Fun Number Games For Children

Mathematics can be fun. The apparent paradoxes of ruthless logic, the twists and tricks of simple arithmetic, are the delightful thorns in every intelligent person's intellectual flesh. The truth is that we're all intellectual snobs, and games of skill and chance (where the skill is ours, the chance the other fellow's) intrigue us all, from nine to ninety, student to dilettante. Here is a mine of brain-teasers and brain-trainers, to while away the idle hour in improving your mathematical skills.

Squares Within Squares Game: The task is to write such numbers in the diagram that the sum of the squares of two adjacent numbers is the same as the sum of the squares of the two on the opposite side of the diagram. For example, put 16 in square A and 2 in square B. 16 2 = 256; 2 2 = 4; 256 + 4 = 260. We said that F 2 4~ G 2 must be the same. Suitable numbers would be 8 and 14, because 8 2 = 64; I4 2 == 196; 64 + 196 = 260.

Similarly B 2 + C 2 must be equal to G 2 + H 2 ; also, A 2 + K 2 = F 2 + E 2 .

What numbers must we write in the empty squares ? Only whole numbers may be used. Since A 2 + B 2 = F 2 + G 2 , then A 2 F 2 = G 2 B 2 ; in other words, the difference between the squares of numbers on the same diagonal must always be the same. In our case, the difference is i6 2 - 8 2 = I4 2 2 2 = 192

Similarly, C 2 - H 2 = 192.

But the difference between the squares of two numbers must be equal to the sum of those numbers multiplied by their difference. Using symbols: (x y) (x -f y) = # 2 jy 2 .

Therefore, we can write: (C + H) (C ~ H) = 192.

The result 192 also tells us that (C + H) and (C H) cannot both be odd numbers; otherwise, their product would not be even. If one (say, C + H) is even, then the other must be as well, because the sum of the difference of the two numbers can be even only if
both numbers, C and H, are even or if both are odd.

Expand 192, using even numbers: 2 X 96, 4 x 48, 6 x 32, 8 x 24, 12 X 16. Therefore: And these numbers then can be written instead of C and H.

C + H = 48
C-H= 4

Further: C + H = 48

H = 22

The Broken Board: When we insert the numbers into their positions, it seems as if it does not matter which we take as I and which as H. However, we must take care. If in one pair the larger number is in the upper half of the diagram, we must be sure that the larger number is in the pair next to it in the lower half, since the sum of the squares of two larger numbers cannot give the same result as the sum of two smaller ones. Continuing in this fashion, we can get the other numbers too.

Susan was very interested in how numbers are related to each other. As soon as she saw a number, her imagination started working until she found something interesting about it. "Look, Claire," she said to her friend. "Look what I have noticed. Can you see that broken board?" Claire said, "Yes, I can see it. What about it? It says 3,025."

"See how two numbers were left when the board was broken, 30 and 25. If we add them together, we get 55. And 55 X 55 (that is, 55*) is 3,025, which is the original number," said Susan proudly.

"Yes, you are right," said Claire. "Let's find other numbers which are similar, and then we can tell the teacher about it at the next math lesson."

So they took pencils and paper and tried out various numbers. Suddenly Claire exclaimed, "Eureka! 9,801." Indeed, 98 + 1 ==99, and 99 x 99 = 9,801.

A few days later at school Susan wrote down the numbers in question on the board. "What do you think?" asked the teacher. "Are there any other numbers of this type?"

"Please," said George, "is there a way of finding such numbers without using a trial-and-error method?" "Yes," said the teacher. "George is thinking about this just as a mathematician does when he keeps trying to find a general rule to cover all possible solutions. Let's have a look at 2,025 : 20 + 25 = 45 and 45 X 45 = 2,025."

"But our numbers are better," shouted Claire.

"What do you mean by better?"

"Well, in our numbers all the digits are different."

"You are right," said the teacher. "But 2,025 cannot be excluded for that reason; let's see how many numbers of this type there are."

They tried and tried, but apart from 3,025 (55 X 55), 9,801 (99 x 99), and 2,025 (45 X 45), they could not find any others. The teacher then explained that there are none.

Why? The four-figure number must be given by the square of a two-figure number; let's call this # 2 . Let us call the two two-digit numbers x and y. We are saying that the two-figure numbers are added, that the result is squared, and that we get back to the original four-figure number. That is: (x +3 ; ) 2 = & r x + y = a an d y = a x.

As we can see from Claire's example, we can think of 01 as a two-figure number, and even oooo is a satisfactory four-figure number.

On the other hand, in the original four-figure number, x can be regarded as the number of hundreds (expressing the thousands as hundreds) and y the units (expressing the tens as units). Consequently, 2 (the original number) can be written as: 1oo# + y = a

We know that jy equals a - #; therefore, substituting: + a - x = a 2 = a 2 a

As we said at the beginning, x must be a whole number. This can only happen if a(a i) can be divided by 99 without a remainder being left (99 can be expressed as 9 X n).# can be a whole number in four cases:
1. a = 99 when the fraction is simplified so we get x = 98 and y = i, giving the four-figure number as 9,801.

2. a i = 99. But then a = 100, which we cannot use, as a 2 == 10,000, which is a five-figure number.

3. a is divisible by 9, and (a i) by n. How do we find a number like that?

Let us write down the one and two-digit numbers which are divisible by nine and the numbers which are one less than these: 8,18,17,27,26,36,35,54,53,63,62,72,71,81,80,90, 99 and 98.

The only pair of numbers which satisfies all our requirements is 45 and 44. In this case, when we simplify our equation, we get: x = 20 and y = 25, giving 2,025 as the four-figure number.