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.
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Thanks Moderator. As usual.
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I concur with Bunuel here. You are not likely to see this, at least not in DS format. You could possibly see something similar in PS format and it will be based on logic, not hit and trial. Hit and trial makes it long, repetitive and cumbersome, things GMAT doesn't mess with. You will have a starting point and there will be a reason why an alphabet will stand for a particular digit.
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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??

From the stem we can assume that E is not 0.
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To add to what Bunuel said, the question stem tells us that when we add three 2 digit numbers, we get a 3 digit number. Had you obtained a 2 digit number as the sum, you would have wirtten the addition as

AD
BD
CD
---
FG

and not EFG.
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That's a relief. Cause I was able to solve the problem but it took over 5 mins.

how long does it take when you did your error and trial? It took me almost 7 minutes

Hi, Why cant it be AD =56; BD =76; CD =96 and EFG =218

In this case the sum is E+F+G = 11 only.

56 + 76 + 96 = 228, not 218 and in this case E and F are not distinct.
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Another version of the same numbers could be
16 +36 + 56 = 108 => 9

No, this does not work: A and E must be distinct. please refer to the highlighted part.
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Interesting problem, luckily the fact the consecutive odd restrictions might actually help to eliminate numbers as Bunuel elegantly decsribed. Interestingly another case, albeit not consecutive odd i.e A,B,C as 7,8,9 also result in an E+F+G as 9. See below.
71
81
+91
243
2+4+3= 9

Hi Bunuel

Why we do not consider 0 as an option for D?
Thanks in advance.

50 + 70 + 90 =220, so E and F are both 2, while we are told that all the digits are distinct.
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What about 58 + 78 + 98 = 234? Would this not qualify since all the letters have a different value?

RedOcean, Bunuel has explained this scenario as well is his first reply. Check it out. The catch however, is we should see the addition through to the end and not just assume there are two possible ways for Statement 1, since both the ways give the same sum of 9!

Hi Bunuel
In first statement ,Case 1,why have we assumed that E can only be =1

If A, B, and C are 1, 3, and 5, then the sum is less than 200.
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