Question

A 32-digit number has all 9’s. Find the remainder when the number is divided by 111.

• A. 9
• B. 11
• C. 1
• D. 99

Answer 1

The correct option is D 99

Any number of the form aaa (3 digits) or aaaaaa (multiple of 3 digits, i.e. 6, 9 .....) is divisible by 111.

Therefore group the above numbers into groups of 3 or multiples of 3.

A 32-digit number can be broken up as (3*10 digits) + (2 digits). The first part is divisible by 111. The last two digits will thus be the remainder = 99.