In the correctly worked addition problem above, A, B, C, D

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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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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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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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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Hello Bunuel

I had a scenario like this

3 4
5 4
+ 7 4
------------
1 6 2

and the other can be

5 8
7 8
+ 9 8
-----------
2 3 4

Then how can A be sufficient?

The question asks "What is the sum of E, F, and G ?". In both valid cases (ii and iii) the sum of E, F, and G is 9. Check the solution it clearly explains this:

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.
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Sorry, I misread the question.

Correct option : A

we have :
1. AD + BD + CD = EFG (D unit is common for all Integer AD, BD, CD)
2. Correctly worked addition problem above, A, B, C, D, E, F, and G are distinct digits.
(Statement :1) A, B, and C are consecutive odd integers Eg. {1,3,5,7,9}
(Statement :2) E = 2

To Find : What is the sum of E, F, and G ?

Statement 1 :
Sum of two digit interger of 3 numbers, gives us 3 digit integer Number
D has to be same for all 3 integer in Unit place.
A, B, C has to be odd integer, keeping unit digit D constant
{11,31,51,71,91} are the set, and unit must give us 3 digit integer
easy : 91+71+51 = 213 = (E + F + G) = (2 + 1+ 3) = 6
Statement 1 sufficient , eliminates option D, and E

Statement 2 : E = 2, this cannot give us Solution but can be used as a check indicator,
Statement 2 is No Sufficieint - Eliminates B, and C

We have:

AD
BD
+CD
------
EFG

We know that A, B, C are consecutive odd: 1,3,5 or 3,5,7 or 5,7,9

We can discard the set 1,3,5.
This is because the carry in this case would be only 1, i.e. E=1, which is not possible since all digits should be distinct.

Case 1: If we use 3,5,7: 3+5+7 = 15
So, there must be a carry from the units place else F=5 will match B=5
To have a carry from the units place, D must be either 4, 6, 8 or 9 (other digits either won't give a carry, or would be equal to one of A, B or C)

# With 4, we have:
34
54
+74
-----
162 => Satisfies all conditions. Thus, E+F+G = 1+6+2 = 9


# 6 cannot be used since the carry would be 1, resulting in F=6, which would be equal to D=6
# 8 cannot be used since the carry would be 2, resulting in F=7, which would be equal to C=7
# 8 cannot be used since we would have G=7, which would be equal to C=7

Case 2: If we use 5,7,9: 5+7+9 = 21
Possible values of D = 6, 7, 8, 9
Note: 1 is not possible since we would have D=F=1
2 is not possible since we would have D=E=2
3 is not possible since we would have G=C=9
4 is not possible since we would have G=E=2
5,7 or 9 is not possible since we would have D equal to one of A, B or C
6 is not possible since we would have F=E=2
8 is the only possible case:

58
78
+98
-----
234 => Satisfies all conditions. Thus, E+F+G = 2+3+4 = 9

Thus, Statement 1 is sufficient since we get a consistent value of E+F+G=9


Statement 2:
We only know that E = 2
There can be many possible scenarios for that:
For example: 97+87+67 = 251
97+87+47 = 231, etc.

Thus, it is not sufficient

Answer A
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