monsoon1 wrote:
If a, b, and c are different non negative digits, which of the following CANNOT be a solution to the addition problem below?
abc +
cba
-------------
A) 929
B) 1,110
C) 1,111
D) 1,322
E) 1,776
Source : Manhattan Advanced Quant Question No. 4
OFFICIAL SOLUTION
Note that many of the answer choices have four digits, which would require “carrying” from the hundredths digit. This observation gives us a good starting place.
If \(a + c < 10\), no carrying would be required from the ones place to the tens place. The resulting sum would have an even tens digit (\(b + b = 2b\)) and could have either three or four digits. A three digit sum will be a palindrome, as both the ones digit and hundreds digit will be the single-digit \(a + c\). A four digit sum occurs for \(a + c < 10\) only when \(a + c = 9\) and \(2b ≥ 10\) (i.e., carry a 1 from the tens place to the hundreds place), and must have the following digits:
1 0 even 9 .
If \(a + c ≥ 10\), we must “carry” a 1 from the ones place to the tens place. The resulting sum would have an odd tens digit (\(1 + b + b = 2b + 1\)) and must have four digits. Both the ones digit and hundreds digit are formed by summing a + c, so these digits will either be the same (if we don’t carry from the tens digit) or differ by exactly 1 (that is, if a 1 is carried from the tens digit).
We now inspect the answer choices:
(A) OK. Even tens digit in a three digit number.
(B) OK. Odd tens digit in a four digit number with hundreds digit = ones digit + 1.
(C) OK. Odd tens digit in a four digit number with hundreds digit = ones digit.
(D) NO. Even tens digit in a four digit number, yet the first two digits are not 1 and 0.
(E) OK. Odd tens digit in a four digit number with hundreds digit = ones digit + 1.
Alternatively, we could try to produce each sum:
(A) 929 = 316 + 613
(B) 1,110 = 258 + 852
(C) 1,111 = 209 + 902
(D) CANNOT
(E) 1,776 = 789 + 987
The correct answer is D.