Bunuel wrote:
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 ?
AD
BD
CD
---
EFG
Notice that E can be only 1 or 2 (no sum of 3 two-digit numbers can give number more than 297).
(1) A, B, and C are consecutive odd integers. 3 cases are possible:
(i) A, B, and C are 1, 3, and 5 (it doesn't matter which is which) --> 1+3+5=9 then E (hundreds digit) can only be 1, which is not possible since we are told that the digits are distinct and we already have 1 (A, B, or C);
(ii) A, B, and C are 3, 5, and 7 (it doesn't matter which is which) --> 3+5+7=15 then E (hundreds digit) can only be 1. So, D can be 0, 2, 4, 6, 8, or 9. After trial and error we can get that only D=4 will give all distinct digits:
34
54
74
---
162
E+F+G=9.
(iii) A, B, and C are 5, 7, and 9 (it doesn't matter which is which) --> 5+7+9=21 then E (hundreds digit) can only be 2. So, D can be 1, 3, 4, 6, or 8. After trial and error we can get that only D=8 will give all distinct digits:
58
78
98
---
234
E+F+G=9.
So, as you can see in both valid cases (ii and iii) the sum of E, F, and G is 9. Sufficient.
(2) E = 2. After some trial and error you can find that several numbers can be found which will give different values for the sum of E, F, and G, for example: 58+78+98=234 and 38+78+98=214. Not sufficient.
Answer: A.
P.S. Though not very hard this question is not likely to appear on the GMAT because of long and boring math.
Hi Bunuel ,
Just curious to know why A=1,B=3,C=5 and sum = 9 and the resulting number 96 cannot be the case. It says any digits so why EFG =096 cant be a possibility??