**252) How many rhombuses?**

For every of the three corners, there are 6 rhombuses consisting of 2 triangles, and there is one rhombus consisting of 8 triangles. So all together there are 3*7=21 rhombuses.

**253) My friend and his granddaughter**

My friend is 75 years old and his granddaughter 27.

3 years ago, he was 72 and his granddaughter 24.

8 years before that, he was 64 and his granddaughter 16.

**255) Old year – New Year**

2015.

**256) The two sisters**

The oldest is 15, the youngest 13.

& years ago they were 8 and 6, together they were 14. One is now older than 14, the other is younger.

**257) Christmas 2015**

0 – TREE

1 – TRET

2 – TEET

3 – TEST

4 – MEST

5 – MAST

6 – MART

7 – MARY

Did you find a solution in a fewer number of steps? Please let us know!

**258) What’s next?**

Combine the digits in groups of 2, starting at the right, and sum them

5 67 89 gives 5 6+7=13 8+9=17

5 13 17 gives 5 1=3=4 1+7=8

548 will give 5 4+8=12

512

53

8

**259) Rabbits eating carrots**

If 5 rabbits eat 5 carrots in 5 minutes, then 1 rabbit eats 1 carrot in 5 minutes. 1 rabbit eats ½ a carrot in 2.5 minutes.

Half of 5 rabbits is 2.5 rabbits, but half a rabbit is a dead half rabbit and doesn’t eat a thing. So effectively we have 2 rabbits, which means they will eat exactly 1 carrot in 2.5 minutes.

**260) The Babuschka and the pearls**

The “babuschka” in the title has a double meaning, because this is a puzzle in a puzzle.

The chances of taking out one of k white balls from a box of n balls is k/n. The chance that the second ball is also white is (k-1)/(n-1), so the chance of taking out 2 balls and both being white is k(k-1) / n(n-1).

The chances are displayed in this spreadsheet. There is no combination that gives a chance of exactly 1:4. So the shopservant must be lying, and the shopowner a TruthSpeaker.

**261) From 5 to 4**

There are several solutions, here is none of them:

**262) How many rhombuses?**

counting per row: bottom row row 2 row 3

size 3 trapeziums: 5 3 1 9

size 4 trapeziums 4 2 0 6

size 5 trapeziums 3 1 4

size 6 trapeziums 2 2

size 7 trapeziums 1 1

spanning 2 rows:

size 5+3 2 1 3

size 7+5 1 1

size 7+5+3 1 1

27

For three directions this makes 3*27-81.

**263) Pattern code square and circle**

A=circle in corner, D=along edge

B=square is horizontal, E=square is rotated 45 degrees

C=Touches square, F=Not touching

Solution: AEF

**264) A square**

The square of 38 is 1444.

**265) Two watches**

As the speed differences between the watches is 3 minutes per hour, it will take 20 hours before they differ exactly 1 hour. That is the easy part. Now you may be tempted to think: Oh, it is 9.05 and 25sec, so 20 hours ago it was 01.05.

But if the watch displayed is the slower one, it will not have progressed 20 hours, but just 19 hours 20 minutes. Likewise, the faster of the two would have progressed 20 hours 20 minutes.

If the watch being displayed is the faster one, the two watches started at 9.05.25 – 20.20 = 1.05.25 – 0.20.00 = 12.45.25

If the watch being displayed is the slower one, the two watches started at 9.05.25 – 19.40 = 2.05.25 – 0.40.00 = 1.25.25

Sam Loyd gives a slightly different solution: … the watches must have started at 45 minutes 25 seconds past midnight, and as the fastest one gets three minutes ahead of the other every hour, it would be one hour ahead in 20 hours; it is therefore 20 minutes ahead of correct time, while the other is 8.45m 25 sec.

Same Loyd therfore seems to assume that the watch being displayed is the fastest of the two watches; something the text suggests but does imho not exactly state.

**266) Lines and spheres**

3 lines between spheres A,

4 lines between spheres B

1 dimensional C,

2 dimensional D

protuding ends E,

No protuding ends F

Solution: ADE

**267) The arrows**

A, B and C indicate the different kind of arrow tips.

D are arrows pointing both ways, E indicates the arrow pointing right, while F means arrow pointing left.

G, H and I are different types of lines: G is solid, H is double and I is triple line.

The solution is : CEI.

**268) The camper at Trafalgar square**

Look at:

Ron Rubbish answered: Sluggy Sarah took it.

Sluggy Sarah said: I’m innocent.

It bis clear that one of the is lying. If only one of the three lies, the third must speak the truth:

Mighty Mike replied: Ron Rubbish did it.

So the inspector should keep Ron Rubbish for further custody.

**270) Letter square**

The numbers are the number of vowels in the 3×3 square surrounding the number. So the question mark should be a “2”.

**272) How many 5’s?**

There are 100 numbers with a 5 in the first position (500-599)

There are 100 numbers with a 5 in the second position (50, 51, .., 150, 151, )

There are 100 numbers with a 5 in the rightmost position (5, 15, 25, … 995)

That makes 300, but there are a couple of doubles:

– there are 10 doubles between the first and 2nd position, 10 doubles between 1rst and 3rd position, and 10 doubles between 2nd and 3rd position, so that makes 270.

– but those doubles have 555 in common, so we have to add 1, giving our solution of 271.

**273) 61 equals 28?**

Turn the 6 upside down so that it’s a 9, the move the 2 to the upper right of the 9 and you get 9^2=81.

**274) Coloured grids (2)**

The missing colour is orange: at every row there are three cells with the same colour.

**278) Bongard problem 4**

The diamonds are arranged as the diamonds on (French) playing cards.

**280) Stefan de Banach**

A person born in 2070 will be 46 in 2116, and 2116=46*46.

**281) Five triangles**

Yes, they can all be done, see the following pictures:

With these two images you can easily expand the series to any number of triangles. I leave it to the reader to prove that all these triangles have sides adjacent to the right angle 1:2.

**282) Billy and the christmas party**

An elementary law of physics = Distance = Time * Speed.

Call the time Billy needs to run back to the entrance of the tunnel t_{1}. Call his speed V_{b}, and the length of the tunnel L. Then we have:

0.5L -32 = t_{1} * V_{b}, or t_{1} = (0.5L -32) / V_{b}.

In the same time, the train, covers a distance of L-1:

L-1 = V_{t} * t_{1} ==> t_{1} = (L-1)/V_{t}

This means we can eliminate the time t_{1}:

(0.5L -32) / V_{b} = (L-1)/V_{t}

We can rewrite this to:

V_{t} = V_{b} * (L-1) / (0.5L -32). Let’s call this equation (I).

We also know that Billy would have been overtaken 20 meters before the exit if he had ran the other way.

He would have ran 0.5L +32 -20 meters to the point in a time which we may call t_{2}:

t_{2} * V_{b} = 0.5L +32 -20 or t_{2} = (0.5L +32 -20) / V_{b}

The train would have driven a distance of 2L-20 = V_{t} * t_{2}, or t_{2} = (2L-20) / V_{t}.

Combining these 2 gives us:

(0.5L +32 -20) / V_{b} = (2L-20) / V_{t}

we can rewrite this to:

V_{t} = V_{b} * (2L-20) / (0.5L + 12) Let’s call this equation (II).

We can now combine the equations (I) and (II) and get rid of the speed of the train:

V_{t} = V_{b} * (2L-20) / (0.5L + 12) = V_{b} * (L-1) / (0.5L -32)

And to our suprise we see we the speed of Billy is cancelled out too:

(2L-20) / (0.5L + 12) = (L-1) / (0.5L -32)

Multiplying both sides gives us:

(2L-20) * (0.5L -32) = (0.5L + 12) * (L-1)

Fractions are always awkward, so let’s multiply both sides with 2:

(2L-20) * (L -64) = (L + 24) * (L-1)

We can work out the multiplications on both sides and bring everything to one side:

L^{2} -171L + 1304 = 0.

This equation has two solutions: L=163 and L=8. The latter clearly does not fill our bill, so the tunnel must have a length of 163.

**283) Clock in equal parts**

The numbers 1 – n have n(n+1)/2 as a sum. So the sum of the digits oon the cloc is 12*13/2=78. This can not be divided into 4 equal parts, but it is equal to 3*26. Two straight lines suffice, they can even be drawn paralel:

**284) Coded germs**

A=First curve left, D=First curve right

B=First and 2nd curve same direction, E=different directions

C=half circle, F=quarter of circle

Solution: DEF

**286) The house with eight kids**

Dorothy is playing draughts, which is a game for two players, so she must be playing with Henry.

**290) My colleague and his daughter**

My colleague is 40 years old and his daughter 6. In 11 years, he will be 51 and his daughter 17.

**292) Knights on a chessboard**

A knight, when moving, always moves to a field of the opposite colour. So when you put 32 knights on all the white squares, there will not be a knight that can take another knight.

To prove this is the maximum, consider an 2×2 subsection. You can check by hand the possible combinations of 4 knights on this 2×4 section. If there can be more than 32 knigyhts, there will have to be a 2×4 subsection with 5 knights. This is impossible, so 32 is the maximum.

**294) 135**

The other numbers are 175=1+7^2+5^3, 518=5+1^2+8^3 and 598=5+9^2+8^3.

**296) Happy 2016**

First of, we observe that the final two digits of 2016 x 2016 are the same as those of 16×16, because the digits for the hundreds or bigger can never influence the final to digits.

So the following table, which lists the powers of 16 (modulo 100 of the previous result) should suffice. We see that 2016^5, 2016^10 and 2016^15 all and with 16. Expanding this pattern gives us 16 as the two final digits of 2016^2015.

16

1 256

2 896

3 1536

4 576

5 1216

6 256

7 896

8 1536

9 576

10 1216

11 256

12 896

13 1536

14 576

15 1216

16 256

**298) Bongard problem 5**

The rule is: There are always 3 dots, and a bloack dot is closest to the white border.

**299) What comes next?**

100. The numbers are the squares of 1, 2, 3, etc, grouped with 3 digits. They are easier to recognize when writter 1 4 9 16 25 than when written 149 162.

**300) 10 glasses**

Label the glasses A to J.

1)

– Move glass B to a position between glasses F and G.

– Move glass G to the vacant spot of glass B

– Move glass D to a position between glasses H and I

– Move glass I to the vacant spot of glass D

2)

– Take glass B and empty the content into glass G

– Take glass D and empty the content into glass I

**301) The optician**

Prices are the sum of:

1=temples to left, 2=temples to right

10=straight bridge, 20=curved bridge

100=circulare eyewires, 200=rectangular eyewires

Solution: 222

**302) The Roman pantheon**

16323

902654

76104

+ 73523

——-

1068604

A=3 D=1 E=0 F=7 G=9 I=6 N=2 S=4 T=8 U=5

**303) The river war**

The answer is 2 hours.

Let:

T_{1} be the time between the start of the Asbestos and its seizure by the Berilyum.

T_{2} be the time needed fromthe moment of seizure to the arrival is Borovitch

V_{a} be the the speed of the asbestos (known to be 5)

V_{r} be the speed of the current in the river.

AD=distance between Astrakhanitch and interception point D.

DB=distance between interception point D and Borovitch.

Then:

(1) AD = T_{1} * (V_{a} + V_{r}) (Distance = time * speed. In T1 hours, the Asbestos, added by the speed of the riverflow, makes it from A to D.)

(2) DB = T_{1} * (V_{a} – V_{r})

(3) DB = T_{2} * (V_{a} + V_{r})

(4) 20 = AD + DB = (T_{1} + T_{2}) * (V_{a} + V_{r})

Combining (2) and (3) gives us:

T_{1} * (V_{a} – V_{r}) = T_{2} * (V_{a} + V_{r}).

==> T_{2} = T_{1} * (V_{a} – V_{r}) / (V_{a} + V_{r}).

Substituion in (4) gives us:

20 = (T_{1} + T_{1} * (V_{a} – V_{r}) / (V_{a} + V_{r})) * (V_{a} + V_{r})

which we can simplify to:

20 = T_{1}*(V_{a} + V_{r}) + T_{1}*(V_{a} – V_{r})

20 = 2T_{1}*V_{a}

Substitute V_{a}=5.

T_{1}=2.

**303) Spot the differences**

F1 and B2.

**304) Bongard problem (6)**

All boxes on the left contain a smiley consisting of mouth and eyes.

**306) Pyramid**

The original tetrahedon has four corners.

Each corner which is cut off gives 3 new corners (provided that the cuts do not intersect). So the resulting figure has 12 corners.

When these corners are cut off, each again generates three new corners (again, if the cuts do not intersect), giving 36 corners total.

Going on, you will quickly see that the general formula is 4 *3^n if we cut off the corners n times.

**310) On which day?**

Solution: He can not.

Assume A is saying “Today is Saturday” and B is saying: “Today is Sunday”. If it is Sunday, both should tell the truth, while obviously one of them is lying. So it can not be Sunday, and B is lying.

If we assume that One is a Furster and the other is a Secunder, we can not have them lying and A will be speaking the truth, hence it is Saturday.

It is possible that both are Fursters and the sentences are spoken on Monday, Tuesday or Wednesday.

It is possible that both are Secunders, and both sentences are spoken on Thursday or Friday.

**312) Alphametic problems**

473309

4218095

+ 41692

——–

4733096

B=8 E=9 L=0 M=5 O=1 P=4 R=2 S=6 U=7 Z=3

**314) The four cards**

The hypothesis is that at the back of every card with a ‘B’ there is a ‘2’.

So you certainly want to turn the card with the ‘B’. If there is a ‘1’ on the other side, the hypothesis is falsified.

Turning the card with a ‘2’ on the other hand, doesnt tell you anything. There may be an A or a B, but in both cases it does not confirm the hypothesis. You do want to turn the card labeled ‘1’, because if there is a ‘B’ on the other side, the hypothesis is false.

**317) 5 coins in a row**

Take one coin from the end of the longest line and put it on top of the coin in the center:

**319) the logicians club**

To be sitting alternating speaking the truth and lying, there must have been an even number of members. So the president was speaking the truth, and the secretary was lying.

**320) 9 trees, 9 rows**

Unfortunately I do not have the plates which came with the original book, but here is the solution I came up with:

If you have a better solution, I love to hear about it!

**321) The lazy comrade**

The party administrator was right. Lets say the field was 1800 furrows.

the first 600 furrows take him 6 days. The next 600 furrows take him 3 days, while he needs 2 days for the last 600 furrows.

That means he needed 11 days, while at a steady pace of 200 furrows he would have needed just 9 days.

**323) Alphametic plants**

495592

275567

+ 92092

——-

863251

A=3 C=4 E=7 I=0 L=6 N=2 O=9 P=8 S=1 T=5

**324) The four cards (cont’d)**

Of course we have to check the card with C face up. If the reverse side comes up with a ‘2’, that confirms the rule. If it come sup with ‘1’, we know the rule is not valid and are done.

Those among us who are not schooled in formal logic, may be tempted to turn the card with ‘2’ face up, because a Con the back side might ‘confirm’ the rule we are testing. But it really doesnt add anything to our knowledge. Suppose there is a B on the other side? Then the rule stays equally valid.

Instead, we have to turn the other two numerical cards. The reason is that, if there is a ‘C’ on the other side, the rule would be invalid. If there is no ‘C’ on the reverse, there is no proof to the contrary poissible and we must assume the rule is true.

**329) Turn pyramid upside down**

Take coins away from the red positions and put them on the green spots.

**331) Cryptarithm metals**

A=7 B=6 E=2 I=3 L=1 M=9 R=4 S=8 T=0 U=5

674359

246359

——-

920718

**332) How many?**

The answer is 3.

Write out the numbers as wqords, and count the number of slanted straight lines:

ONE – 1 (in letter N)

TWO -2 (both in W)

THREE – 1 (in R)

FOUR -1

FIVE -2

SIX – 2

SEVEN – 3

EIGHT – 0

NINE – 2

TEN – 1

ELEVEN – 3

**333) Matchsticks digital numbers**

**334) Coffee**

My wife had added sugar, which she tasted right away.

**335) Find a number**

2017

**337) The evil castle lord and the young gardener**

**338) Bongard problem 33**

The black symbol is vertically in the middle.

**341) Flags (1)**

Here is a list of the countries and their capitals:

Belgium – Brussels

China – Bejing

Columbia – Bogota

Germany – Berlin

Hungary – Budapest

Lebanon – Beirut

Mali – Bamako

Romania – Bucharest

Serbia – Belgrade

Slovakia – Bratislava

The three flags you can choose from, are from:

Slovenia – Ljubljana

Guinea-Bissau – Bissau

Bulgaria – Sofia

I don’t think I need to point out the single correct choice.

**342) What does not belong in this series? (1)**

The logic in this series is: Vowel, prceding letter, and so on. Only the letter J does not fit the bill.

**343) Bongard (7)**

The rule is that all figures on the left can be drawn without lifting your pencile from the pater and without drawing a line twice.

**344) Prime numbers**

3

7

11

13

17

19

23

29

31

37

41

43

47

59

61

67

71

73

79

83

89

97

**345) Ex****ample mini Sudoku**

**348) The island of Odders and Eveners**

Consider the following scheme:

Wednes-

day

Friday

Sunday | Monday | Tuesday | Thursday | Satur- day |
ODDERS | T | L | T | L | T | L | T | EVENERS | L | T | L | T | L | T | L |

An Odder will speak the truth on Sunday, and will truthfully state that on Saturday.

An Evener will lie on Sunday, and lie on Saturday that he will speak the truth on Sunda.

**349) Cryptarithm: cookies**

A=0 C=7 E=8 I=2 K=3 M=5 O=1 R=6 S=9 T=4

506289 + 1681 + 6198448 + 406871 = 7113289

**350) On the first day of Christmas**

You may have been tempted to to sum 1 + 2 + 3 + 4 + 5 + … + 12 = 78. The general formula is sum(n) = n(n+1)/2.

But that’s what the true love is bringing on the twelfth day.

The total number of items brought on the 12 days is 364. If you want to count the pear tree as an extra item, add 11.

**352) What does not belong in this series? (2)**

In the series 3, 4, 8, 9, 10, 18, 19, 38 we have an alternating +1 and x2. The only term that does not fit in is 10 – if we omit it, 18 is twice the previous term 9.

**353) Bongard (9)**

In every figure on the left, all *curves* are *outward*. Every figure on teh right has at least one *curve* which is *inward*.

**357) Cryptarithm: cookies**

B=7 C=1 E=5 H=0 I=6 K=8 L=9 O=2 S=4 W=3

861059

367595

——-+

1228654

**357) On the second day of Christmas**

42

**360) Remove 5 letters**

Solutions:

1) Thursday

2) Council

3) Playful

4) Progress and sprigged

5) Courage and Burglar

**361) Alphametic – untrue (1)**

6428 + 98285 = 104713

**362) The odometer**

The next prime number consisting of consecutive digits is 1234567. 1234567-65432 = 1169135.

**364) The three triangular blocks**

The left block is labeled with an ‘A’, and thus can not bear a ‘B’. We do not have to rotate that.

If the left back of the right block would be labeled ‘B’, the rule would be false, so we do want to rotate this block twice.

The middle block is a bit more complicated. Let’s rotate it once clockwise. If we get a colour, we are done: the block can not falsify the rule we want to verify. The same applies if we get a side labeled ‘A’.

But if we get a B, we want to rotate once more, to see if the rightback is Purpur.

**362) Matchstick pattern**

Solution: GNT

**363) The twenrty one trees**

Dudeney gives two pleasing arrangements of the trees. In each case there are twelve straight rows with five trees in every row.

**371) Alphametic – untrue (2)**

761 + 70966 + 5386 + 626861 = 703974

(E=6 F=5 H=0 I=3 L=2 N=1 R=9 T=7 V=8 Y=4 )

**372) The two torches**

Suppose the torches are X cm long at start.

Sam’s torch burns X/3 cm/hour, so after N hours Sam burnt NX/3 cm and has X-NX/3 cm left.

Moshe’s torch burns X/4 cm/hour, so after N hours he burnt NX/4 cm and has X-NX/4 cm left.

It is clear that Sam’s torch is burning faster and will be the shorter one.

If N is the time needed to get to the point where one torch is 3 times as long a the other, we have:

X – NX/4 = 3(X – NX/3)

multiply both sides with 4:

4X – NX = 12X – 4NX

X(4-N) = X(12-4N)

4-N=12-4N

3N = 8

N = 8/3 hour = 2 2/3 hour = 160 minutes

**374) Bongard problem (2)**

Every figure has an even number of corners.

**375) The online certification**

I missed 223-201=22 points, which is caused by 2 questions of 5 points and 4 questions of 3 points.

I can not have missed 4 questions of 5 points each, because 4×5=20; 22-20=2 which is not a multiple of 3.

I can not have missed 3 questions of 5 points each, because 3×5=15; 22-15=7 which is not a multiple of 3.

I CAN have missed 2 questions of 5 points each, because 2×5=10; 22-10=12, which IS a multiple of 3.

I can not have missed 1 question of 5 points each, because 1×5=5; 22-5=17, which is not a multiple of 3.

**376) On the third day of Christmas**

19 vowels and 35 consonants

**377) On the third day of Christmas**

A=2 C=6 D=7 E=3 H=4 L=1 R=8 S=0 T=9

63728 + 12864 + 21738 = 98330

and

A=7 C=2 D=3 E=1 H=8 L=5 R=4 S=6 T=9

21374 + 57428 + 15314 = 94116

**381) Alphametic – untrue (3)**

71299 + 8042 + 8659 + 716273 = 804273

(E=9 F=8 H=1 I=6 O=0 R=2 T=7 U=4 V=5 Y=3)

**382) Cryptarithm**

4093 + 93723 + 987653 = 1085469

A=7 C=2 E=3 H=4 I=6 O=0 P=9 R=8 S=5 W=1

**383) The two squares (1)**

A=2nd square starts at nearly same height as left square,

C=2nd square start at half the height of 1rst square

B=both squares are filled

F=right square is empty

D=non touching

E=overlapping

Solution: CEF

**384) Bongard problem (1)**

The triangles are arranged as spots on an ordinary six sided die.

**385) Five blouses**

Call the prices of the five different blouses a, b, c, d and e, with a<b<c<d<e.

Assming that all prices and in 95 cents, we calculate them as:

b+c+d+e=66

a+c+d+e=62

a+b+d+e=59

a+b+c+e=58

a+b+c+d=55

This means:

b=a+4

c=b+3

d=c+1

e=d+3

a+b+c+d=55

a+b+2c+1=55

a+b+2c=54

a+b+2b+6=54

a+3b=48

4a+12=48

4a=36

a=9

Calculating prices back to end with 0.95, we get at a price of 74.75 for all 5. So the woman had 74.70 in her purse.

**385) The twelve mince pies**

If you ignore the four black pies in our illustration, the remaining twelve are in their original positions. Now remove the four detached pies to the places occupied by the black ones, and you will have your seven straight rows of four, as shown by the dotted lines.

**386) Ages**

When he is 73, his wife will be 71. 144 = 8*18

Should he become 82 and his wife 80, 162 = 9*18.

**387) Cryptarithm**

A=2 G=3 H=1 M=5 N=9 O=6 S=0 T=7 U=8 Y=4

524

524

524

524

283807

283807

——-

569710

**393) Cryptarithm : metals**

A=8 D=7 E=0 I=4 L=6 M=5 N=9 S=1 T=3 U=2

349 + 6087 + 497425 = 503861

**396) The two squares (2)**

A=smaller square is at left side

D=smaller square is at right side;

B= smaller suqare is at bottom side

C=smaller square is at top side

E=smaller square is inside

F= smaller square is at outside

Solution:BDE

**398) The party of logicians**

There are 100 persons who visit a party and shake hands with some or all of the other guests. Every guest speaks either the truth all of the time or lies all of the time. At the end of the party, every guests is asked with how many truth tellers he or she has shaken hands with. All answers 0, 1, 2, … 98, 99 were given exactly once.

Let’s have a look at the person who answered 99.

There are two possibilities: either this person speaks the truthg, or he lies.

If he speaks the truth, all persons who answered 0 – 98 were truth tellers. That includes the peron who answered 0. If the person who answered 99 speaks the truthg, person 0 must have spoken the truth. Person 0 claims to have spoken with 0 persons who spoke the truth. That means that person 99 was lying, which means we have a contradiction. So the person who answered 99 was a liar.

We can apply the same reasoning to the person who answered 98.

If this person is a person who speaks the truth, he or she spoke with 98 truth tellers. No 99 can not be one of them, as we have already established that he is a liar. So these 98 persons must have been the persons who answered 0-97.

That includes the person who said 0, who must have spoken the truth according to the statement above. But number 0 spoke to no truth teller, so we again have a contradiction. So the person who answered 98 must be a liar too.

This logic can be applied to all persons 97 right down to 1. They all must be liars.

The person who said 0 is a special case. We already established that the persons saying 1 – 99 are all liars. So the persons who said ‘0’ shook only hands with liars, and not with a single person telling the truth. So the person who replied ‘0’ is the only one who spoke the truth.

**399) **

A=3 G=5 H=7 M=9 N=8 O=4 S=6 T=2 U=1 Y=0

930

930

930

315162

315162

315162

——-

948276

**402) What’s next?**

1,000 – thousAnd

1,000,000,000 – Billion

20 – sCore

100 – hunDred

1 – onE

4 – Four

8 – eiGht

3 – tHree

The numbers in the sequence are the lowest numbers where a letter in the alphabet appears in when spelled out.

The next number would 5 – fIve.

**403) Chaos sudoku**

By R. A. Nonenmacher – Created by, GFDL, https://commons.wikimedia.org/w/index.php?curid=4788885

**404) 6-digit number**

When someone is born in 2042, he will be 8 in 2050, while 2+0+4+2=8

**405) Alphametic**

4093 + 932206 = 1035784

A=3 D=9 E=0 G=2 N=8 O=7 P=5 R=6 S=4 W=1

**407) The ten coins**

The answer is that there are just 2,400 different ways. Any three coins may be taken from one side to combine with one coin taken from the other side. I give four examples on this and the next page. We may thus select three from the top in ten ways and one from the bottom in five ways, making fifty. But we may also select three from the bottom and one from the top in fifty ways. We may thus select the four coins in one hundred ways, and the four removed may be arranged by permutation in twenty-four ways. Thus there are 24 × 100 = 2,400 different solutions.

As all the points and lines puzzles that I have given so far, excepting the last, are variations of the case of ten points arranged to form five lines of four, it will be well to consider this particular case generally. There are six fundamental solutions, and no more, as shown in the six diagrams. These, for the sake of convenience, I named some years ago the Star, the Dart, the Compasses, the Funnel, the Scissors, and the Nail. (See next page.) Readers will understand that any one of these forms may be distorted in an infinite number of different ways without destroying its real character.

In “The King and the Castles” we have the Star, and its solution gives the Compasses. In the “Cherries and Plums” solution we find that the Cherries represent the Funnel and the Plums the Dart. The solution of the “Plantation Puzzle” is an example of the Dart distorted. Any solution to the “Ten Coins” will represent the Scissors. Thus examples of all have been given except the Nail.

On a reduced chessboard, 7 by 7, we may place the ten pawns in just three different ways, but they must all represent the Dart. The “Plantation” shows one way, the Plums show a second way, and the reader may like to find the third way for himself. On an ordinary chessboard, 8 by 8, we can also get in a beautiful example of the Funnel—symmetrical in relation to the diagonal of the board. The smallest board that will take a Star is one 9 by 7. The Nail requires a board 11 by 7, the Scissors 11 by 9, and the Compasses 17 by 12. At least these are the best results recorded in my note-book. They may be beaten, but I do not think so. If you divide a chessboard into two parts by a diagonal zigzag line, so that the larger part contains 36 squares and the smaller part 28 squares, you can place three separate schemes on the larger part and one on the smaller part (all Darts) without their conflicting—that is, they occupy forty different squares. They can be placed in other ways without a division of the board. The smallest square board that will contain six different schemes (not fundamentally different), without any line of one scheme crossing the line of another, is 14 by 14; and the smallest board that will contain one scheme entirely enclosed within the lines of a second scheme, without any of the lines of the one, when drawn from point to point, crossing a line of the other, is 14 by 12.

**408) Flags(2)**

All the flags in the group contain blue.

**414) Billy at the Christmas party**

12, when spelled out as twelve, has 6 letters.

6, when spelled out as six, has 3 letters.

and as 4, when spelled out as four, has four letters, that should have been his answer.

**415) 6-digit number**

From Johns statement we met conclude that either Jim or Jack is the thief

Suppose Jack is the thief. Then Jacks first statement is a lie, but both Jims statements are true, which is impossible, so Jack is not the thief.

Suppose Jim is the thief. Then his second statement is true, and his first statement is false. Johns statements are both false. This situation is possible.

So we may conclude that Jack is the thief.

**416) Pattern code matchstick triangle**

Solution: GNT

**420) The prisoner**

It can not be done. To understand this, cover the prison with a checkerboard pattern.

If the prisoner starts day 1 on a yellow square,

he will be

on day 2 on a brown square

on day 3 on a yellow square

on day 4 on a brown square

and so on till he is on a yellow square again on the 81th day

and he will not be able to move to his initial cell on a yellow square as yellow squares are never adjacent.

**421) The stolen toupet**

Suppose that Theo is the thief. Than Petes statements are True, True end False respectively. But he cann’t speak two sentences out of three which are true, so Theo can not be the thief.

Suppose that Ray is the thief. That means that Theo’s statements are all true, which is equally impossible.

So the only solution is that Pete is the thief, and if you check the statement you will find that at most one of three sentences is true.

**422) The floor boards**

The answer is 0.

**423) Spot the differences**

The upper right and middle left images are identical.

**424) Bongard problem (8)**

All lines on the left have one end in the enclosed area and one end outside. All lines on the right either have both ends of the line outside the enclosed area or both inside.

**426) The Burmese plantation**

The arrangement on the next page is the most symmetrical answer that can probably be found for twenty-one rows, which is, I believe, the greatest number of rows possible. There are several ways of doing it.

**427) Japanese tangram – In and around the water**

**428) Matchsticks triangles – surface**

**430) The windmills**

Passing a windmill doesn’t take time, passing the distance between them does. There are three distances between the first four windmills. There are 6 distances between the group of seven windmills, so this group will take twice as long, thus 20 seconds.

**440) The clock**

There are five periods between the six strikes. There are 10 periods between the 11 strikes. So striking 11 will take twice as long, that is, 24 seconds.

**441) Matchsticks**

Add one matchstick before the figure an turn it upside down.

**442) The Turks and Russians**

The main point is to discover the smallest possible number of Russians that there could have been. As the enemy opened fire from all directions, it is clearly necessary to find what is the smallest number of heads that could form sixteen lines with three heads in every line. Note that I say sixteen, and not thirty-two, because every line taken by a bullet may be also taken by another bullet fired in exactly the opposite direction. Now, as few as eleven points, or heads, may be arranged to form the required sixteen lines of three, but the discovery of this arrangement is a hard nut. The diagramPg 192 at the foot of this page will show exactly how the thing is to be done.

If, therefore, eleven Russians were in the positions shown by the stars, and the thirty-two Turks in the positions indicated by the black dots, it will be seen, by the lines shown, that each Turk may fire exactly over the heads of three Russians. But as each bullet kills a man, it is essential that every Turk shall shoot one of his comrades and be shot by him in turn; otherwise we should have to provide extra Russians to be shot, which would be destructive of the correct solution of our problem. As the firing was simultaneous, this point presents no difficulties. The answer we thus see is that there were at least eleven Russians amongst whom there was no casualty, and that all the thirty-two Turks were shot by one another. It was not stated whether the Russians fired any shots, but it will be evident that even if they did their firing could not have been effective: for if one of their bullets killed a Turk, then we have immediately to provide another man for one of the Turkish bullets to kill; and as the Turks were known to be thirty-two in number, this Pg 193 would necessitate our introducing another Russian soldier and, of course, destroying the solution. I repeat that the difficulty of the puzzle consists in finding how to arrange eleven points so that they shall form sixteen lines of three. I am told that the possibility of doing this was first discovered by the Rev. Mr. Wilkinson some twenty years ago.

**443) Cryptarithm – authors**

A=7 C=6 E=4 H=8 M=5 O=9 R=0 S=1 T=3 U=2

6872640 + 84114 + 59904 + 89540 + 132703 = 7238901.

**450) The adventurous snail**

At the end of seventeen days the snail will have climbed 17 ft., and at the end of its eighteenth day-time task it will be at the top. It instantly begins slipping while sleeping, and will be 2 ft. down the other side at the end of the eighteenth day of twenty-four hours. How long will it take over the remaining 18 ft.? If it slips 2 ft. at night it clearly overcomes the tendency to slip 2 ft. during the daytime, in climbing up. In rowing up a river we have the stream against us, but in coming down it is with us and helps us. If the snail can climb 3 ft. and overcome the tendency to slip 2 ft. in twelve hours’ ascent, it could with the same exertion crawl 5 ft. a[Pg 246] day on the level. Therefore, in going down, the same exertion carries it 7 ft. in twelve hours—that is, 5 ft. by personal exertion and 2 ft. by slip. This, with the night slip, gives it a descending progress of 9 ft. in the twenty-four hours. It can, therefore, do the remaining 18 ft. in exactly two days, and the whole journey, up and down, will take it exactly twenty days.

**452) Firsters and Secunders**

The reply is Friday

On Thursday, a Firster will speak the truth and admit that he lied on Wednesday.

On Thursday, a Secunder will lie about Wednesday and claim he lied on Wednesday.

So the Inspector is reading the report on Friday.

**453) Cryptaithm: days**

A=7 D=3 E=0 N=5 O=2 S=8 T=6 U=1 W=9 Y=4

6108374

815374

——–

6923748

**453) What comes next?**

The numbers represent

**454) Matchsticks – hexagon**

There are probably other solutions

**460) The snail**

Answer: he crawls up 40 cm during a full 24 hr period. At the end of the 5th night he is at a height of 2 meter. At the end of the 6th day he is at 3 meter, and does not slide back.

**461) 15 differences**

the number of free polyominoes:

Number of monominoes: 1

Number of dominoes: 1

Number of triominoes: 2

Number of tetrominoes: 5

Number of pentominoes: 12

Number of hexominoes: 35

Number of heptominoes: 108

Number of octominoes: 369

**462) Sudoku**

Author Oceanh assumed. License CC BY 2.5, https://commons.wikimedia.org/w/index.php?curid=2489372

**463) Cities**

Each of the cities in the group has a square the name of which starts with the letter T:

Tianmen square, Bejing

Tahrir square, Cairo

Taksim Square – Istanbul

Times square, New york

Taylor Square – Sydney

Taunton Green – Taunton, Massachusetts

Of course, London belongs in this group:

Trafalgar square, London,

Amsterdam has the Dam and the Beestenmarkt, Berlin has Alexanderplatz, Breitscheidplatz, Gendarmenmarkt, John-F.-Kennedy-Platz, Platz der Republik, Potsdamer Platz,

and Schlossplatz. but none of these starts with a T.

**470) The beggar and the cigarettes**

Answer: from the first 49 butts he makes 7 cigarettes and smokes them. At that moment he has 7 new butts plus 55-49=6 found buts. The 7 new ones give him one more butt, and when he smokes it, one more but. Together with the cigarette buts he had found but not yet used, it gives him another but. That nmakes a total of 7 + 2 = 9 cigarettes.

**471) The hiker, the cyclist and the moped**

Lets shorten the names to A, B and C.

C. Takes the moped for the first 100/3 km, where she leaves it behind for B. That takes her just 5/3 hrs. She walks the remaining 80/3 km, and arrives at her destination after 5/3 + 16/3 = 7 hours.

B. Cycles 10 km (taking her 1 hr) and leaves the bike behind for A. She hikes the next 70/3 km in 14/3 hr, and then picks up the moped left behind by C. The remaining 80/3 km take 80/3/20 = 4/3 hr, making her arrive after 1 + 14/3 + 4/3 = 7 hrs.

A. Hikes the first 10 km (taking her 2 hours) and then picks up the bicycle left behind by B. She walks the remaining 50 km, which takes her 5 more hours, making her arrive after 7 hours.

There are other solutions. I have the feeling that for any combination of three people with three speeds there is a solution in which they all three arrive at the finish in the same time, but I have no proof for this.

**479) Cryptarithm – dyas**

A=4 D=7 M=6 N=3 O=0 S=8 T=1 W=2 Y=9

603749

603749

——-

1207498

**480) The farmer**

Answer:

furrow 1 – turn 1 at end

furrow 2 – turn 1 at start

furrow 3 – turn 2 at end

furrow 4 – turn 2 at start

furrow 5 – turn 3 at end

furrow 6 – turn 3 at start

furrow 7

**481) Three threes**

(3 + 3 / .3 = 20

**482) Rearrange matchsticks and have 3 squares**

**483) Bongard 34**

In the figures on the left, all curves are outward. The figures on the right side has at least one inward curve

**491) Round about**

squeaked

unmarked

magazine

mineral

exchange

recreate

trailing

immortal

majority

emigrate

the first letters make the word summertime

**492) Bongard problem 13**

The intersecting lines create 3 triangles.