das.ashmita wrote:The three-digit positive integer n can be written as ABC, in which A, B, and C stand for the unknown digits of n. What is the remainder when n is divided by 37?
(1) A+B10+C100=B+C10+A100
(2) A+B10+C100=C+A10+B100

OA: D

Important point: The value of a 3-digit integer xyz is 100x + 10y + z
Example: 723 = (7)(100) + (2)(10) + 3

Statement 1: A+B10+C100=B+C10+A100
On the left-hand side, A represents the units digit of the sum, and on the right-hand side, B represents the units digit of the sum.
Since both sums (left and right) are equal, we can conclude that A=B

Also, on the left-hand side, B represents the tens digit of that sum, and on the right-hand side, C represents the tens digit of that sum.
So, we can conclude that B=C

From this, we can conclude that A=B=C
In other words, all 3 digits of the number ABC are equal.
This means that the number could be 111, 222, 333, ...,888, or 999

It turns out that all of these possible numbers yield the same remainder when divided by 37. That remainder is 0. In other words, each of the 9 possible numbers is divisible by 37.

Aside: 111 is divisible by 37. Since the other numbers (222, 333, etc) are all multiples of 111, they too are divisible by 37.

So, statement 1 is SUFFICIENT


Statement 2: A+B10+C100=C+A10+B100
Using the same logic as above, we can conclude that A=B=C
So, statement 2 is SUFFICIENT

Answer = D

Cheers,
Brent