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

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In the correctly worked addition problem above, A, B, C, D [#permalink]  12 Mar 2012, 06:56
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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
[Reveal] Spoiler: OA

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Re: In the correctly worked addition problem above, A, B, C, D [#permalink]  12 Mar 2012, 12:22
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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 ?

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.

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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Re: In the correctly worked addition problem above, A, B, C, D [#permalink]  12 Mar 2012, 13:12
Expert's post
Thanks Moderator. As usual.
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Re: In the correctly worked addition problem above, A, B, C, D [#permalink]  12 Mar 2012, 23:12
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carcass 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 ?

(1) A, B, and C are consecutive odd integers

(2) E = 2

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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Status: May The Force Be With Me (D-DAY 15 May 2012) Joined: 06 Jan 2012 Posts: 291 Location: India Concentration: General Management, Entrepreneurship Followers: 1 Kudos [?]: 128 [0], given: 16 Re: In the correctly worked addition problem above, A, B, C, D [#permalink] 12 Mar 2012, 23:21 Ya hopefully shall never see this on the GMAT. Was dumbfounded _________________ Giving +1 kudos is a better way of saying 'Thank You'. GMAT Club Legend Joined: 09 Sep 2013 Posts: 6187 Followers: 345 Kudos [?]: 70 [0], given: 0 Re: In the correctly worked addition problem above, A, B, C, D [#permalink] 17 Oct 2013, 22:09 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Current Student Joined: 07 Jan 2013 Posts: 45 Location: India Concentration: Finance, Strategy GMAT 1: 570 Q46 V23 GMAT 2: 710 Q49 V38 GPA: 2.9 WE: Information Technology (Computer Software) Followers: 1 Kudos [?]: 16 [0], given: 23 Re: In the correctly worked addition problem above, A, B, C, D [#permalink] 18 Oct 2013, 08:34 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?? _________________ Help with Kudos if I add to your knowledge realm. Math Expert Joined: 02 Sep 2009 Posts: 29194 Followers: 4741 Kudos [?]: 50147 [0], given: 7529 Re: In the correctly worked addition problem above, A, B, C, D [#permalink] 20 Oct 2013, 12:20 Expert's post adg142000 wrote: 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?? From the stem we can assume that E is not 0. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5867 Location: Pune, India Followers: 1482 Kudos [?]: 7988 [0], given: 190 Re: In the correctly worked addition problem above, A, B, C, D [#permalink] 20 Oct 2013, 19:27 Expert's post adg142000 wrote: 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?? 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: In the correctly worked addition problem above, A, B, C, D [#permalink]  01 Aug 2014, 15:34
VeritasPrepKarishma wrote:
carcass 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 ?

(1) A, B, and C are consecutive odd integers

(2) E = 2

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.

That's a relief. Cause I was able to solve the problem but it took over 5 mins.
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Re: In the correctly worked addition problem above, A, B, C, D [#permalink]  01 Aug 2014, 15:35
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 ?

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.

P.S. Though not very hard this question is not likely to appear on the GMAT because of long and boring math.

how long does it take when you did your error and trial? It took me almost 7 minutes
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Re: In the correctly worked addition problem above, A, B, C, D [#permalink]  05 Jul 2015, 22:27
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.
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Re: In the correctly worked addition problem above, A, B, C, D [#permalink]  05 Jul 2015, 23:55
Expert's post
karthikv606 wrote:
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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Re: In the correctly worked addition problem above, A, B, C, D [#permalink]  06 Jul 2015, 20:35
Another version of the same numbers could be
16 +36 + 56 = 108 => 9

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 ?

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.

P.S. Though not very hard this question is not likely to appear on the GMAT because of long and boring math.
Math Expert
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Posts: 29194
Followers: 4741

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Re: In the correctly worked addition problem above, A, B, C, D [#permalink]  07 Jul 2015, 00:07
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KUDOS
Expert's post
kelvind13 wrote:
Another version of the same numbers could be
16 +36 + 56 = 108 => 9

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 ?

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.

P.S. Though not very hard this question is not likely to appear on the GMAT because of long and boring math.

No, this does not work: A and E must be distinct. please refer to the highlighted part.
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Re: In the correctly worked addition problem above, A, B, C, D   [#permalink] 07 Jul 2015, 00:07
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