A Coin Trick

Take a couple of coins -- 5 dollar and 2 dollar (any two similar coins of 
18 mm.  and 25 mm. in diameter will do).
Then, on a sheet of paper, cut out a circle equal to the circumference of the 2 dollar coin.

Do you think the 5-dollar coin will get through this hole?

There is no catch to the problem; it is genuinely geometrical.

Similar Figures

This problem is meant for those who understand geometrical similarity.

Answer the following two questions:

1) Are the two triangles in the picture similar?

2) Are the outer and inner rectangles of the picture frame similar?

The Height of a Tower

There is a very big tower in your town, but you do not know its height. You have, however, a photograph of the tower.

Can it help you to find the real height?

Two Watermelons

A man is selling two watermelons. One is one-quarter bigger than the other, but costs one and a half times more. Which one would you buy?

A Brick

A regular-size brick weighs 4 kilogrammes.
How much will a similar toy brick, four times smaller but made of the same material, weigh?

The Way of the Fly

On the wall inside a cylindrical glass container, three centimeters from the upper circular base,
there is a drop of honey. On the lateral surface, diametrically opposite it, there is a fly.

Show the fly the shortest route to the honey.

The diameter of the cylinder is 10 centimetres and the height 20.

Don’t expect the fly to find this way itself and thus facilitate the solution of the problem: for that it would have to be well versed in geometry, and that is something beyond a fly’s ability.

Find a Plug (3)

And here is yet another problem of the same type.
Find a plug for the three apertures in the picture

*** click here for answer ***

Find a Plug (2)

If you have solved the previous problem, try this also.

You are given a small plank with three holes.
Can you make one plug that would fit all the three apertures?

Find a Plug

You are given a small plank with three holes: square, triangular and circular.

Can you make one plug that would fit all the three apertures?

A Carpenter's Level

You have probably seen a carpenter’s level with a glass tube with a bubble that deviates from the centre when placed on a sloping surface. The bigger the slope, the more does the bubble deviate from the mark.



It moves because, being lighter than the liquid in the tube, it rises to the surface. If the tube were straight, the bubble would move to the end of the tube, that is, to its highest point. A level like that, as it may easily be seen, would be very inconvenient. That is why the tube is usually arched.
When the level is horizontal, the bubble, situated at the highest level point of the tube, is in the centre;
if the level is sloped, the highest point is then not its centre, but some point next to it, and the bubble moves from the mark to another part of the tube.*

The problem is to determine how many millimetres the bubble will move away from the mark if the level is sloped 1/2° and the radius of the arch of the tube is 1 metre.


* It would be more correct to say that "the mark moves from the bubble", because the latter really remains in its place while the tube and the mark glide past.

A Match Trick

Out of 12 matches you can build the figure of a cross, the area to equal five "match" squares.

Can you rearrange the matches in such a way as to cover an area equal to only four "match" squares?
The use of measuring instruments is forbidden

Trough a Magnifying Glass

How big will the angle of 1.5° seem if you look at it through a glass that magnifies things four times?

A Cart

Why does the front axle of the cart wear out faster than the rear?

The Missing Digit

Tell your friend to write any multidigit number, but no ending in noughts, say, 847. Ask him to add up these three digits and then subtract the total from the original.

The result will be:
847 – 19 = 828

Ask him to cross out any one of the three digits and tell you the remaining ones. Then you tell him the digit he has crossed out, although you know neither the original nor what your friend has done with it.

How is this explained?

Very simply:

All you have to do is to find the digit which, added to the two you know, will form the nearest number divisible by 9. For instance, if in the number 828 he crosses out the first digit (8) and tells you the other two (2 and 8), you add the and get 10. The nearest number divisible by 9 is 18. The missing number is consequently 8.

How is that?

No matter what the number is, if you subtract from it the total number of its digits, the balance will always be divisible by 9. Algebraically we can take a for the number of hundreds, b for the number of tens and c for the number of units. The total number of units is therefore:

100a + 10b + c

From this number we subtract the sum total of its digits a + b + c and we obtain:

100a + 10b + c – (a + b + c) = 99a + 9b = 9 (11a + b)

But 9(11a+b) is, of course, divisible by 9. Therefore, when we subtract from a number by the sum total of its digits, the balance is always divisible by 9.

It may happen that the sum of the digits you are told is divisible by 9 (for example, 4 and 5). That shows that the digit your friend has crossed out is either 0 or 9, and in that case you have to say that the missing digit is either 0 or 9.

Here is another version of the same trick:

Instead of subtracting from the original number the sum total of its digits, ask your friend to subtract the same number only transposed in any way he wishes.
For instance, if he writes 8,247, he can subtract 2,748 (if the number transposed is greater that the original, subtract the original).

The rest is done as described above
8,247 – 2,748 = 5,499

If the crossed-out digit is 4, then knowing the other three (5, 9, and 9), you add them up and get 23. The nearest number divisible by 9 is 27.
Therefore, the missing digit is 27 – 23 = 4.

Grandfather and Grandson

“In 1932 I was as old as the last two digits of my birth year. When I mentioned this interesting coincidence to my grandfather, he surprised me by saying that the same applied to him too. I thought that impossible...”

“Of course that's impossible,” a young woman said.

“Believe me, it's quite possible and grandfather proved it too. How old was each of us in 1932?”

Hundred Dollars for Five

A stage magician once made the following attractive proposal to his audience.

“I shall pay 100 dollars to anyone who gives me 5 dollars in 20 coins – 50 cent, 20 cent, and 5 cent coins. One hundred for five! Any takers?”

The auditorium was silent. Some people armed themselves with paper and pencil and were evidently calculating their chances. No one, it seemed, was willing to take the magician at his word.

“I see you find it too much to pay 5 dollars for 100.” The magician went on. “All right. I’m ready to take 3 dollars in 20 coins and pay you 100 dollars for them. Queue up!”

But no one wanted to queue up. The spectators were slow in taking up this chance of making “easy” money.

“What?! You find even 3 dollars too much. Well, I’ll reduce it by another dollar – 2 dollars in 20 coins. How’s that?”

And still there were no takers.

The magician continued:
“Perhaps you haven’t any small change? It’s all right. I’ll trust you. Just write down how many coins of each denomination you’ll give me.”

Who Has It?

This clever trick requires three little things that can be put in one’s pocket – a pencil, a key, and a penknife will do very well. In addition to that, place a plate with 24 nuts – draughts or domino pieces or matches will do just as well – on the table.

Having completed these preparations, ask each of your three friends to put one of the three things into his pocket – one the pencil, the second the key, and the third the penknife. This they must do in your absence, and when you return to the room, you guess correctly where each object is.

The process of guessing is as follows: on your return (i.e., after each has concealed the object) you ask your friends to take care of some nuts – you give one nut to the first, two to the second, and three to the third. Then you leave the room again, telling them that they must take more nuts – the one who has the pencil should take as many nuts as he was given the first time; the one with the key twice as many as he has been given; and the one with the penknife four times the number. The rest, you tell them, should remain in the plate.

When they have done that, they call you into the room. You walk in, look at the plate and announce what each of your friends has in his pocket.

The trick is all the more mystifying since you do it solo, so to speak, without any assistant who could signal to you secretly.

There is really nothing tricky in the riddle – the whole thing is based on calculation. You guess who has each of the things from the number of nuts remaining in the plate. Usually there are not many of them left – from one to seven – and you can count them at once glance.

How, then, do you know who has what thing?

Simple. Each different distribution of the three objects leaves a different number of nuts in the plate. Here is how it is done.

Let us call your three friends Dan, Ed, and Frank, or simply D, E, F. The three things will be as follows: the pencil – a, the key – b, and the penknife – c. The three objects can be distributed among the trio in just six ways:

D E F
-----
a b c
a c b
b a c
b c a
c a b
c b a

There can be no other combinations – the table above exhausts all of them.

Now let us see how many nuts remain after each combination:

DEF    number of nuts taken         total remainder
abc     1+1=2; 2+4=6; 3+12=15    23     1
acb     1+1=2; 2+8=10; 3+6=9      21     3
bac     1+2=3; 2+2=4; 3+12=15    22     2
bca     1+2=3; 2+8=10; 3+3=6      19     5
cab     1+4=5; 2+2=4; 3+6=9        18     6
cba     1+4=5; 2+4=6; 3+3=6        17     7

You will see that the remainder is each time different. Knowing what it is, you will easily establish who has what in his pocket. Once again, for the third time, you leave the room, look at your notebook into which you have written the table above (frankly speaking, you need only the first and last columns). It is difficult to memorize this table, but then there is really no need for that. The table will tell where each thing is.

If, for example, there are five nuts remaining in the plate, the combination is bca, i.e.,
Dan has the key,
Ed has the penknife, and
Frank has the pencil.

If you want to succeed, you must remember how many nuts you gave each of your three friends (the best way is to do so in alphabetic order, as we have done here).

Matches

The man emptied a box of matches on the table and divided them into three heaps.

“You aren't going to start a bonfire, are you?” someone quipped.

“No, they're for my brain-teaser. Here you are – three uneven heaps. There are altogether 48 matches. I won't tell how many there are in each heap. Look well. If I take as many matches from the first heap as there are in the second and add them to the second, and then take as many from the second as there are in the third, and add them to the third, and finally take as many from the third as there are in the first and add them to the first – well, if I do all this, the heaps will all have the same number of matches.

How many were there originally in each heap?”

An Arithmatical Trick

“I'll give you an arithmatical trick and ask you to explain it. One of you – you, Professor, if you like – write down a three-digit number, but don't tell me what it is.”

“Can we have any noughts in it?”

“I set no reservations. You can wirte down any three numerals you want.”

“All right, I've done it. What next?”

“Write the same number alongside. Now you have a six-digit number.”

“Right.”

“Pass the slip to your neighbor, the one farther away from me, and let him divide this six-digit number by seven.”

“It's easy for you to say that, and what if it can't be done?”

“Don't worry, it can.”

“How can you be so sure when you haven't seen the number?”

“We'll talk after you've divided it.”

“You're right. It does divide.”

“Now pass the result to your neighbor, but don't tell me what it is. Let him divide it by 11.”

“Think you'll have your own way again?”

“Go ahead, divide it. There'll be no remainder.”

“You're right again. Now what?”

“Pass the result on and let the next man divide it, say, by 13.”

“That's a bad choice. There are very few numbers that are divisible by 13... You're certainly lucky, this one is!”

“Now give me the slip, but fold it so that I don't see the number.”

Without unfolding the slip, the man passed it on to the professor.
“Here is your number. Correct?”

“Absolutely.” The professor was surprised. “That is the number I wrote down.”

What Are the Digits?

Here is another similar problem.

In the following multiplication more than half of the digits are expressed by X’s.
The task is to find the missing digits.


  xx5
  1xx
  ---
 2xx5
13x0
xxx
-----
4x77x

Resolution of Digits

In the following multiplication more than half of the digits are expressed by X’s.


   x1x
   3x2
   ---
   x3x
 3x2x
x2x5
------
1x8x30

Can you restore the missing digits?

Thirty

The number 30 may easily be written by three 5’s:
5 X 5 + 5

It is harder to do it by using three other identical digits. Try it.

You may find several solutions.

Twenty-Four

It is very easy to write 24 by using three 8’s:
8 + 8 + 8

Can you do that by using three other identical digits?

There is more than one solution to this problem.

One Thousand

Can you write 1,000 by using eight identical digits (in addition to digits you may use signs of operation)?

About "Figures for Fun"

Figures for Fun is a book by Yakov Perelman.
It's a book of math puzzles, brainteasers, stories and conundrums.

It's a very old book (it was published in 1960's), and it is hard to find it at any bookstore now.

I will post some best puzzles here so that anyone who doesn't have access to this book can still enjoy it.

Here is the answer

and here is the answer...