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Re: In the correctly worked addition problem above, A, B, C, D
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12 Mar 2012, 13:22
18
18
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.
_________________
Re: In the correctly worked addition problem above, A, B, C, D
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13 Mar 2012, 00:12
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2
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.
_________________
Re: In the correctly worked addition problem above, A, B, C, D
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18 Oct 2013, 09:34
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??
_________________
Re: In the correctly worked addition problem above, A, B, C, D
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20 Oct 2013, 13:20
1
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.
_________________
Re: In the correctly worked addition problem above, A, B, C, D
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20 Oct 2013, 20:27
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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
Re: In the correctly worked addition problem above, A, B, C, D
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01 Aug 2014, 16:34
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.
_________________
......................................................................... +1 Kudos please, if you like my post
Re: In the correctly worked addition problem above, A, B, C, D
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01 Aug 2014, 16:35
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 ?
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.
how long does it take when you did your error and trial? It took me almost 7 minutes _________________
......................................................................... +1 Kudos please, if you like my post
Re: In the correctly worked addition problem above, A, B, C, D
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06 Jul 2015, 21:35
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 ?
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.
Re: In the correctly worked addition problem above, A, B, C, D
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07 Jul 2015, 01:07
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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 ?
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.
No, this does not work: A and E must be distinct. please refer to the highlighted part.
_________________
Re: In the correctly worked addition problem above, A, B, C, D
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02 Oct 2017, 21:43
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
(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
Why we do not consider 0 as an option for D? Thanks in advance.
(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
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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19% (02:37) correct 
