That leaves only Albert and Bill as suspects.
If Bill would be the thief, both Albert and Charles would be speaking the truth, again an impossibility, so Bill must be innocent.
If Albert would be the thief, Albert and Bill would both be lying, while only Charles speaks the truth. This means that Albert must be guilty.
18) The merchant of Baghdad
The hoggshead contains 63 gallons of water, and the barrel 31.5 of wine.
1-3. Fill the three 10-gallon bottles with wine from the 31.5 gallon barrel.
(hgh: 63wa, b:1.5wi, 10g1:10wi, 10g2:10wi, 10g3:10wi, 2g:0, 4g:0)
4. Pour the remaining 1.5 g of wine in the 2 gallon measure.
(hgh: 63wa, b:0, 10g1:10wi, 10g2:10wi, 10g3:10wi, 2g1.5wi, 4g:0)
5-20. By means of the 4g measure, fill the barrel from the hoggshead, eventually leaving 1/2g in the 4 gallon measure.
(hgh: 31wa, b:31.5wa, 10g1:10wi, 10g2:10wi, 10g3:10wi, 2g1.5wi, 4g:0.5wa)
21. Give this 1/2 gallon to camel no 1.
(hgh: 31wa, b:31.5wa, 10g1:10wi, 10g2:10wi, 10g3:10wi)
22-35. By means of 4g measure, return 28 gallons of water from the barrel to the hoggshead.
(hgh: 59wa, b:3.5wa, 10g1:10wi, 10g2:10wi, 10g3:10wi, 2g:1.5wi, 4g:0)
36. Pour 1.5 g of wine from the 2 g measure into the 4 g measure.
(hgh: 59wa, b:3.5wa, 10g1:10wi, 10g2:10wi, 10g3:10wi, 2g:0, 4g:1.5wi)
37-38. Pour 2g water from the barrel into the 2g measure and return to hoggshead.
(hgh: 61wa, b:1.5wa, 10g1:10wi, 10g2:10wi, 10g3:10wi, 2g:0, 4g:1.5wi)
39-40. Draw off remaining 1.5g of water from the barrel and give it to camel 2.
(hgh: 61wa, b:0, 10g1:10wi, 10g2:10wi, 10g3:10wi, 2g:0, 4g:1.5wi)
41. Pour 1.5g of wine from 4g measure into 2g measure.
(hgh: 61wa, b:0, 10g1:10wi, 10g2:10wi, 10g3:10wi, 2g:1.5wi, 4g:0)
42-448. Repeat the moves 5-37 11 more times (407 moves), so that six camels shall have each received two ½ gall. drinks, and another six camels two 1½ gall. drinks. Eight camels have now received 3 gall. each, and
four camels 1 gall. each, and there is 35 gall. of water in hhd.
(hgh: 35wa, b:0, 10g1:10wi, 10g2:10wi, 10g3:10wi, 2g:1.5wi, 4g:0)
449-464. Fill barrel from hogshead, using 4-gall. measure
(hgh: 1wa, b:31.5wa, 10g1:10wi, 10g2:10wi, 10g3:10wi, 2g:1.5wi, 4g:0.5wa)
465. Give ½ gall. over to camel No. 13
466. Draw 3 gall. from hogshead into 4-gall. measure
467-478. Return all wine to hogshead. Empty barrel into three 10-gall. bottles, and draw remaining 1½ gall. into 2-gall. measure. Return contents of three bottles to barrel, pour 1½ gall. from 2-gall. measure into bottle No. 1 (12 manipulations).
479-491. Fill the 2-gall. measure from 4-gall., leaving 1 gall. in 4-gall. Fill barrel from 2-gall. measure, and give remaining ½ gall. to camel No. 13. Give five camels 2 gall. each, all the camels have now been served (13 manipulations).
492-496. Fill the two empty bottles from barrel, and draw remaining 1½ gall. into bottle No. 1. Return contents of bottles Nos 2 and 3 to barrel (5 manipulations).
496-506. Pour 1 gall. from 4-gall. measure into No. 2 bottle.
Put 6 gall. wine in bottle No. 3, using 2-gall. and 4-gall. measures.
Empty the 1 gall. from No. 2 into 4-gall. measure and fill up that measure with wine from bottle No. 3.
Pour contents of 4-gall. measure into bottle No. 2.
Draw 2 gall. water from barrel and put into bottle No. 2
The thirteen camels have now each received 3 gall. of water, one of the 10-gall. bottles contains 3 gall. of water, another 3 gall. of wine, and the third 3 gall. of wine and 3 gall. of water mixed. The hogshead contains 25½ gall. of wine, and the barrel 18 gall. of water. Total number of manipulations: 506.
(footnote by Martin Gardner in his Dover reprint: In an interview published in The Strand magazine, April 1926, Henry Dudeney, England’s great puzzlist, disclosed that Loyd once appealed to him for help on this problem. Loyd had offered cash prizes to his readers for the best solution and was anxious to avoid giving them by having an answer of his own that topped all those received. Dudeney worked out a solution in 521 moves which he later reduced to the 506 given above. This did the trick and Loyd always claimed that Dudeney had saved him thousands of dollars)
20) Next number
The next number is 25. Alternately the series halves the previous number and subtracts 10.
22) The answer is of course still 3. As soon as you have 3 socks, there must be 2 of them with the same colour.
23) The monk at Koningsberg
Label the bridges as indicated:
If the monk starts at the south shore, the possible combinations are:
|Left shore||Right shore|
|Farmer Fox Chick Spd Ctp Lettuce|
|1||Farmer Chicken Ctp ?||Fox Spd Lettuce||Farmer Chicken Ctp|
|2||? Farmer||Fox Spd Lettuce Farmer||Chicken Ctp|
|3||Farmer Spd Lettuce ?||Fox||Chicken Ctp Farmer Spd Lettuce|
|4||? Farmer Chicken Ctp||Fox Farmer Chicken Ctp||Spd Lettuce|
|5||Farmer Fox ?||Chicken Ctp||Spd Lettuce Farmer Fox|
|6||? Farmer||Chicken Ctp Farmer||Fox Spd Lettuce|
|7||Farmer Chicken Ctp ?||Farmer Fox Chick Spd Ctp Lettuce|
It is no coincidence that the number of moves is identical to the original problem. The solution follows the same mechanism. The mechanism can be generalized to a boat and a farmer with n objects, with 2n+1 objects to be transported, while object i is at odds with the objects i-1 and i+1.
25) Animals in the house
Dodgson gives as the solution: I always avoid a kangaroo. I admit I did not check this solution, and I welcome any confirmations or, or course, other findings.
26) The white waste-paper basket
Mr Brown declares: Both Green and I are guilty.
Mr Green declares: Brown stole it.
As one of them is lying and one of them is speaking the truth, it is clear that Brown is lying. So Green is speaking the truth, and Brown stole it.
28) Next number
The next number is 28. We have two alternating series, one adds 8, the other doubles.
29) Jimmy’s lemonade
There 2 solutions:
12588 6478 6478 98783 183 ------+ 124510
31) Five pirates
Let´s call the pirates A, B. C, D and E, with E having the highest rank.
There is a powerful heuristic (general problem solving method) which suggests to start at the end. The end is in this puzzle:± suppose there is only 1 pirate left. That would be A. A would have all 100 gold pieces.
What when there are 2 left?
B can make any offer he wants, because if the votes are equally divided, his offer will be accepted. So B can offer 0 to A, and 100 to himself.
So B’s offer will be:
B’s proposal: 0 to A, 100 to B.
What if there are 3 pirates left? There is no way C can convince B to vote for his proposal, but A will already be satisfied with 1 gold piece, because B would offer him nothing.
C’s proposal: 1 to A, 0 to B, 99 to C.
What if there are 4 pirates? D needs only 1 other vote besides his own. Obviously offering 1 gp to b will suffice.
D’s proposal: 0 to A, 1 to B, 0 to C, 99 to D.
And now we get to our exercise: 5 pirates.
E has to get 2 votes aside from his own. Offering 1 gp to both A and C suffices.
E’s proposal: 1 to A, 0 to B, 1 to C, 0 to D and 98 to E.
32) With 4 socks you can still have 4 of the same colour. To make sure you have at least two colours, you must take at least 5 socks.
|Left shore||Right shore|
|Warrior Lion Tiger Leopard|
|Warrior Lion Tiger Leopard|
|Warrior Lion Tiger ?|
|? Warrior Tiger|
|Warrior Leopard Tiger ?|
34) Birds in the aviary
The most complicated (in the sense of number of steps needed) valid conclusion is: No bird in this aviary lives on mince-pies.
35) Next number
The next number is 3. Again we have two alternating sequences, one subtracting 5, the other dividing by 2.
36) The way
If he would point at either way and ask straight way: “Is this the way to Utopia?” the answer might be a yes or no, but he would have no clue if it was true or not.
If he would ask: do you speak the truth, the answer would always be yes, and he would gain nothing.
But if he points to one of the two ways and asks: “Would the other say that this is the way to Utopia?”, there would always be one lie included in the answer. That is, if the guard he asks it tells the truth, the other would lie about it, and the guard whom he asked the question would truthfully say what the lier would answer. If the other guard always tells the truth, he would truthfully answer yes or no, but the guard being asked would lie about it. So if the answer is No, the hiker should take that road, and if the answer is Yes, he should not.
37) Price labels
The relation between price and formula is price = weight x (weight+1). So 20=4×5, and the answer is 4.
39) Captain Brownbeard’s hourglasses
He does not need to make any preparations in advance, except having both hourglasses ready when the captain retires.
0: Start both glasses
3: End of 3 hour glass. Turn the 3 hour glass
5: End of 5 hour glass. Turn the 5 hour glass
6: End of 3 hour glass. The 5 hour glass now has 1 hour of sand. Turn the 5 hour glass!
7: End of 5 hour glass.
0. Start both glasses.
3. 3 hour glass ends. Turn the 3 hour glass
5. 5 hour glass ends. The 3 hour glass has 1 hour left. Turn the 3 hour glass, so that it will run again for 2 hours.
7. The 3 hour glass ends.
The second method is 1 step shorter, but both methods take just 7 hours.
41) Next number
The next number is 20. The second number is twice the first number, the third number is 2 less than the second number, and the operations x2 and -2 are applied repeatedly.
42) This reverses the situation. Suppose there are N socks in both colours. The number you must pick up to have 2 socks of the same colour is 3. The number to get socks in 2 different colours is N+1. So N+1=3, which means that N=2.
43) The soldiers and the 2 boys.
Denote the boys by b and the soldiers by S.
|Nr||Left shore||Movement||Right shore|
44) Showy talkers
Showy talkers are not really well informed.
45) The seventeen horses
The puzzles states that the brothers should receive the proportions of 1/2, 1/3 and 1/9th. It does not state that they should receive the fractions of seventeen. So if they receive, 9, 6 and 2 the will is carried out exactly. Therefor the ridiculous old method does carry out the will exactly.
46) The blue towel
It will be clear that we will have to do with Dirty Dave’s statements: “I am innocent. Big Barry did it.”. We know that at least one of them must be a lie. If his first statement would be true, his second statement would also be true, which contradicts what we know, so the first statements can not be true. If it is false, his second statement is also false, which is a situation allowed by the description. So his first statement is a lie, and he is guilty.
47) The Wassail bowl
The division of the twelve pints of ale can be made in eleven
manipulations, as below. The six columns show at a glance the quantity
of ale in the barrel, the five-pint jug, the three-pint jug, and the
tramps X, Y, and Z respectively after each manipulation.
Barrel. 5-pint. 3-pint. X. Y. Z.
7 .. 5 .. 0 .. 0 .. 0 .. 0
7 .. 2 .. 3 .. 0 .. 0 .. 0
7 .. 0 .. 3 .. 2 .. 0 .. 0
7 .. 3 .. 0 .. 2 .. 0 .. 0
4 .. 3 .. 3 .. 2 .. 0 .. 0
0 .. 3 .. 3 .. 2 .. 4 .. 0
0 .. 5 .. 1 .. 2 .. 4 .. 0
0 .. 5 .. 0 .. 2 .. 4 .. 1
0 .. 2 .. 3 .. 2 .. 4 .. 1
0 .. 0 .. 3 .. 4 .. 4 .. 1
0 .. 0 .. 0 .. 4 .. 4 .. 4
And each man has received his four pints of ale.
48) Four fours
4×4 / (4+4)=2
(4! +4) / 4 + 4=11
4*(4-4/4) = 12
4×4 – 4/4 = 15
4×4 / (4/4) = 16
4×4 + 4/4 = 17
4×4 + 4 – sqrt(4) = 18
42 +4 – 4/4 = 19
4*(4+4/4) = 20
4×4 +4 +sqrt(4)=22
4! – (4 / (sqrt(4) + sqrt(4)))=23
4×4 +4+4 = 24
4! + (4 / (sqrt(4) + sqrt(4)))=25
4! + (4+4)/4=26
4! + 4-4/4=27
(4+4/4)! / 4 = 30
4! +4 +sqrt(4) +sqrt(4) = 32
For those who are interested, you can make all even numbers in the range 0-100.
51) Next number
The next number is 20. Alternatively, we add 2 and add 3 to the previous number.
53) Officer and men
53) If you answered 4*537+1=2149, you forgot the officer. The solution is of course similar to nr 43. They need 4 trips for every soldier, and 4 trips for the officer, to get everyone across, for a total of 4*538+1=2153 trips.
54) Five pirates again
If there is only 1 pirate, this will be A, and he receives all gold.
A’s offer would be: 100 to A.
If there are two pirates, A and B, A will reject whatever B proposes, as this will not only give him all god but also B’s death.
If there are three pirates, C will neven be able to satisfy A, but B will be glad to stay alive, even with no gold.
So C can offer: 0 to A, 0, to B, and 100 to C.
If there are 4 pirates, D must get support from two other pirates to get his proposal accepted. Those two pirates must be A and B, as C’s position can not be topped.
So D’s offer would be: 1 to A, 1 to B, 0 to C and 98 to D.
If there are 5 pirates, E needs 2 other votes vesides his own.
So E’s best proposal would be:
2 to A, 0 to B, 1 to C, 0 to D and 97 to E.
55) Scented flowers
The valid conclusion is that all flowers grown in the open are scented.
56) The stolen washing-glove
If All is guilty, Bill speaks the truth, so All lies and Charley tells the truth. That makes two truth-tellers, so All is innocent
If Bill is guilty, Bill lies, All lies, and Charley tells the truth. So only one of them tells the truth, which makes Bill the thief.
Just as a check:
Suppose Charley did it. Then Bill tells the truth and Al;l tells the truth, which makes two people tells the truth, which cvan not be.
58) Next number
The next number is 95. Each number is twice the previous number+1.
The surgeon was his mother.
This is of course an incredibly easy puzzle, and relies heavily on cultural bias. I once read that even many feminists have problems to come to the right conclusion quickly.
There are socks in 5 different colours. Having N different colours simply means that you must take out N+1 socks.
63) Family Softleigh
– The two boys go over, 1 boy returns.
– Dad goes over, other boy returns.
– two boys cross, and Ma gets over on the same way as dad.
– one of the boys takes over the dog
– the two boys cross
64) Next number
The next number is 8. To get a new number, divide the previous one by 2 and add 3.
None of your sons are fit to serve on a jury.
66) The blue eye paper envelope
Mr Black: “I am innocent, inspector. It was White who stole the envelope.”
Mr Green: “Black is innocent, inspector. Black is lying when he says White is guilty.”
Mr White: “Black is innocent. Green is innocent.”
Suppose White is guilty. Then White would be telling the truth twice, which is impossible. So White is innocent. That Leaves Black and Green as suspects.
Suppose Black is guilty. Then both Blacks sentences are lies. Green tells at least one lie when he claims Black is innocent. So does White.
Suppose Green is guilty. Then Blacks second sentence is a lie. The first two sentences of Green and White are true. So Green is telling the truth twice, which can not be.
67) Divide 24 liters in three equal portions
Call these A, B and C respectively and call the 25 tank T.
postions: T A B C
start: 24 0 0 0
nr. 1: 11 0 0 13
nr. 2: 11 5 0 8
nr. 3: 11 5 8 0
nr. 4: 16 0 8 0
nr. 5: 3 0 8 13
nr 6: 3 5 8 13
Henry Dudeney remarks there are several solutions with 6 moves.
8 x 2 = 46416
6 + 3 = 23618
15 : 3 = 513545
9 – 3 = 38127
each result starts with a/b and ends with axb. In between you find the product of these two.
12 x 4 = 3 (=12/4) concatenated with 144 (=3×48) concatenated with 48 (=12×4).
70) Japanese cryptarithm
72) Next number
The next number is 112. To get a new number, add 2 and double the previous number.
73) The cargoship, 2 crewmen, peanuts, pigs & lilies.
|John, pea||Jack, ship, pig||Lil|
|1||Ship+Jack ? NP||John, Jack, Ship, pea||Pig,||Lil|
|2||Ship+Jack+John ? SP||Pea||Pig, Ship, John, Jack,||Lil|
|3||Ship+John ? NP||Pea, ship, John||Pig, Jack||Lil|
|4||Ship+John+pea ? Island||–||Pig, Jack||Lil, pea, ship, john.|
|5||Ship+John + Lil? SP||–||Pig, lil, jack, john, ship||Pea|
|6||Ship+John+Jack->island||–||Pig, lil||Pea. Ship, Jack, John|
|7||Ship+Jack ? SP||–||Pig, lil, ship, Jack||Pea, John|
|8||Ship+Jack+Pig ? NP||Pig, ship, jack||Lil||Pea, John|
The conclusion is, not surprisingly, that wasps are unwelcome.
75) New coats
Conclusion d, All my new coats are closed with zips, is the only valid conclusion.
76) The stolen chocolate
Suppose Cindy stole it. Then Anna and Belinda both tell the truth. Cindy herself lies, and both Diana and Elizabeth tell the truth.
If Belinda stole the chocolate, then Anna, Belinda, and Cindy lie. Diana tells the truth, and Elizabeth lies.
78) Two ropes
At time=0, light one rope at both ends. Light the other rope at one end.
After 30 minutes, the first rope has burnt out completely. We don’t know where, but it has taken 30 minutes. Now also light the other end of rope 2, which still has 30 minutes left. Burning at both ends, it will burn for 15 minutes.
79) Next number
The next number is 8. To get a new number, add 10 and divide by 3.
80) Number squares
The number in the center is the difference between the products of the corners and the product of the numbers along the edges.
So the answer is 150.
81) Gbrainy 1
So the correct solution is ACE
82) Classic variation train shunting
1-3: The engine moves left, picks up wagon A, moves down over the right half of the circle, and pushes A on the lower straight track.
4-5: The engine moves left, picks up wagon B, pulls it to the middle of the right half of the circular track, leaves it there and drives on through the tunnel to the lower straight track, connecting to A.
6-7: The engine picks up A, takes it to the left, then moves right to the right half of the circle, connecting wagon A with wagon B.
8-9: The engine takes both wagons left, then moves them to the right on the straight track. The sequence from left to right now is: engine – A – B.
10-11: The engine takes wagon A, and parks it in the middle of the right half of the circular track
12: The engine moves down, through the tunnel to up, and then pushes B down to the lower part of the circular track.
13-15:The enige moves right, picks up wagon B, moves back, pushes it over the right half of the circular track to the upper part, and moves alone to the straight upper track.
|Left shore||Right shore|
|Farmer Dog Fox Goose Grain|
|1||Farmer Fox ?||Dog Goose Grain||Farmer Fox|
|2||Farmer ?||Farmer Dog Goose Grain||Fox|
|3||Farmer Goose ?||Dog Grain||Farmer Goose Fox|
|4||Farmer Fox ?||Farmer Dog Fox Grain||Goose|
|5||Farmer Dog ?||Fox grain||Farmer Goose Dog|
|6||Farmer ?||Farmer Fox Grain||Goose Dog|
|7||Farmer Grain ?||Fox||Farmer Dog Goose Grain|
|8||Farmer ?||Farmer Fox||Dog Goose Grain|
|9||Farmer Fox ?||Farmer Dog Fox Goose Grain|
84) Next number
The next number is 18. Each number is the sum of the two previous numbers.
85) Old jacks
The correct answer is c. Some leather jacks are brown. From A you can not deduce that all old jacks are made of leather, while new leather jacks may have different colours.
86) Crack the code
87) Cross-sum: Physics
4 7 3
8 2 4
2 5 7
88) Mastermind: 4 colours, 3 spots
<img src="https://justpuzzles.files.wordpress.com/2011/02/mastermind-2013-11-07-4-on-3-solution.jpg" alt="Mastermind 2013-11-07 4 on 3 solution" width="144"
height=”188″ class=”alignleft size-full wp-image-2950″ />
90) Three glasses – again
Tilt the three oz glass so that it remains exactly half full. This will be when the water touches the lower point of the open end and the top of the bottom. 1,5 remains.
Do the same with the 5 oz glass – 2,5 remains
Now empty the contents of the 5-glass in the three glass – exactly 1 oz will remain.
91) Missionaries and Cannibals
Denote the cannibals by C’s, and the missionaries by M’s.
|Left shore||Right shore|
|5||MM –>||MC||MMCC||6||MC <–||MMCC||MC||7||MM –>||CC||MMMC||8||C <–||CCC||MMM||9||CC –>||C||MMMCC||10||C <–||CC||MMMC||11||CC –>||MMMCCC|
Note that the demands can be sharpened by requiring that only one of the missionaries and only one of the Cannibals can row.
92) The pastor and the penitence
(solution still under construction)
Label the bridges A through G as in the following figure.
Starting with bridge E, possible routes are:
Aside from this, the combinations starting with bridge F are:
The routes starting with bridge G are similar to those starting with bridge F. So there are 12 + 2*14=40 ways to cross all bridges once, starting on the southern riverbank and ending on the northern shore.
93) Geometry pattern
All circles are marked with A, all ellipses with B.
There are two sizes, small ones have F, large ones have E.
If the three shapes are in a horizontal or vertical line, they are marked with C, if they are on a diagonal line they are marked with a D.
So the correct solution is BDF.
94) Next number
The next number is 27. The first one is 1×1+2, the second one 2×2+2, the third one 3×3+2, and so on.
95) Summer tourists
In that complicated puzzle of the summer tourists who had to cross a stream in a boat which would hold but two, and where certain personal feeling added to the difficulties of the strained relations, it can be shown that the entire party can be ferried across the stream in seventeen trips as follows:
First – Mr. and Mrs. C. cross over.
Second – Mr. C. returns alone.
Third – Mr. C. takes over a lady.
Fourth – Mr. C. returns with his wife.
Fifth – Mr. C. takes over another lady.
Sixth – Mr. C. returns alone.
Seventh – The two gentlemen cross over.
Eighth – Gentlemen and wife return
Ninth – Mr. and Mrs. C. cross over
Tenth – Gentlemen and wife return
Eleventh – Two gentlemen cross over.
Twelth – Mr. C. comes back alone.
Thirteenth – Mr. C. takes lady over.
Fourteenth – Mr. and Mrs. C. return
Fifteenth – Mr. C. takes lady over.
Sixteenth – Mr. C. returns alone.
Seventeenth – Mr. C and wife go over and the entire party have been trasnported to the other side.
96) Dirty pigs.
All conclusions are valid.
97) Cross-sum: Bible
4 0 9
1 5 3
8 8 1
98) The brook
Call the vessels A and B
0 1 – 3rd move
0 2 – 7th move
0 3 – 11th move
and so on. It will take 4n-1 moves to measure n pints, or 23 moves for 8 pints if he wants to bring back the other empty.
101) Three jealous couples
Call the man A, B and C, and their respective girl friends a,b and c.
|Left shore||Right shore|
|5||AC –>||Bb||ACac||6||Aa <–||ABab||Cc||7||AB –>||ab||ABCc||8||c <–||abc||ABC||9||ab –>||c||ABCab||10||b <–||bc||ABCa||11||bc –>||ABCabc|
102) What comes next?
Next is 78. These are the speeds of music records.
104) Next number
The next number is 33. The first difference is 4, the second difference is 5, the third difference is 6, and so on.
105) Geometric figure: four angles
All figures with right angles have an A, those without have a B.
The figures with four sides of equal length all have a C, those without have a D.
The figures with a horizontal side have letter E, those without have letter F.
So the correct solution is ADF.
106) The switch problem
First engine F passes alone through the switch via C, B, A (two moves),
pulls engine E to D and once more passes through switch via C, B, A (total five moves);
pulls car D to D, pushes E out to the right; passes again through switch (eight moves);
pulls C* to D, pushing others out to right, engine goes through switch again (eleven moves);
pulls B to D, and passes through switch as before (fourteen moves);
pulls A to D and passes through switch (seventeen moves);
goes to right, the draws A, B, C, D, E, G to left and backs G onto switch (twenty moves);
draws A, B, C, D, E to left, backs them to right (twenty-two moves);
goes to left alone, backs up on switch at A and takes G to left (twenty-four moves);
goes to right, then pulls everything out to left, backs H, I onto switch (twenty seven moves);
pulls G, A, B, C, D, E out to left, backs them to right, connecting them to H, I** (thirty-one moves);
and is now prepared to go ahead on the thirty second move.
There are some minor inaccurancies in Loyds solution, namely:
*) Loyd had G instead of C here.
**) Loyd had here: “pulls G, A, B, C, D, E out to left, backs them to right, then takes F to right and backs up to switch and connects G to H, I”
107) Cross-sum: games
3 2 4 6
6 4 2 3
2 3 6 4
4 6 3 2
108) 3, 4 and 5
1. fill 3q can (result: 3,0)
2. empty 3q into 5q can (result: 0,3)
3. Fill 3q can (result: 3,3)
4. Fill 5q can from 3q can (result: 1,5)
5. Empty 5q can into tank (result: 1,0)
6. empty 11 from 3q can into 5q can (result: 0,1)
7. fill 3q can (result: 3,1)
8. add 3q can to 5q can (result: 0,4)
Suppose Arrogant Andy is guilty. Then his first statement is a lie, and his second statement is true
The two statemente by Boasting Billy would be both lies, a perfectly possible situation.
Suppose Chirping Charles is guilty. Then both Andy’s statements are lies. Billy’s statements would both be true, which is not possible. Hence Chirping Charles is innocent, and Boasting Billy must be the thief.
111) A boat for three
It is obvious that there must be an odd number of crossings, and that if the five husbands had not been jealous of one another the party might have all got over in nine crossings. But no wife was to be in the company of a man or men unless her husband was present. This entails two more crossings, eleven in all.
The following shows how it might have been done. The capital letters stand for the husbands, and the small letters for their respective wives. The position of affairs is shown at the start, and after each crossing between the left bank and the right, and the boat is represented by the asterisk. So you can see at a glance that a, b, and c went over at the first crossing, that b and c returned at the second crossing, and so on.
|Left shore||Right shore|
|1||abc –>||ABCDE de||abc|
|2||bc <–||ABCDEbc de||a|
|3||bcd –>||ABCDE e||abcd|
|4||d <–||ABCDE de||abc|
|5||ABC –>||DE de||ABCabc||6||Cc <–||CDE cde||AB ab||7||CDE –>||cde||ABCDEab||8||b <–||bcde||ABCDE a||9||bcd –>||e||ABCDEabcd||10||bc <–||bc e||ABCDEa d||11||bce –>||ABCDEabcde|
There is a little subtlety concealed in the words “show the quickest way.”
Everybody correctly assumes that, as we are told nothing of the rowing capabilities of the party, we must take it that they all row equally well. But it is obvious that two such persons should row more quickly than one.
Therefore in the second and third crossings two of the ladies should take back the boat to fetch d, not one of them only. This does not affect the number of landings, so no time is lost on that account. A similar opportunity occurs in crossings 10 and 11, where the party again had the option of sending over two ladies or one only.
To those who think they have solved the puzzle in nine crossings I would say that in every case they will find that they are wrong. No such jealous husband would, in the circumstances, send his wife over to the other bank to a man or men, even if she assured him that she was coming back next time in the boat. If readers will have this fact in mind, they will at once discover their errors.
113) Primitive Railroading.
Sam Loyd gives the following solution:
- Back the R engine far out to te right.
- Run the R engine on to the switch.
- Run the L engine with three cars onto the right.
- R engine back to the main track.
- R. engine out to the left, with three cars to the left of the switch.
- L engine pn to the switch.
- R engine and cars to the right.
- R engine pulls seven cars to the left.
- L engine runs to the main track.
- L. engine backs the train (more accurate: L. engine backs till it hooks on to the train).
- L engine pulls five cars to the right of the switch.
- L engine backs rear car on to the switch
- L engine draws four cars to right.
- L engine backs four cars to left.
- L engine alone goes to right.
- L engine backs to switch.
- L engine pulls car from switch to track.
- L engine backs to left.
- L engine goes forward with six cars.
- L engine backs rear cars on to switch.
- L engine goes to right with five cars.
- L engine backs five cars to left.
- L engine goes to right with one car.
- L engine backs on to switch.
- L engine goes to right with two cars.
- L engine backs to left of switch.
- L engine draws seven cars to right of switch.
- L engine backs end car on to switch.
- L engine goes to right.
- R train backs to right.
- R train picks up its four cars and skips.
- L train backs to switch.
- L train picks up its third car and goes its way rejoicing.
114) Next number
The next number is 41. Starting with the third number in the row, each number is the sum of the two previous numbers + 1.
115) The cabdrivers
As the answer to the question “Are you a TruthTeller” is Koa, it is clear that Koa means Yes, as the answer to this question would be Yes for both TruthTellers and LieSpeakers.
As the second driver answers Loao, this word can only mean No, and it implies that the two man are different types: one of the two must be a TruthTeller, the other a LieSpeaker.
As the second cab driver would always say that he is a TruthTeller, and the first cab driver says Loao=no, the first cab driver must be a LieSpeaker and the second cab driver a TruthTeller.
116) Coffee with milk, please
The following solution comes from Pratik Poddar:
Let the total milk consumed be m and the total coffee consumed be c
m/4 + c/6 = 1 cup
Total number of cups consumed = m+c = t
t – c/3 = 4
t + m/2 = 6
Since t is an integer, and t > 4 and t < 6
We get t = 5. Total number of people in the family = 5
A. A dog who playes basketball 1 Air Bud
B. Son of Gray Wolf and Kazan 3 baree
C. Laura Ingalls’ second dog in Little House on the Prairie 2 Bandit
D. Fictional collie dog character created by Eric Knight 5 lassie
E. First dog in Married … With children 4 buck
A. Hendrik Lorentz 4. Physics
B. Euclides 5. Mathematics
C. Ivan Pavlov 1. Physiology or Medicine
D. Carolus Linnaeus 3. Botany
E. George Bernard Shaw 2. Literature
Only two things are infinite, the universe and human stupidity, and I’m not sure about the former. – Albert Einstein
Faith is taking the first step even when you don’t see the whole staircase. – Martin Luther King
Blood, sweat, toil and tears – Winston Churchill
By all means, marry. If you get a good wife, you’ll become happy; if you get a bad one, you’ll become a philosopher. – Socrates
Love is a fruit in season at all times, and within reach of every hand – Mother Teresa
A: G. K. Chesterton – father Bronw
B: Agatha Christi – hercule Poirot
C: Ruth Rendell – Chief Inspector Reginald Wexford
D: Isaac Asimov – Henri
E: Dorothy Leigh Sayers – Lord Peter Wimsey
A: tri – 3
B: edno – 1
C: chetiri – 4
D: dve – 2
E: pet – 5
I: Antonio Lucio Vivaldi 1678
II: Johannes Brahms 1833
III: John Lennon 1940
IV: Johann Christian Bach 1735
V: Wolfgang Amadeus Mozart 1756
I: Maple Tree 3. maple syrup
II: black willow 5. substitute for quinine
III: Pará rubber tree 2. rubber
IV: Scots Pine 4. Turpentine
V: spruce trees 1. Whitewood
V England 5 Beef in Beer
IV germany 3 sauerbraten
III greece 4 horta
II Italy 1 pizza
I spain 2 tapas
I Sawao Kato – 4 Gymnastics
II Michael Phelps – 2 swimming
III Elisabeta Lipa – 1 rowing
IV Larisa Lazutina – 3 Cross-country skiing
V Anton Geesink – 5 Judo
118) The barrel puzzle
Tilt the barrel till the water reaches the point where it just doesnt stream out of the barrel. If you can see the bottom, it is less then half full. If you can only see the degs, it is more than half full.
121) Two bigamists
Call the man A and B, and their respective wives a, a, b, and b.
|Left shore||Right shore|
|8||aB <–||aa||ABbb||9||aa <–||AaaBbb|
As you can see, A and a have been underlined. A short analysis of the solution will show you that these are the only two persons which are required to row.
122) What comes next?
The logic behind the sequence is Ten, twenty, Thirty, Forty, Fifty, Sixty, Seventy. So next is E of Eighty.
123) The contractor on the northern shore
The secret is that the northern shore and the island at the right, with the pub, should remain having an uneven number of connections, while the other two should be brought in a state with an even number of connections. There is only 1 solution:
124) Next number
The difference between two consecutive numbers are 3, 5, 7, 9, so the next number will be 26+11=37.
125) The four elopements
The following text come straight from Sam Loyds cyclopedia:
Contrary to the published answers to the famous puzzle of the four couples,who had to cross a river in a boat which would carry but two persons at a time, the feat can be performed in 17 trips, instead of 24. Utilizing the island in the middle of the stream and complying with the conditions that no young lady was to be in the company of any gentlemen unless her ffinacee was present, and no man was to be alone in a boat, if any young lady was left alone, except to the one to whom she was engaged.
Describing the young men as ABCD, and the young ladies as abcd, the 17 trips van readily be followed.
Now the men begin rowing
126) A farmer went to the market
A farmer buys 100 animals for 100 dollars but lost his receipt. Cows are $10 each, pigs are $3 each and chicks are $.50 each. How many of each did he buy?
Abbreviate the number of pigs as p, the number of the number of cows as c, and the number of chicks as h (hens).
Then c+p+h=100, and 10c+3p+h/2=100.
Combining these 2 equations gives us: 120c+6p+100-c-p=200, or 19c+5p=100. –> p=20-19c/5, or p=1, c=5 and thus h=47.
127) A prohibition poser
First fill and empty the 7-quart measure 14 times and you will have thrown away 98 and leave 22 quarts in the barrel in 28 transactions (Filling and emptying are two transactions).
Then, fill 7 quart; fill 5-qrt from 7-quart leaving 2 in 7-qrt;
empty 5-qrt; transfer 2 from 7-qrt to 5-qrt;
fill 7-qrt; fill up 5-qrt from 7-qrt; leaving 4 in 7-qrt;
empty 5-qrt; transfer 4 to 5-qrt;
fill 7 qrt; fill up 5-qrt from 7-qrt, leaving 6 in 7 qrt;
empty 5-qrt; fill 5-qrt from 7-qrt leaving 1 in 7-qrt;
draw off remining 1 qrt from barrel into 5 qrt;
and the thing is done in 14 more transactions, making, with the 28 transactions above, 42 transactions.
Or you can start by wasting 104 and leaving 16 in barrel. These 16 can be dealt with in 10 transactions, and the 104 require 32 transactions to empty them – 12×7 and 4×5.
129) The runners
James, Jill, Jessica, John, Jane
130) The milkman
Sam Loyd gives the folowing solution: Let us call the 10-gallon cans A and B, and proceed as follows to show how the milkman supplied his two customers with two quarts each:
– Fill 5 qt pail from can A
– Pour 5 qt pail into 4 qt pail
– Empty 4 qt pail into can A.
– Pour 5 qt pail into 4 qt pail
– Fill 5 qt pail from can A
– Fill 4 qt pail from 5 qt pail
– Empty 4 qt pail into can A
– Fill 4 qt can from can B
– Pour 4 qt pail in can A.
131) A family affair
The following solution will do the job, and is found elsewhere on the web:
- The Thief and the Policeman cross over.
- The Policeman comes back (The Thief does not run away when left alone!).
- The Policeman and a boy cross over.
- The Policeman and the Thief come back.
- The Father and the other boy cross over.
- The Father comes back.
- The Father and the Mother cross over.
- The Mother comes back.
- The Thief and the Policeman cross over.
- The Father comes back.
- The Father and the Mother cross over.
- The Mother comes back.
- The Mother and a daughter cross over.
- The Thief and the Policeman come back.
- The Policeman and the other daughter cross over.
- The Policeman comes back.
- The Policeman and the Thief cross over.
132) The monk at Koningsberg
Call the bridges ABCD and E. A connects the two riverbanks, BC connect the northern riverbank with the island, d and E connect the island with the southern riverbank.
If the monk starts on the southern riverbank, the possible combinations are:
133) Next number
3=1x1x1+2, 10=2x2x2+2, 29=3x3x3+2, so the next one is 4x4x4+2=64.
134) Two wolves and a dog
Let’s abbreviate the wolves to v, the dog to d, the goat to g, the bag of grain to b and the farmer to f.
|Left shore||Right shore|
Call the two man A and B, and their respective wives a, a a, b, b and b.
|Left shore||Right shore|
|10||a <–||aaa||ABbbb||11||aa –>||a||AaaBbbb|
|12||a <–||aa||AaBbbb||13||aa –>||AaaaBbbb|
The number is 4789, 4*7*8*9=2016.
137) sum with 2 pairs of swapped digits
138) Three glasses puzzle
Call the glasses with 3oz, 5oz and 8oz A, B and C respectively.
Then a solution is:
0. start : 2 3 5
1. C -> A: 3 3 2
2. A -> B: 1 5 2
139) The stairs of sheik Oil Well
The number of steps is 2014.
Each word has one letter more then the preceding one.
141) Four jealous couples, an island and a picknick
(this solution is still under construction)
Solution: represent the boys with ABCD,
the girls with abcd. The bag with picknick food is represented by p.
142) Geometry pattern squares
The correct solution BCED
A = the square is divided into four identical parts, D = the four parts are not identical
B = straight lines, E = curved lines
C = identical colours are in opposite parts, F = identical colours are adjacent.
- Move engine to the right past the switch.
- Move engine to left to A.
- Move engine + wagon A to the right, adjacent to B.
- Move engine + wagon A left unto the straight track
- Move engine to the right to B
- Move the engine left over the switch up, through the tunnel, and have it push B over the switch up.
- The engine moves back to the right
- The engine moves left adjacent to A
- The engine pulls A to the right
- The engine pushes A to the left, and on the switch moves A up adjacent to B.
- The engine pulls A and B to the right
- The engine pushes B and A onto the straight track
- The engine pulls A to the right track.
- The engine drives through the tunnel and completes nearly a circle, ending adjacent to A again.
- The engine goes left onto the straight track, and connects to B
- The engine pulls B to the right.
- The engine pushes B to the left and up.
- The engine moves to the right
- The engine moves back to it’s initial position.
144) Pattern code: smileys
Eyes:A=left, B=Up, C=right
Nose:D=straight, E=left, F=right
Mouth: G=glad, H=neutral, J=sad
So solution is BDG.
Art: I am innocent. Chuck is guilty.
Bert: I am innocent. Chuck is guilty.
Chuck: I am innocent. Art is a LieSpeaker.
Suppose Art is guilty. Then Berts statement 1 is True, his second statement is false. That is impossible, because as a native he always lies or always speaks the truth.
Suppose Bert is guilty. Then Arts first statement is true, and his second statement is a lie. Again this is impossible for someone who always tells the truth or always lies.
So Chuck is guilty.
146) Geo pattern: ballpoints
A=Clip to front, D=Clip to right.
B=writing tip out, E=writing tip in
C=button and writing tip correspond, F=button and writing tip are out fo sync.
So the solution is CDE
147) Sum with 2 pairs of swapped digits
148) 4, 5 , 10 and 10
10-quart. 10-quart. 5-quart. 4-quart.
10 .. 10 .. 0 .. 0
5 .. 10 .. 5 .. 0
5 .. 10 .. 1 .. 4
9 .. 10 .. 1 .. 0
9 .. 6 .. 1 .. 4
9 .. 7 .. 0 .. 4
9 .. 7 .. 4 .. 0
9 .. 3 .. 4 .. 4
9 .. 3 .. 5 .. 3
9 .. 8 .. 0 .. 3
4 .. 8 .. 5 .. 3
4 .. 10 .. 3 .. 3
149) What next?
One useful heuristic for solving series is subtracting the numbers between the different terms, and write them on the line below the sequence line. For example, if the series is 7, 13, 23, 38, 58, …
you would get:
As you can see, the pattern becomes much clearer.
If we apply this to the problem stated in this exercise, we see:
Looking at it this way, you will notice that the differences are multiples of 4. It is clear that the missing numbers are 13, 21 and 25.
150) Pattern code: Chemistry.
OH-group: A No OH-group:C
CL: B No CL: F
Circle: D straight line: E
151) The farmer, the kids and the pets.
To the alert reader it will be clear that this puzzle is identical to the previous one, with the following equivalences:
- Farmer – policemen
- Dog – thief
- Son – father
- Daughter – mother
- Rabbits – girls
- hamsters – boys
With this in mind, you can check the solution at 131
The solution is as follows. You may accept the invitation to “try to do
it in twenty moves,” but you will never succeed in performing the feat.
The fewest possible moves are twenty-six. Play the cars so as to reach
the following positions:—
|————||= 10 moves.|
|————||= 2 moves.|
|123 87 4|
|————||= 5 moves.|
|E312 87 4|
|————||= 9 moves.|
The general pattern is the same in each row:
a/b c/a b/c
So the answer is 6/3
155) The scepter of dignity
Let me repeat the two statements here:
Peter: Paul is a Thief. But he did not steal the scepter.
Paul: Peter is a Thief. And Peter stole the scepter.
Suppose Peter is guilty. Then he is a Thief and thus speaks the truth, so Paul is a Thief (and thus tells the truth) but innocent. So far so good.
In that case Paul speaks the truth, and Peter is a Thief, and thus speaks the truth. Also his second sentence is true. That till gives a consistent set of statements.
Suppose Paul is guilty. Then Paul is a Thief and thus tells the truth. That makes Peter the thief of the scepter. That contradicts with our suppostion that Paul stole it, so this is impossible. So IF one of the two is guilty, it must be Peter.
156) What’s next
In every column, the center number is the sum of the top and bottom number + 1: 17=7+9+1, 16=12+3+1, so 18=5+12+1.
157) Sum with 2 pairs of swapped digits
158) 3, 5 and 8 – divide equally
Call them A, B and C.
The easiest solution is:
1. C -> A: 3 0 5
2. A -> B: 0 3 5
3. C -> A: 3 3 2
4. A -> B: 1 5 2
5. B -> C: 1 0 7
6. A -> B: 0 1 7
7. C -> A: 3 1 4
8. A -> B: 0 4 4
But these are 8 moves, while it can be doen in seven:
1. C -> B: 0 5 3
2. B -> A: 3 2 3
3. A -> B: 0 2 6
4. B -> A: 2 0 6
5. C -> B: 2 5 1
6. B -> A: 3 4 1
7. A -> C: 0 4 4
159) Boasting Billy
Suppose Billy is guilty.
Then both statements by Arrogant Andy are true. That is not possible, as we know he lies at least once in every two sentences. Hence Boasting Billy is innocent.
160) Geometry pattern circle
A=Line through center of circle
B=No line through center of circle
C=Image is point symmetrical
D=Image is line symmetrical
E=Image is A-symmetrical
162) pattern code: Lines
A= Opening at top
B=Opening at bottom
E=Besides each other
F=On top of each other
Hence solution is `BCE
163) Odd letter out
N. All other letters are line symmetrical around their vertical axis.
164) Gratte ciel 4×4
165) The five suspects
suppose CC is guilty. Then the statements of the 5 suspects are: F T F T F
suppose DD is guilty. Then the statements of the 5 suspects are: F F F T F
suppose HH is guilty. Then the statements of the 5 suspects are: T T F T T
suppose JJ is guilty. Then the statements of the 5 suspects are: F T T F F
suppose LL is guilty. Then the statements of the 5 suspects are: F T F T F
Hence DD is guilty if only 1 person speaks the truth, and HH is guilty if only one person lies.
166) Pattern code books
E=Title starts reading at top,
F=Title starts at bottom
Hence the solution is BDE.
167) Sum with 2 pairs of swapped digits
169) Pattern code books
B=Branched, E=not branched
So the solution is ABC
170) Geometry pattern: lines
A=there is a line without intersection
B=there is a line with 1 intersection
C=there is a line with 2 intersections
D=there are lines that are parallel
E=there are lines that intersect at a right angle
So correct code is BCE
|Smoke||Kools||Chesterfield||Old Gold||Lucky Strike||Parliament|
172) More and more pirates
First we have a look at the first problem. We can summarize the proposals as:
A: 100 to A.
B: 0 to A, 100 to B.
C: 1 to A, 0 to B, 99 to C.
D: 0 to A, 1 to B, 0 to C, 99 to D.
E: 1 to A, 0 to B, 1 to C, 0 to D and 98 to E.
We can summarize this as:
173) Odd digit out
7. 7 consists of two straight lines. All other digits are drawn with curved lines
174) Gratte ciel 5×5
176) Pattern code Circles
A=Straight line, B=Curved lines
C=Lines through center, D=not through center
E=Line symmetry, F= a-symmetric
177) Sum with 2 pairs of swapped digits
178) The christmas turkey
Sam Loyds solution is:
Santa Clausstarted off with his left foot to chase that turlkey and if you follow in his tracks in the snow counting left foot, right foot, etc., you will find he has gained one step somewhere. This can only be done by going round the first circle twice, so he has made four complete turns to arrive at his present position!
Frankly, I’m not sure I agree with Sam Loyd here. If we take the upper side of the illustration to be north, santa started out be running north-west. He makes one complete turn in the firdst circle, a second one in the second circle, and ends up running south-west. That makes 2.75 turns, not 3.
181) Sultan Oil Well
let’s have a look at the number of pearls in jar 1.
If there was 1 pearl at start, after 1 round there are 6 at the end of the circle.
If their were 2 pearls at start in jar 1, the number after 1 round would be 7.
If their were 3 pearls at start in jar 1, the number after 1 round would be 8.
If their were 4 pearls at start in jar 1, the number after 1 round would be 9.
If their were 5 pearls at start in jar 1, the number after 1 round would be 10.
So obviously the answer is 5 pearls in jar 1, 6 in jar 2, etc.
182) The Dutchmens wives
For the solution, first we observe that every one pays a square amount of shillings.
Second, using the fact that every husband pays 63 shilling more thanh his wife, a little diligent calculation shows us that there are only three squares that are another square +63:
If Hendrick buys 23 more hogs than Katrun, Hendrick must have bought 32 and Katrun 9.
If Elas buys 11 more hogs than Gurtrun, Elas must have bought 12 and Gurtrun 1.
That leaves Cornelius with 8 and Anna with 31.
183) Odd number out
Starting with the third number, every number except 13 is the sum of the previous two numbers.
184) Gratte ciel 5×5
2 4 1 3 5
4 5 3 2 1
1 2 5 4 3
3 1 4 5 2
5 3 2 1 4
185) Lines through 3×3 dots
Here are the 5 solutions my daughter Margreet and I could come up with, and I don’t say there are more::
A) Use an exceptionally broad line, covering 3 dots at once.
B) The dots have some width. Use this width to draw a line going slightly up through the three bottom dots, continue it a the back of the paper, continue at the front through the three middle dots, etc.
C) Same as 2, but lay the piece of the paper down on the floor, continue the line around the globe, come back at the middle three dots, etc,
D) Cut the paper in 3, and lign them up so that all 9 dots are in 1 row.
E) You can fold the paper in such away that the 9 dots get on one line (http://www.archimedes-lab.org/How_to_Solve/9_dots.html)
Solution C) is disputable, because to go around the earth the pencil needs to leave the paper, and because going around the globe does not make a straight line.
186) Pattern code: signalling people
A=Head upright, D=head tilted
B=both arms have same position, E=arms have different position
C=legs straight, F=legs bent
So correct solution = BCD
187) Where does the zero go?
The numbers are arranged into alphabetical sequence, so the zero goes to the end.
190) Post its
C=Text in capitals
D=Text in lower case
A=straight horizontal line
B=straight vertical line
C=From Top left to Bottom right
D=From top right to bottom left
The bronze medal costs 155, the silver 260 and the golden 365. That makes the prize of the fourth set 780.
193) Odd L out
The fourth L is a mirror image, the other three are just rotations.
194) Edgeview 4×4
196) Trinairo 6×6
197) What comes next?
The digits are the numbers of letters in the numbers spelled out, starting with zero:
zero – 4
one – 3
two – 3
five – 4
six – 3
The next one will be eleven, which has 6 letters.
198) How many straight cuts?
Here are 2 solutions from Dudenys'”Amusement in Mathematics”:
199) Two Russian women
Call the two starting points A and B. Denote the distance between them by d, and the time between sunrise and noon t0
Call the walking speed of the woman starting at A va, and the speed of the woman starting at B vb.
Woman A takes t0+ 4 hrs to cover distance d, so d=va * (t0+ 4).
Woman B takes t0+ 9 hrs to cover distance d, so d=vb * (t0+ 9).
So we have va * (t0+ 4) = vb * (t0+ 9). We can rewrite that too:
va = vb * (t0+ 9) / (t0+ 4).
We also know that they meet at noon. So at noon, the distance they covered together is d.
va * t0 + vb * t0 = d
Now it may look that we have 3 equations with 4 unknowns, but wait:
va * t0 + vb * t0 = va * (t0+ 4)
vb * t0 = 4 * va
vb * t0 = 4 * vb * (t0+ 9) / (t0+ 4)
(t0+ 4) * vb * t0 = 4 * vb * (t0+ 9)
(t0+ 4) * t0 = 4 * (t0+ 9)
t02 = 36
t0 = 6
200) Book covers
201) The boxes
The boxes contain 3, 4, 5 and 6 pearls respectively.
Cube E is different – two colours have been swapped.
203) Odd series out
The third series is the only one which has to do with arithmetic, all other series are based on geometrical properties.
204) Edgeview 5X5
206) Trinairo 9×9
B C C B A C A B A
A A B C A C B C B
C B A B A B C A C
A A C B C A B C B
C C B C B B A A A
A C B A C C A B B
B A A C B B C A C
C B A A B A C B C
B B C A C A B C A
210) 100 prisoners
One prisoner is indicated as the leader. If he finds the light on, he knows that someone has been at the light bulb, and switches it off.
All other prisoners switch it on the FIRST time only they get a chance to do so. If they find it on, they leave it that way. If they find it off and they have already switched it on before, they leave it that way.
The leader now simply has to count how often he switches the light off.
211) Men in a circle with shillings
Lewis Carrol gives the following solution:
Let m= the number of men, k= number of shillings possessed by the last=poorest man.
After one circuit, each is a shilling poorer, and the moving heap contains m shillings. Hence, after k circuits, each is k shillings poorer, the last man now having nothing, and the moving heap contains mk shillings.
Hence the thing ends when the last man is again called on to hand on the heap, which then contains (mk+m-1 shillings, the penultimate man man now having nothing, and the first man having (m-2) shillings.
It is evident that the first and last man are the only neighbours whose possessions can be in the ratio 4:1.
The first one gives mk=3m-7, i.e. k=3-7/m, which evidently gives no integral values other than m=7, k=2.
The second gives 4mk=2-3m, which evidently gives no positive integral values.
Hence the answer is ‘7 men, 2 shillings’.
So far for Lewis Carrols solution. I must admit that the latter was not evident to me right away. It does become clearer when you rewrite the last equation to:
4mk=2-3m means :
212) Triangle sum
The numbers in the bottom row are: 9, 1, 7, 4, 9
- The white train moves left, entirely past the sidetrack, pushing the black train far to the left.
- The white train drives back, pushing its rightmost 4 wagons unto the side track.
- The white train moves left to the main track, leaving its rear 4 wagons on the side track
- The black train moves right past the right end of the side track, pushing the white train far to the right
- The black train moves left, pushing its last wagon to the white wagons on the side track
- The black train moves right, pulling all wagons on the main track, pushing the white train even further to the right
- The white train moves left past the side track, pushing the entire black train the the left.
- The white train moves right onto the side track, so that the last wagon is nearly at the main track again
- The black train moves to the right, so that the last wagon is just past the right switch
- The white train moves slightly to the right, and connects to its other four wagons, which are disconnected from the black train
- The white train moves to the left onto the main track
214) Edgeview 6X6
216) Trinairo 9×9
A C B C A A B C B
A B A B C C A B C
B C A A B C C A B
A C B C A B B C A
C B A A B B C A C
B A C C A A B B C
C B B A C A A C B
C A C B B B C A A
B A C B C C A B A
218) The christmas geese
Squire Hembrow, from Weston Zoyland—wherever that may be—proposed the following little arithmetical puzzle, from which it is probable that several somewhat similar modern ones have been derived: Farmer Rouse sent his man to market with a flock of geese, telling him that he might sell all or any of them, as he considered best, for he was sure the man knew how to make a good bargain. This is the report that Jabez made, though I have taken it out of the old Somerset dialect, which might puzzle some reader in a way not desired. “Well, first of all I sold Mr. Jasper Tyler half of the flock and half a goose over; then I sold Farmer Avent a third of what remained and a third of a goose over; then I sold Widow Foster a quarter of what remained and three-quarters of a goose over; and as I was coming home, whom should I meet but Ned Collier: so we had a mug of cider together at the Barley Mow, where I sold him exactly a fifth of what I had left, and gave him a fifth of a goose over for the missus. These nineteen that I have brought back I couldn’t get rid of at any price.” Now, how many geese did Farmer Rouse send to market? My humane readers may be relieved to know that no goose was divided or put to any inconvenience whatever by the sales.
220) Three students
The answer is Yes. Many people at first think that you dont have sufficient information.
We know that Alex is an art0student anf=d that Charles is not. We dont know about Bert.
If Bert is an art-student, Bert, an art-student, sends a message to Charles, who is not.
If Bert is not, then Alex, who is an art-student, sends a message to Bert, who is not.
So in both cases an art-student sends a message to someone who is not an art-student.
221) Triangle sum
The numbers in the bottom row are: 9, 9, 1, 4, 8
223)Quickie numbers spelled out
Two Hundred and six.
226) Trinairo 9×9
A B C A B C A B C
B B A C B A C C A
B A C B A C B C A
A B A C C C A B B
C A B B A B C A C
A C B B B A A C C
C A A C A B C B B
B C C A C B B A A
C C B A C A B A B
228) Under the mistletoe bough
Everybody was found to have kissed everybody else once under the mistletoe, with the following additions and exceptions: No male kissed a male; no man kissed a married woman except his own wife; all the bachelors and boys kissed all the maidens and girls twice; the widower did not kiss anybody, and the widows did not kiss each other. Every kiss was returned, and the double performance was to count as one kiss. In making a list of the company, we can leave out the widower altogether, because he took no part in the osculatory exercise.
7 Married couples 14
3 Widows 3
12 Bachelors and Boys 12
10 Maidens and Girls 10
Total 39 Persons
Now, if every one of these 39 persons kissed everybody else once, the number of kisses would be 741; and if the 12 bachelors and boys each kissed the 10 maidens and girls once again, we must add 120, making a total of 861 kisses. But as no married man kissed a married woman other than his own wife, we must deduct 42 kisses; as no male kissed another male, we must deduct 171 kisses; and as no widow kissed another widow, we must deduct 3 kisses. We have, therefore, to deduct 42+171+3=216 kisses from the above total of 861, and the result, 645, represents exactly the number of kisses that were actually given under the mistletoe bough.
230) Weather forecast
A=Even number of suns, B=odd number of suns
C=Even number of clouds, D=off number of clouds
E=Even number of showers, F=odd number of showers
So the solution is ADE
231) Triangle sum
The numbers in the bottom row are: 7, 3, 5, 4, 2
The solution in BDE.
A=largest brick has length 4
B=largest brick has length 3
E=all bricks have same size
F=Bricks have different sizes
233) Lines through 4×4 dots
This is one of many possible solutions with 7 lines.
If you drop the requirement that every dot may be visited only once, it is possible to solve it with 6 lines.
235) Crack the code
The numbers on the left consist of an odd number of letters, those on the right have an even number of letters.
237) Eleusis 2
after a black card, play an even card, after a red card, play an odd card
A = 3 circles
B = 2 circles
C = 1 circles
D = 4 lines cross
E = 6 lines cross
F = 8 lines cross
I = 3 squares
H = 2 squares
G = 1 squares
241) Triangle sum
The numbers in the bottom row are: 7, 1, 10, 9, 8
242) Geo pattern snakes
A= straight line on top
B= straight line at bottom
C=2 half circles
D=3 half circles
E=Start going up
F=Start going down
244) Two sons and their medals
The first son has a 1/4 th chance of winning a medal. The second medal wouldnt make any difference for the question.
The 2nd son has a 1/3 chance to win a medal.
So i have a chance of 1/4 + 3/4*1/3 = 1/2 (50%) that one of them wins a medal.
That may sound like a calculation which would yield a different result if we start with the second son, but that gives the same result:
1/3 + 2/3 * 1/4 = 1/3 +1/6 = 1/2.
246) Language patterns
Each acronym is noted by its last letter instead of its first. So European Union becomes NN.
247) Billy’s Big Christmas Party
The numbers are all numbers as appear on a digital clock. The numbers are the sum of the lines which make up the digits on a digital clock. So the answer = 5+5+4+5=19.
248) Eleusis 1
249) Solutions to 100 Bongard problems
01: empty vs something
02: large figure vs small
03: white vs black
04: convex vs not convex
05: straight edges vs curved edges
06: triangle vs quadrangle
07: vertical vs horizontal
08: along right side vs along left side
09: smooth line vs sawed line
10: triangle vs quadrangle
11: oblong vs compact
12: eloganted shape, may be curtved, but still ‘long’ and with ends more or less towards edges.?
13: vertical rectange or horizontal ellipse
14: large vs small
15: closed line vs not closed
16: clockwise vs anti clockwise
17: edges pointing inwards sharp vs rounded
18: two ends thicker than middle
19: middle section horizontal
20: two diots on one end vs on opposing ends
21: at least one small onject
22: all objects same size
23: one vs two
24: at least one circle
25: black triangle
26: black triangle
27: more black than white objects
28: more black circles than white corcles
29: more inside than outside
30: one enclosed area
31: one line vs two lines
32: sharp point pointing out
33: sharp point vs rounded
34: large wghite inside black
35: elongated shapes in same drection vs perpendicular
36: triangle above circle
37: triangle above circle
38: triangle bigger than circle
39: three parallel lines
40: three dots in one line
41: Left: three white circles on 1 line
42: three circles inside figure on 1 line
43: increasing amplitudes vs decreasing amplitudes
44: circles on different curves
45: white shapes over black shapes vs black shapes over white shapes
46: triangle over circle vs circle over triangle
47: triangle in circle vs circle inside triangle
48: black shapes on top, whites shapes lower
49: three small circles in larger shape, with three small circles outside shape grouped together
50: vertical axis of symmetry
51: at least two grouped together
52: arrows in opposing directions
53: smaller number of edges inside larger number of edges
54: Sequence going with the clock is: circle – triangle – plus
55: Object to left side of indent
56: everything either white or black vs mixed colours
57: two objects with same shape and size
58: two black squares of identical size vs two different sizes
59: two shapes of same shape but different size
60: two figures with same shape but different size
61: Equal number of objects on both sides of the line vs Unequal number of objects on opposing sides of the line
62: The two ends of the line are far apart vs The two ends of the line are close together
63: Shadow at right side vs Shadow at left side
64: small circle is not on axis of ellipse vs small circle is on axis of ellipse
65: triangles are on a horizontal line vs triangles are arranged vertically
66: not connected dots are on a horizontal line vs not connected dots are on a vertical line
67: the right ‘leaf’ is attached above the left ‘leaf’ vs the left leaf is attached above the right leaf
68: the top of the right ‘leaf’ is extends above the top of the left ‘leaf’ vs the top of the left leaf ends above the top of the right leaf
69: the dot is on top of the central line vs the dot is not on top of the central line, but on top of one of the side ‘leaves’
70: only the central lines has laves vs the leaves split
71: Three levels of nesting vs Two levels of nesting
72: The end lines are parallel with each other vs The end lines are perpendicular to each other.
73: Ellipse and rectangle are perpendicular vs Ellipse and rectangle are parallel
74: Body is widest at base vs Body is widest at top
75: Triangle is located at hollow side of arc (inside) vs Triangle is located “outside” the arc
76: The long sides are curved inward vs Long sides are curved outward
77: line intersect corner into two angles of same size?
78: three lines intersect in 1 point
79: black circle is closer to white circle than to triangle
80: draw a straight line between the two dots, and project the plus on it. Than it will be between the dots.
81: Both the white objecxts and the black objects are grouped together vs they are mixed
82: there is an equilateral triangles on the left
83: the white circle is between the crosses
84: the square is outside the figure formed by the small circles
85: the figure consists of three lines
86: one point with exactlye three lines
87: 4 lines
88: three ellipses
89: three groupes
90: three groups of white ellipses
91: three lines, three objexts, etc, vs four
92: one line, no branches vs branched line
93: dot at intersection is white vs black
94: black dot not at end of line vs black dot at end of line
95: vertical lines vs horizontal lines
96: triangle vs quadrangle
97: triangle vs corcle
98: triangle vs quadrangle
99: loops of triangles and circles intersect vs no intersevtion
100: A vs B
Douglas Hofstadters problems:
101: Both indentations are horizontal vs 1 horizontal and 1 vertical
102: Both arrows point from center outwards
105: connecting string touches end shapes at same side (for example, both below)
107: three normal sides
108: point smaller than base
109: white circle is in right half vs left half
110: 5 geometric shapes, four of them circles
111: there is a triangle in the middle
113: one line is connected to another line vs three lines meet
114: 4 crossings vs 2
115: innermost and outermost shapes are connected
117: triangle points towards center of circle
118: 1 end point vs loop
230: graphs of order 7 (7 points)
250) A lightbulb and three switches
Switch on the left switch. Keep it on for a minute or so. The flip it off and turn on the middle. Now walk to the room with the bulb.
– if the bulb is burning, it is controlled by the middle switch;
– if the bulb is off but hot, it is controlled by the left switch;
– if it is off and cold, it is controlled by the switch you have not yet used;