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# Let a, b, c, d and e be distinct integers such that (6-a)(6-

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Director
Joined: 29 Nov 2012
Posts: 758
Let a, b, c, d and e be distinct integers such that (6-a)(6-  [#permalink]

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Updated on: 03 Oct 2013, 23:44
4
13
00:00

Difficulty:

75% (hard)

Question Stats:

63% (02:07) correct 37% (01:34) wrong based on 249 sessions

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Let a, b, c, d and e be distinct integers such that (6-a)(6-b)(6-c)(6-d)(6-e) = 45. What is a+b+c+d+e?

A) 5
B) 17
C) 25
D) 27
E) 30

Originally posted by fozzzy on 03 Oct 2013, 23:39.
Last edited by Bunuel on 03 Oct 2013, 23:44, edited 1 time in total.
Edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 52164
Re: Let a, b, c, d and e be distinct integers such that (6-a)(6-  [#permalink]

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03 Oct 2013, 23:52
5
2
fozzzy wrote:
Let a, b, c, d and e be distinct integers such that (6-a)(6-b)(6-c)(6-d)(6-e) = 45. What is a+b+c+d+e?

A) 5
B) 17
C) 25
D) 27
E) 30

45=3*3*5

Since a, b, c, d and e are distinct integers, then (6-a), (6-b), (6-c), (6-d), and (6-e) must also be distinct integers, thus (6-a), (6-b), (6-c), (6-d), and (6-e) must be -1, 1, -3, 3, and 5.

(6-a) + (6-b) + (6-c) + (6-d) + (6-e) = 30- (a+b+c+d+e) = -1+1-3+3+5 = 5
a+b+c+d+e = 30 - 5 = 25.

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Manager
Joined: 10 Jan 2011
Posts: 141
Location: India
GMAT Date: 07-16-2012
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Re: Let a, b, c, d and e be distinct integers such that (6-a)(6-  [#permalink]

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04 Oct 2013, 00:10
Since 45 = 3 * 3* 5. The whole product should result into 3, 3, and 5. as 6-1= 5 we assume that a=1, also 6-3 = 3 so let say b=3, now we need one more 3 which can be 6-9 = -3 (as a, b,c,d,e are distinct we can not have c = 3), so c= 9. the whole product is now negative, so we have to balance it, so we say d= 7 (resulting into 6-7 = -1) and last must be 5 so that we stick to our product. so e= 5 (6-5 =1).

The total = 1+3+9+7+5= 25. Answer is c
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-------Analyze why option A in SC wrong-------

Director
Joined: 29 Nov 2012
Posts: 758
Re: Let a, b, c, d and e be distinct integers such that (6-a)(6-  [#permalink]

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04 Oct 2013, 21:00
Since the prime factorization is 3*3*5

The constraint in the question is that they are distinct so the only possible way is 3,-3,5,1,-1

6-a = 3 => a=3
6-b = -3 => b=9
6-c = 5 => c=1
6-d = 1 => c=5
6-e = -1 => e = 7

Answer is C = 25
Intern
Joined: 20 Jan 2017
Posts: 5
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Let a, b, c, d and e be distinct integers such that (6-a)(6-  [#permalink]

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18 Sep 2017, 18:17
The question is exclusively asking for value of a+b+c+d+e so that should be 5.
why are we subtracting it from 30.
Math Expert
Joined: 02 Sep 2009
Posts: 52164
Re: Let a, b, c, d and e be distinct integers such that (6-a)(6-  [#permalink]

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18 Sep 2017, 20:20
rghvaggarwal wrote:
The question is exclusively asking for value of a+b+c+d+e so that should be 5.
why are we subtracting it from 30.

Check here: https://gmatclub.com/forum/let-a-b-c-d- ... l#p1274086
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Re: Let a, b, c, d and e be distinct integers such that (6-a)(6-  [#permalink]

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24 Dec 2018, 03:14
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Re: Let a, b, c, d and e be distinct integers such that (6-a)(6- &nbs [#permalink] 24 Dec 2018, 03:14
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# Let a, b, c, d and e be distinct integers such that (6-a)(6-

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