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p, q, and r are positive

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by vipulgoyal » Tue Apr 07, 2015 9:22 pm
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

[spoiler]D[spoiler][/spoiler]
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Source: — Problem Solving |

by sanju09 » Wed Apr 08, 2015 1:37 am
vipulgoyal 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

[spoiler]D[spoiler][/spoiler]
It implies that p, q, r are digits only, say 2, 3, and 4; and then look for the factors by dividing 234234 by 234, and we get 234234 = 234 × 1001, where the factor 234 is optional but the factor 1001 is a must.

Since 1001 = 7 × 11 × 13, out of which, only [spoiler]7 is in the options, hence (D)[/spoiler].
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by GMATGuruNY » Wed Apr 08, 2015 2:31 am
vipulgoyal 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
Integer PQRPQR can be rephrased as follows:
P(10�) + Q(10�) + R(10³) + P(10²) + Q(10) + R

= P(10�) + P(10²) + Q(10�) + Q(10) + R(10³) + R

= 10²P(10³ + 1) + 10Q(10³ + 1) + R(10³ + 1)

= (10³ + 1)(10²P + 10Q + R)

= (1001)(10²P + 10Q + R)

= (7*11*13)(10²P + 10Q + R).

Thus, 7 must be a factor of PQRPQR.

The correct answer is D.
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