If n, m, and p are distinct, five-digit positive integers and if p = n

Official Explanation

To understand this question without going variable-crazy, we can draw a diagram of adding two five-digit numbers with blank spaces, like this,



or whatever helps to visualize the addition. We're being asked whether the digits in the thousands position--the highlighted cells, in this figure--add up independently. Just as we learned in grade school, it boils down to whether we carried a 1 from adding the hundreds digits. If those digits add up to 10 or more, then we'll carry a one and the answer to this question will be "no"; otherwise, it will be "yes." On to the data statements, separately first.

Statement (1) tells us about the wrong column. It's saying nothing about whether the hundreds column will carry over to the thousands. Insufficient.

Statement (2) allows us to determine that there is no carrying of a digit to the thousands column, so we can answer the question definitively (in the affirmative). We have sufficient information.

The correct answer is (B).

Video Explanation

If n, m, and p are distinct, five-digit positive integers and if p = n + m, is the thousands digit of p equal to the sum of the thousands digits of n and m?

Let n, m & p be expressed as n1n2n3n4n5, m1m2m3m4m5 & p1p2p3p4p5 respectively.
p1p2p3p4p5 = n1n2n3n4n5 + m1m2m3m4m5

Q. Is p2 = n2 + m2 ?

(1) The tens digit of p is equal to the sum of the tens digits of n and m.
p2 = n2 + m2
Since p2 = n2 + m2; there is no carry over of sum of unit digts; p1 = n1 + m1
Since there is no carry over of sum of tens digits;
p3 = n3 + m3; or p3 +1 = n3 + m3
It is not certain whether p2 = n2 + m2
NOT SUFFICIENT

(2) The hundreds digit of p is equal to the sum of the hundreds digits of n and m.
p3 = n3 + m3
There is no carry over of sum of hundredth digit
p2 = n2 + m2
SUFFICIENT

IMO B

30% of the people thought the way I thought so I am going to ask this- What if the thousands digits of m and n are 5 or bigger? say 7 and 6? In that case the thousands digit of P is NOT equal to the thousands digit of M + N. (6 + 7 = 13 not 3).
I know it sounds dumb but the question needs to somehow clarify this.

I thought the same way. My answer remains E!

Exactly, mi answe remains E