Question
A 32-digit number has all 9’s. Find the remainder when the number is divided by 111.
Answer: Option D
:
D
option d
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
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:
D
option d
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
Was this answer helpful ?
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