Manhattan GMAT Challenge Problem of the Week – 13 Dec 2010
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Question
In the correctly worked addition problem above, A, B, C, D, E, F, and G are distinct digits. What is the sum of E, F, and G ?
(1) A, B, and C are consecutive odd integers
(2) E = 2
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed.
Answer
(1) SUFFICIENT: This statement yields three possibilities for A, B, and C, in some order: {1, 3, 5}, {3, 5, 7}, {5, 7, 9}; it makes no difference which of A, B, and C is which digit, because they are all added together.
- If A, B, and C are 1, 3, and 5 (in whatever order), then the digit E must be 1. This means the digits arent distinct, so its impossible.
- If A, B, and C are 3, 5, and 7 (in whatever order), then E = 1. That means D is one of the leftover digits 0, 2, 4, 6, 8, 9. If these values are plugged into D one at a time, the only one that yields all distinct digits is D = 4 (34 + 54 + 74 = 162).
- If A, B, and C are 5, 7, and 9 (in whatever order), then E = 2. That means D is one of the leftover digits 0, 1, 3, 4, 6, 8. If these values are plugged into D one at a time, the only one that yields all distinct digits is D = 8 (58 + 78 + 98 = 234).
- Therefore, there are only two possible values of EFG: 162 and 234. In both cases, the sum of E, F, and G is 9, so (1) is sufficient.
(2) INSUFFICIENT: If E = 2, then trial and error will yield a number of working sets of digits yielding different values for EFG. Examples: 58 + 68 + 78 = 204 (E + F + G = 6); 41 + 71 + 91 = 203 (E + F + G = 5); 49 + 69 + 89 = 207 (E + F + G = 9); and several others. Therefore, (2) is insufficient.
The correct answer is A; statement (1) alone is sufficient.
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