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p, q, and r are positive integers. If p, q, and r are assembled into t

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p, q, and r are positive integers. If p, q, and r are assembled into t [#permalink]

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p, q, and r are positive integers. If p, q, and r are assembled into the six-digit number pqrpqr, which one of the following must be a factor of pqrpqr?

(A) 23
(B) 19
(C) 17
(D) 7
(E) none of the above
[Reveal] Spoiler: OA

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Re: p, q, and r are positive integers. If p, q, and r are assembled into t [#permalink]

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New post 28 Oct 2014, 01:09
Nice question. i was wondering what would be the shortest way to solve this problem.

Bunuel, your help is required ?
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Re: p, q, and r are positive integers. If p, q, and r are assembled into t [#permalink]

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One short way -
pqrpqr = 1000pqr + pqr = (1000+1)pqr = 1001pqr

Therefore any factor of 1001 is a factor of pqrpqr
7 is a factor of 1001

So D

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Re: p, q, and r are positive integers. If p, q, and r are assembled into t [#permalink]

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New post 13 Oct 2017, 06:42
JohanH wrote:
One short way -
pqrpqr = 1000pqr + pqr = (1000+1)pqr = 1001pqr

Therefore any factor of 1001 is a factor of pqrpqr
7 is a factor of 1001

So D


How to know that 7 is a factor of 1,001?:
1. Take the last digit of 1,001 and multiply it by 2: 1x2 = 2
2. Calculate the difference between the other digits and the last result: 100 - 2 = 98, is 98 a factor of 7? Yes (98 : 7 = 14). Then, answer D is correct.

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Re: p, q, and r are positive integers. If p, q, and r are assembled into t [#permalink]

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New post 19 Oct 2017, 21:37
JohanH wrote:
One short way -
pqrpqr = 1000pqr + pqr = (1000+1)pqr = 1001pqr

Therefore any factor of 1001 is a factor of pqrpqr
7 is a factor of 1001

So D




Why are we doing 1000 pqr???
just wondering what the logic is

Thanks

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p, q, and r are positive integers. If p, q, and r are assembled into t [#permalink]

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New post 21 Oct 2017, 04:36
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PareshGmat wrote:
p, q, and r are positive integers. If p, q, and r are assembled into the six-digit number pqrpqr, which one of the following must be a factor of pqrpqr?

(A) 23
(B) 19
(C) 17
(D) 7
(E) none of the above


Another, more clear method of doing this.

pqrpqr => 100000p+10000q+1000r+100p+10q+r =>(100000+100)p +(10000+10)q+(1000+1)r =>100100p + 10010q +1001r => 1001 (100p+10q+r)

So pqrpqr = 1001 (100p+10q+r) = 1001 (pqr)

1001 is always divisible by 7.

hence answer is D
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Re: p, q, and r are positive integers. If p, q, and r are assembled into t [#permalink]

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New post 21 Oct 2017, 04:40
zanaik89 wrote:
JohanH wrote:
One short way -
pqrpqr = 1000pqr + pqr = (1000+1)pqr = 1001pqr

Therefore any factor of 1001 is a factor of pqrpqr
7 is a factor of 1001

So D




Why are we doing 1000 pqr???
just wondering what the logic is

Thanks


Hi Zanaik,

Here, is how it is derived.

pqrpqr => 100000p+10000q+1000r+100p+10q+r =>(100000+100)p +(10000+10)q+(1000+1)r =>100100p + 10010q +1001r => 1001 (100p+10q+r)

So pqrpqr = 1001 (100p+10q+r) = 1001 (pqr)
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Kudos are the only way to tell whether my post is useful.

GMATPREP1: Q47 V36 - 680
Veritas Test 1: Q43 V34 - 630
Veritas Test 2: Q46 V30 - 620

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Re: p, q, and r are positive integers. If p, q, and r are assembled into t [#permalink]

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New post 21 Oct 2017, 23:51
Jabjagotabhisavera wrote:
zanaik89 wrote:
JohanH wrote:
One short way -
pqrpqr = 1000pqr + pqr = (1000+1)pqr = 1001pqr

Therefore any factor of 1001 is a factor of pqrpqr
7 is a factor of 1001

So D




Why are we doing 1000 pqr???
just wondering what the logic is

Thanks


Hi Zanaik,

Here, is how it is derived.

pqrpqr => 100000p+10000q+1000r+100p+10q+r =>(100000+100)p +(10000+10)q+(1000+1)r =>100100p + 10010q +1001r => 1001 (100p+10q+r)

So pqrpqr = 1001 (100p+10q+r) = 1001 (pqr)


Excellent explanation!!

Thank u very much for the reply...Really appreciate it!!

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Re: p, q, and r are positive integers. If p, q, and r are assembled into t   [#permalink] 21 Oct 2017, 23:51
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