Question

Manish took a 3 digit number and formed two other 3 digit numbers using the digits of the original number. He then added the three resulting numbers and divided their sum by 37. He concluded that the result was exactly divisible by 37, no matter what number he chose. Is Manish's statement true or false?

• A. True
• B. False

Answer 1

The correct option is A

True

Let the number be abc.

abc = 100a + 10b + c.

By rearranging the digits, let us say the two other numbers formed are cab and bca.

Expressing these two numbers in the expanded form,

cab = 100c + 10a + b

bca = 100b + 10c + a

abc + cab + bca
= (100a + 10b + c) + (100c + 10a + b) + (100b + 10c + a)
= 111(a + b + c)
= 37 $\times$ 3 $\times$ (a + b + c), since 111 is the product of 37 and 3.

Hence the sum is divisible by 37.