Solutions to puzzles 501-750


501) 2020
1)
0*0*2*2=0 Figure can be drawn without lifting your pencil from the paper and without drawing a line twice
0+0+2/2=1
0+0*2+2=2
2*2+0-0!=3
2*2+0+0=4
2*2+0!+0=5
2*2+0!+0!=6
2^(2+0!)-0!=7
2^(2+0!)-0!=8
2^(2+0!)+0!=9

2) “This century” means the first two digits are “20”. So which years can we make up with 2 and 0? Those years are 2000, 2002, 2020 and 2022.

3) “This millennium” means the first digit is a “2”. Add 2200, 2202, and 2220 to the previous list to have all years this millennium with a 2 and a zero. That makes 7 years if the second digit is a zero. But the zero may also be one of the other numbers 1-9. So we have 9*7 is 63 different years.

502) The Christmas Geese
1) Farmer Rouse sent exactly 101 geese to market. Jabez first sold Mr. Jasper Tyler half of the flock and half a goose over (that is, 50-½ + ½, or 51 geese, leaving 50); he then sold Farmer Avent a third of what remained and a third of a goose over (that is, 16-2/3 + 1/3, or 17 geese, leaving 33); he then sold Widow Foster a quarter of what remained and three-quarters of a goose over (that is, 8-1/4 + 3/4 or 9 geese, leaving 24); he next sold Ned Collier a fifth of what he had left and gave him a fifth of a goose “for the missus” (that is, 4-4/5 + 1/5 or 5 geese, leaving 19). He then took these 19 back to his master.

510) Bongard problem 10
All figures on the left can be drawn without lifting your pencil from the paper and without drawing a line twice.

503) Ikura






511) 1 to 9


512) a pack of milk
His mother exclaims: “Why did you buy 6 milk???”

513) The card frame puzzle
The sum of all the pips on the ten cards is 55. Suppose we are trying to get 14 pips on every side. Then 4 times 14 is 56. But each of the four corner cards is added in twice, so that 55 deducted from 56, or 1, must represent the sum of the four corner cards. This is clearly impossible; therefore 14 is also impossible. But suppose we came to trying 18. Then 4 times 18 is 72, and if we deduct 55 we get 17 as the sum of the corners. We need then only try different arrangements with the four corners always summing to 17, and we soon discover the following solution:—

The final trials are very limited in number, and must with a little judgment either bring us to a correct solution or satisfy us that a solution is impossible under the conditions we are attempting. The two centre cards on the upright sides can, of course, always be interchanged, but I do not call these different solutions. If you reflect in a mirror you get another arrangement, which also is not considered different. In the answer given, however, we may exchange the 5 with the 8 and the 4 with the 1. This is a different solution. There are two solutions with 18, four with 19, two with 20, and two with 22—ten arrangements in all. Readers may like to find all these for themselves.

520) Bongard problem 11
All figures on the left have one end outside, and one end inside.

521) Polar bears
Polar bears gather around ice holes (the pip in the middle of a dice, and the number of polar bears are the number of the pips around the ice hole:

So there are 4 polar bears around the ice hole.

522) Tasting the Plum Puddings.



“Everybody, as I suppose, knows well that the number of different Christmas plum puddings that you taste will bring you[Pg 91] the same number of lucky days in the new year. One of the guests (and his name has escaped my memory) brought with him a sheet of paper on which were drawn sixty-four puddings, and he said the puzzle was an allegory of a sort, and he intended to show how we might manage our pudding-tasting with as much dispatch as possible.” I fail to fully understand this fanciful and rather overstrained view of the puzzle. But it would appear that the puddings were arranged regularly, as I have shown them in the illustration, and that to strike out a pudding was to indicate that it had been duly tasted. You have simply to put the point of your pencil on the pudding in the top corner, bearing a sprig of holly, and strike out all the sixty-four puddings through their centres in twenty-one straight strokes. You can go up or down or horizontally, but not diagonally or obliquely; and you must never strike out a pudding twice, as that would imply a second and unnecessary tasting of those indigestible dainties. But the peculiar part of the thing is that you are required to taste the pudding that is seen steaming hot at the end of your tenth stroke, and to taste the one decked with holly in the bottom row the very last of all.

523) The cross of cards
There are eighteen fundamental arrangements, as follows, where I only give the numbers in the horizontal bar, since the remainder must naturally fall into their places.
5 6 1 7 4
3 5 1 6 8
3 4 1 7 8
2 5 1 7 8
2 5 3 6 8
1 5 3 7 8
2 4 3 7 8
1 4 5 7 8
2 3 5 7 8
2 4 5 6 8
3 4 5 6 7
1 4 7 6 8
2 3 7 6 8
2 4 7 5 8
3 4 9 5 6
2 4 9 5 7
1 4 9 6 7
2 3 9 6 7
It will be noticed that there must always be an odd number in the centre, that there are four ways each of adding up 23, 25, and 27, but only three ways each of summing to 24 and 26.

530) Bongard problem 14
The start and end lines both slope down.

531) Seals
Like polar bears, seals concentrate at ice holes, as they have to breath, but normally live in the water, under the ice: take the dice with an ice hole (1, 3 and 5), and total the pips on the bottom side. Dice total 7 on opposing sides, so a 1 scores 6 seals, a 3 scores 4 and a 5 scores 2.
Solution = 8.

532) Domino laydown 1

533) Card triangles
The following arrangements of the cards show (1) the smallest possible sum, 17; and (2) the largest possible, 23.

It will be seen that the two cards in the middle of any side may always be interchanged without affecting the conditions. Thus there are eight ways of presenting every fundamental arrangement. The number of fundamentals is eighteen, as follows: two summing to 17, four summing to 19, six summing to 20, four summing to 21, and two summing to 23. These eighteen fundamentals, multiplied by eight (for the reason stated above), give 144 as the total number of different ways of placing the cards.

540) Bongard problem 12
The lines intersect exactly twice not once, thrice, zero or four times.

541) Rectangle and lines


542) Domino problem 2

543) The “T” card puzzle
If we remove the ace, the remaining cards may he divided into two groups (each adding up alike) in four ways; if we remove 3, there are three ways; if 5, there are four ways; if 7, there are three ways; and if we remove 9, there are four ways of making two equal groups. There are thus eighteen different ways of grouping, and if we take any one of these and keep the odd card (that I have called “removed”) at the head of the column, then one set of numbers can be varied in order in twenty-four ways in the column and the other four twenty-four ways in the horizontal, or together they may be varied in 24 × 24 = 576 ways. And as there are eighteen such cases, we multiply this number by 18 and get 10,368, the correct number of ways of placing the cards. As this number includes the reflections, we must divide by 2, but we have also to remember that every horizontal row can change places with a vertical row, necessitating our multiplying by 2; so one operation cancels the other.

550) Bongard problem 19
In the siX figures on the left, there are always 2 dots in the left half and 1 dot in the right half. In the six squares on the right, there are always 2 dots in the right half and one in the left half.

551) Fish
Fish live under the ice, so again we count the pips at the bottom of the dice. But seals eat fish, so the fish are only where there are no seals, hence where there is no ice hole: under the 2, 4 and 6.
Solution: 9.

552) Domino problem 3

553) Eleusis
after a black card, play a high one (7,8,9,10,J,Q,K). After a red card, play a low one (A-7).

560) Bongard problem S3
In the six squares on the left, the two small dashes are in one line with the end-points of the connector, while in the six squares to the right at least one is perpendicular to the ends of the connecting line.

561) Dice puzzle from G&P 70
Add the even numbered dice and subtract the odd numbered dice.
0.

570) Bongard problem 36
In the 6 shapes on the left, the begin and endpoints are on the same height.

541) 4 rolls of 4 dice
group the dice as ab-cd.
Then 31-26=5.
25-24=1
44-12=32
53-44=9

580) Bongard problem 37
The letter types on the left are serif, those on the right sand serif.
.

581) The Dice Numbers
The sum of all the numbers that can be formed with any given set of four different figures is always 6,666 multiplied by the sum of the four figures. Thus, 1, 2, 3, 4 add up 10, and ten times 6,666 is 66,660. Now, there are thirty-five different ways of selecting four figures from the seven on the dice—remembering the 6 and 9 trick. The figures of all these thirty-five groups add up to 600. Therefore 6,666 multiplied by 600 gives us 3,999,600 as the correct answer.

Let us discard the dice and deal with the problem generally, using the nine digits, but excluding nought. Now, if you were given simply the sum of the digits—that is, if the condition were that you could use any four figures so long as they summed to a given amount—then we have to remember that several combinations of four digits will, in many cases, make the same sum.

10 11 12 13 14 15 16 17 18 19 20
1 1 2 3 5 6 8 9 11 11 12
 
21 22 23 24 25 26 27 28 29 30
11 11 9 8 6 5 3 2 1 1

Here the top row of numbers gives all the possible sums of four different figures, and the bottom row the number of different ways in which each sum may be made. For example 13 may be made in three ways: 1237, 1246, and 1345. It will be found that the numbers in the bottom row add up to 126, which is the number of combinations of nine figures taken four at a time. From this table we may at once calculate the answer to such a question as this: What is the sum of all the numbers composed of our different digits (nought excluded) that add up to 14? Multiply 14 by the number beneath t in the table, 5, and multiply the result by 6,666, and you will have the answer. It follows that, to know the sum of all the numbers composed of four different digits, if you multiply all the pairs in the two rows and then add the results together, you will get 2,520, which, multiplied by 6,666, gives the answer 16,798,320.

The following general solution for any number of digits will doubtless interest readers. Let n represent number of digits, then 5 (10n – 1) ) 8! divided by (9 – n)! equals the required sum. Note that 0! equals 1. This may be reduced to the following practical rule: Multiply together 4 × 7 × 6 × 5 … to (n – 1) factors; now add (n + 1) ciphers to the right, and from this result subtract the same set of figures with a single cipher to the right. Thus for n = 4 (as in the case last mentioned), 4 × 7 × 6 = 168. Therefore 16,800,000 less 1,680 gives us 16,798,320 in another way.

590) Bongard problem 38
The number of white triangles is 1 less than the number of other objects

591) The Montenegrin dice game
The players should select the pairs 5 and 9, and 13 and 15, if the chances of winning are to be quite equal. There are 216 different ways in which the three dice may fall. They may add up 5 in 6 different ways and 9 in 25 different ways, making 31 chances out of 216 for the player who selects these numbers. Also the dice may add up 13 in 21 different ways, and 15 in 10 different ways, thus giving the other player also 31 chances in 216.

600) Bongard problem 41
Each number on the left can be written as a square + 2

601) Playing card puzzle 1
Count jack=11, queen=12, king = 13.
All numbers used are prime numbers

610) Bongard problem 41
The digits in the two numbers differ 5.

611) Playing card puzzle 5


620) Square nr 1


621) Playing card puzzle 2
The sums of all rows are even.

622) Bongard dates(2)
nr 3: Month is 01-04-07-10 has a remainder of 1 when divided by 3
nr 4: som of three numbers is 2044 vs 2054

630) Five fives
(5*5 – 5*5)/5 = 0
(5+5)/5 – 5/5 = 1
5 – (5+5+5)/5 = 2
(5+5)/5 + 5/5 = 3
5*5/5 – 5/5 = 4
5*5/5 * 5/5 = 5
5*5/5 + 5/5 = 6
5 + 5/5 + 5/5 = 7
5+5 – (5+5)/5 = 8
(55-5-5)/5 = 9
55/5 – 5/5 = 10
55/5 * 5/5 = 11
55/5 + 5/5 = 12
(55+5+5)/5 = 13
(5! – 5*5)/5 -5 = 14
5 +5 +5 +5 -5= 15
(55+5*5) / 5 = 16
(55+5)/5 + 5 = 17
(5!-5*5-5)/5 = 18
5*5-5-5/5 = 19
5*5-5-5+5 = 20
5*5 +5/5 = 21

631) Playing card puzzle 3
All diagonals add up to 30.

632) Bongard dates(3)
nr 7: spring vs summer on the northern hemisphere
Nr 8: Easter vs first day of ramadan (in Mecca)

640) Square nr 1


641) Playing card puzzle 4
67-49=18, 34-16=18, and so on.

642) Bongard dates(4)
nr 9: closed area vs no closed area in digits
Nr 10: leading zero’s vs no leading zero’s

651) Playing cards – valid hands
The first letter of the the cards in each hand make a familiar English word:
Taste state
Staff staff
Tea jest
Of the three hands presented, only “snake” is a valid word.

660) Square nr 1


662) 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[Pg 209] 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.

670) Ages
John and Jill are 20 and 36, while Bill and Bess are 8 and 36.

671) A trick with dice
All you have to do is to deduct 250 from the result given, and the three figures in the answer will be the three points thrown with the dice. Thus, in the throw we gave, the number given would be 386; and when we deduct 250 we get 136, from which we know that the throws were 1, 3, and 6.

The process merely consists in giving 100a + 10b + c + 250, where a, b, and c represent the three throws. The result is obvious.

680) Bongard dates 1
.All dates are in the first half of the year.

680) Bongard dates 2
The days of the dates on the left are odd, those on the right are even.

680) 7 7’s
(7-7)*(7+7+7+7+7) = 0
7/7 +(7-7)*(7+7+7) = 1
7/7 + 7/7 + (7-7)*7 = 2
(7+7)/7 +(7+7-7)/7 = 3
(7+7+7+7)/7 * 7/7 = 4
7 – 7/7 – 7/7 * 7/7 = 5
7 – 7/7 * 7/7 * 7/7 = 6
7 + 7/7 – (7+7)/(7+7) = 7
7 + 7/7 * 7/7 * 7/7 = 8
7 + 7/7 + 7/7 * 7/7 = 9
7 + 7/7 + 7/7 + 7/7 = 10
7 + (7+7)/7 + (7+7)/7 = 11
7+7 – 7/7 * (7+7)/7 = 12
7+7 -(7+7)/7 + 7/7 = 13
7+7 +(7-7) * (7+7+7) = 14
7+7 +(7+7)/7 – 7/7 = 15
7+7 + 7/7 * (7+7)/7 = 16
7+7 + 7/7 + (7+7)/7 = 17
7+7 + (7+7+7+7)/7 = 18
7+7+7 – 7/7 – 7/7 = 19
7+7+7 – (7+7-7)/7 = 20
7+7+7 – (7+7)/(7+7) = 20
7+7+7 * (7+7)/(7+7) = 21
7+7+7 – (7+7)/(7+7) = 22
(7*7+7+7)/7 +7+7 = 23
7+7+7 +(7+7+7)/7=24

77/7 +77/7 +7 = 29

690) Bongard dates 1
1) f(n) = n * f(n-2)
or written another way:
f(n) = n * (n-2) * (n-4) * … * 1
Example: the 11th term

2) f(n) = n * f(n-3)