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Games with Numbers

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Games with Numbers:

There are several tricks with numbers to test divisibility by another numbers.

(i) Reversing the digits - two digit number:

Now, let us see if we can explain Monica’s “trick”.

Suppose Jack chooses the number ab, which is a short form for the 2-digit number 10a + b.

On reversing the digits, he gets the number

⇒ ba = 10b + a.

When he adds the two numbers he gets

⇒ (10a + b) + (10b + a) = 11a + 11b = 11 (a + b).

So, the sum is always a multiple of 11, just as Monica had claimed.

Similarly, we can try reversing the digits for dividing by 9.

Selecting a two-digit number: 85

Reversing the digits: 58

Subtract: 85 - 58 = 27

Divide by 9: `27/9` = 3

Remainder is 0.

If the tens digit is larger than the ones digit (that is, a > b),
he does: (10a + b) – (10b + a) = 10a + b – 10b – a = 9a – 9b = 9(a – b).

If the ones digit is larger than the tens digit (that is, b > a), he does:
(10b + a) – (10a + b) = 9(b – a).

  • And, of course, if a = b, he gets 0.
    In each case, the resulting number is divisible by 9. So, the remainder is 0.

(ii) Reversing the digits – three digit number:

Let us see how this trick works.

Let the 3-digit number chosen by Monica be abc = 100a + 10b + c.

After reversing the order of the digits, she gets the number

cba = 100c + 10b + a.

On subtraction:

  • If a > c, then the difference between the numbers is
    (100a + 10b + c) – (100c + 10b + a)
    = 100a + 10b + c – 100c – 10b – a
    = 99a – 99c
    = 99(a – c).
  • If c > a, then the difference between the numbers is
    (100c + 10b + a) – (100a + 10b + c)
    = 99c – 99a
    = 99(c – a).
  • And, of course, if a = c, the difference is 0.
    In each case, the resulting number is divisible by 99. So the remainder is 0.

(iii) Forming three-digit numbers with given three-digits:

For a given three-digit number, rearrange the digits such that all the three numbers are distinct and add them all. Then dividing it by 37, the remainder is always 0.

For example, assuming a three-digit number 489.

On rearranging the digits: 894 and 948.

Add all the three numbers: 489 + 894 + 948 = 2331

Divide 2331 by 37: 2331/37 = 63; remainder is 0.

Let us see.

abc = 100a + 10b + c

cab = 100c + 10a + b

bca = 100b + 10c + a

abc + cab + bca = 111(a + b + c) = 37 × 3(a + b + c), which is divisible by 37

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