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In the correctly worked addition problem above, A, B, C, D 23 Feb 2025, 13:05
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Bunuel
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.


If A, B and C are 5, 7 and 9, and D = 8, then E = 2, F = 3, and G = 4, so there are 2 cases that give different digits, so wouldn't that make the right answer Both Statements together are not sufficient?

The question asks: What is the sum of E, F, and G ?

For both valid cases, the sum of E, F, and G from (1) is 9.
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Thank you so much!! I completely lost track of what the stem asked, and I wouldn't have reached that far without the solution given by you, so thanks again!
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In the correctly worked addition problem above, A, B, C, D 08 Apr 2025, 12:16
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Your question has no set of options, how are we supposed to know which letter to click on?

carcass

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

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In the correctly worked addition problem above, A, B, C, D 08 Apr 2025, 12:18
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nicolereimertb
Your question has no set of options, how are we supposed to know which letter to click on?

carcass

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

Show Spoiler
Attachment:
challprobadd.jpg
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Hi,

This is a data sufficiency question. Options for DS questions are always the same and usually omitted on the site.

The data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise), you must indicate whether—

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D. EACH statement ALONE is sufficient to answer the question asked.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Hope this helps.­
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In the correctly worked addition problem above, A, B, C, D 25 Oct 2025, 06:52
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KarishmaB why cant D be zero when we are taking numbers, 5,7,9 -> A,B,C
then we have 210 which is all distinct??
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In the correctly worked addition problem above, A, B, C, D 25 Oct 2025, 08:19
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KarishmaB why cant D be zero when we are taking numbers, 5,7,9 -> A,B,C
then we have 210 which is all distinct??
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If D = 0, then the sum is 210 which means G = 0 too. All digits are distinct.
Try another Alphametic here: https://anaprep.com/puzzles-corner-solving-alphametics/
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In the correctly worked addition problem above, A, B, C, D 05 Mar 2026, 03:29
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Great question!
Mixing noting down with some intuition this question becomes straightforward.

Write down all numbers first.
Eliminate options for D --> 0/5 cant come because adding either 3 times will end up with same digit and wont be distinct

Statement 1:
Write down all combos of odd consec integers along with their sum

1,3,5 --> 9
3,5,7 --> 15
5,7,9 -->21

It is fascinating to note 2nd set is eliminated because 5,5 repeats once
Among 1st and third
If we choose first:
Here we use a bit of intuition.

Now,
We need to be weary of carry forward, if we have a carry of 1:
9 becomes 10 and we have 1 already in 1st set --> NOT POSSIBLE
If we have carry of 2:
9 becomes 11 and even worse we already have a double 1 --> NOT POSSIBLE
So hmm we cant have a carry and we add D 3 times. Can we actually do that?

We already discussed 0/5 cant be D,
Here 1,3,5 are also taken.
The only smallest number to add to avoid a carry = 4
BUT 4+4+4 = 12 WE GET CARRY!

Hence by this deduction 1,3,5 --> NOT POSSIBLE COMBOS OF A,B,C

We are left with 5,7,9 --> 21 as sum

Okay, now it is important to note that this 21 could vary depending on carry, we cant have 22 BUT we can have 23 with 2 as carry
Lets check no carry first:

We have used up 5,7,9,2,1
D is not 0/5
Therefore D can be 3,4,6,8
Lets add them up:
3+3+3=9
4+4+4=12
6+6+6=18
8+8+8=24

In this case 3 is eliminated because it leads us back to 9 already used!
Remaining all 3 cases leads us to a carry! hence Having no carry is also NOT POSSIBLE

Lets now check for 2 as a carry and having 23 as E and F:
This leads us with 8+8+8 only (only combo with 2 as carry) = 24

Now we have used up:
5,7,9,2,3(carry of 2 with 1=3),8(D value) and 4 (G value)

Its fascinating to notice that this combo perfectly fits!
And we have eliminated every other case and subcase hence A is enough!
SUFFICIENT.


Statement 2:
There are ways to achieve an E=2 and we have no other constraints hence not enough
NOT SUFFICIENT

Answer: Option A

Bunuel is my approach right?

Carcass

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

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