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Reversing the Digits of a 3-digit Number and Subtracting them

The general form of the original 3-digit number abc is (100a + 10b + c) and its reversed number is
cba = (100c + 10b + a). Now, subtract the smaller number from the larger number.
• If a > c, (100a + 10b + c) – (100c + 10b + a) = 99a – 99c = 99(a – c)

• If c > a, (100c + 10b + a) – (100a + 10b + c) = 99c – 99a = 99(c – a)
• If a = c, the difference will be 0.

In each case, the difference between the original number abc and the reversed number cba is
always divisible by 99. The quotient is the difference between the extreme digits, a and c, (a  c).
For example, let’s take a 3-digit number 542, its reversed number is 245.
And difference = 542 – 245 = 297

∴ 297 ÷ 99 = 3, remainder = 0
Forming and Adding All Possible 3-digit Numbers

Let us take three digits a, b, and c and form all possible 3-digit numbers (without repetitions). The
expanded form of all such numbers is as follows:

Digits: a, b, c

abc cab bca acb cba bac

100a + 10b + c 100c + 10a + b 100b + 10c + a 100a + 10c + b 100c + 10b + a 100b + 10a + c

By adding these numbers, we have
abc + cab + bca + acb + cba + bac = (100a + 10b + c) + (100c + 10a + b) + (100b + 10c + a)

+ (100a + 10c + b) + (100c + 10b + a) + (100b + 10a + c)
= 222a + 222b + 222c = 222(a + b + c) = 2 × 3 × 37(a + b + c)

Hence, the sum of all three-digit numbers formed by rearranging the digits a, b, and c is divisible
by 2, 3, 37, 111, and 222 and the sum of its digits.
Example 1: Check whether the sum of three numbers formed by interchanging the digits of the
number 117 is divisible by 111 or not.
Solution: Given number is 117

The three numbers formed by interchanging the digits of 117 are 711, 171, 117
Sum of these numbers = 711 + 171 + 117 = 999 = 9 × 111

Hence, the sum is divisible by 111.
Example 2: Check whether the sum of all the numbers formed by the digits of the number 235 is
divisible by 222 and 37 or not.
Solution: Given number is 235.

The numbers formed by interchanging the digits of 235 are 523, 325, 253, 532, and 352
Sum of these numbers = 235 + 523 + 325 + 253 + 532 + 352 = 2220 = 222 × 10 = 37 × 60

Hence, the sum is divisible by both 37 and 222.

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