IMO choice E.
Let x be the number, we can get x=31n+29. Since 31 is an odd number, 31n could be odd or even (it's odd when n is odd, it's even when n is even). So 31n+29 can be either odd or even. We can easily deduct: when odd numbers are divided by even number 16, there must be remainders; when even numbers are divided by 16, remainders could be 0. So data not sufficient.
Another way is to try with real numbers.
When n=1, remainder of (31n+29)/16 is 12;
When n=2, remainder of (31n+29)/16 is 11.
So data not sufficient.
confused divisor
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maihuna
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Great thank you. OA is correct.
There is another way to look at it:
Number = 31q+29
If 31q was having some common factor a multiple of 16 we know the remainder, here we cant so E
There is another way to look at it:
Number = 31q+29
If 31q was having some common factor a multiple of 16 we know the remainder, here we cant so E
