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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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Joined: 02 Sep 2009
Posts: 51280
If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b  [#permalink]

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01 Aug 2018, 22:15
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Difficulty:

15% (low)

Question Stats:

82% (00:46) correct 18% (02:18) wrong based on 55 sessions

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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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Location: India
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WE: Information Technology (Computer Software)
Re: If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b  [#permalink]

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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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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

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Location: India
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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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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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Joined: 25 Jul 2017
Posts: 94
Re: If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b  [#permalink]

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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
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Joined: 18 Jun 2018
Posts: 255
If a*b*c*d = 390, where a, b, c and d are positive integers, and a < b  [#permalink]

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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
$$(1,2,3,65);(1,2,5,39);(1,3,5,26);(1,2,13,15);(1,3,10,13);(1,5,6,13)$$
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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