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If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b

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If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b  [#permalink]

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New post 01 Aug 2018, 22:15
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If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b < c < d, which of the following could be the value of d − a?

A. 9
B. 10
C. 11
D. 13
E. 15

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Re: If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b  [#permalink]

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New post 01 Aug 2018, 22:26
Bunuel wrote:
If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b < c < d, which of the following could be the value of d − a?

A. 9
B. 10
C. 11
D. 13
E. 15



a*b*c*d = 390
Upon prime factorization we get the below,
2*3*5*13 = 390
Given : a<b<c<d, therefore
2<3<5<13

d-a = 13 - 2 = 11

Hence C
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Re: If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b  [#permalink]

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New post 01 Aug 2018, 22:27
Bunuel wrote:
If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b < c < d, which of the following could be the value of d − a?

A. 9
B. 10
C. 11
D. 13
E. 15



390 = 2*3*5*13

Given

a<b<c<d

a-d = 13 - 2 = 11

The best answer is C.
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Re: If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b  [#permalink]

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New post 01 Aug 2018, 22:29
Bunuel wrote:
If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b < c < d, which of the following could be the value of d − a?

A. 9
B. 10
C. 11
D. 13
E. 15


The easiest way to do this problem is prime-factorize 390 which is 2*3*5*13

We can have 2*3*5*13 as a possible value for a*b*c*d s.t a < b < c < d.

Therefore, the possible difference between the value of d - a is 13 - 2 = 11(Option C)
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Re: If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b  [#permalink]

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New post 01 Aug 2018, 22:45
Bunuel wrote:
If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b < c < d, which of the following could be the value of d − a?

A. 9
B. 10
C. 11
D. 13
E. 15


390 = 2*3*5*13.
Also given a < b < c < d. Therefore, d=13 & a =2.
Now d-a = 13-2 =11
C is the answer.
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If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b  [#permalink]

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New post 01 Aug 2018, 23:35
Bunuel wrote:
If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b < c < d, which of the following could be the value of d − a?

A. 9
B. 10
C. 11
D. 13
E. 15


OA: C

\(a*b*c*d = 390\)
\(390 =2^1*3^1*5^1*13^1\)
As given \(a < b < c < d\), one of value of \(a\) can be \(2\) and of \(d\) can be \(13\).
in that case \(d-a =13-2=11\). it matches one of option so it is OA.

Not related to OA
\(390 =1*2^1*3^1*5^1*13^1\)
other possibilities of value of \((a,b,c,d)\) are
\((1,2,3,65);(1,2,5,39);(1,3,5,26);(1,2,13,15);(1,3,10,13);(1,5,6,13)\)
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If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b &nbs [#permalink] 01 Aug 2018, 23:35
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If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b

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