abhasjha wrote:If m, n, and p are three-digit integers and m + n = p, is the sum of the units digits of m and n at least 2 more than the sum of the tens digits of m and n?
(1) The tens digit of p is greater than the sum of the tens digits of m and n.
(2) The tens and units digits of p are equal.
Let m = ABC, n = DEF, and p = GHI, where the letters A through I represent digits.
Since m + n = p, we get the following sum:
ABC
DEF
GHI
Question stem, rephrased: Is (C+F) - (B+E) ≥ 2?
Statement 1: The tens digit of p is greater than the sum of the tens digits of m and n.
Since H > B+E, the MAXIMUM possible value of B+E is 8, in which case H=9:
A
3C
D5F
G
9I
In order that H > B+E, the LEAST possible value for C+F is 10, so that a "1" is carried over from the units place to the tens place.
To illustrate:
A3
8
D52
G9
0
In this case, (C+F) - (B+E) = (8+2) - (3+5) = 2.
If the value of B+E decreases, the difference between C+F and B+E will INCREASE.
Thus, it must be true that (C+F) - (B+E) ≥ 2.
SUFFICIENT.
Statement 2: The tens and units digits of p are equal.
The sum could look like this:
111
111
2
22
In this case, (C+F) - (B+E) = (1+1) - (1+1) = 0.
The sum could look like this:
179
109
2
88
In this case, (C+F) - (B+E) = (9+9) - (7+0) = 11.
Since (C+F) - (B+E) is LESS THAN 2 in the first case but GREATER THAN 2 in the second case, INSUFFICIENT.
The correct answer is
A.
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