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

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Let a, b, c, d, and e be distinct integers such that (6 − a)(6 − b)(6  [#permalink]

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New post Updated on: 14 Mar 2019, 02:58
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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 04 Oct 2013, 00:39.
Last edited by Bunuel on 14 Mar 2019, 02:58, edited 2 times in total.
Edited the question.
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Re: Let a, b, c, d, and e be distinct integers such that (6 − a)(6 − b)(6  [#permalink]

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New post 04 Oct 2013, 00:52
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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.

Answer: C.
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Re: Let a, b, c, d, and e be distinct integers such that (6 − a)(6 − b)(6  [#permalink]

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New post 04 Oct 2013, 01:10
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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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Re: Let a, b, c, d, and e be distinct integers such that (6 − a)(6 − b)(6  [#permalink]

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New post 04 Oct 2013, 22:00
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
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Re: Let a, b, c, d, and e be distinct integers such that (6 − a)(6 − b)(6  [#permalink]

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New post 18 Sep 2017, 19:17
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.
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Re: Let a, b, c, d, and e be distinct integers such that (6 − a)(6 − b)(6  [#permalink]

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New post 18 Sep 2017, 21:20
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Re: Let a, b, c, d, and e be distinct integers such that (6 − a)(6 − b)(6  [#permalink]

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Re: Let a, b, c, d, and e be distinct integers such that (6 − a)(6 − b)(6   [#permalink] 24 Dec 2018, 04:14
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