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getgyan
Joined: 11 Jul 2012
Last visit: 27 Nov 2017
Posts: 378
Own Kudos:
Given Kudos: 269
Affiliations: SAE
Location: India
Concentration: Strategy, Social Entrepreneurship
Schools: Schulich '15 (A) Neeley '15 (M$)
GMAT 1: 710 Q49 V37
GPA: 3.5
WE:Project Management (Energy)
12 Sep 2012, 22:54
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A no when divided by D gives remainder 7 and when divided by 3D gives remainder 20.
What will be the remainder when twice the number is divided by 3D?
a.1
b.40.
c.27
d.13
e.can' be determined
What will be the remainder when twice the number is divided by 3D?
a.1
b.40.
c.27
d.13
e.can' be determined
13 Sep 2012, 01:13
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getgyan
Denote our number by N. Then we can write:
N = DQ + 7, where Q is a non-negative integer. Necessarily 7 < D.
N = 3Dq + 20, where q is also a non-negative integer, and 20 < 3D.
From the above two equations, we can deduce that DQ + 7 = 3Dq + 20, or D(Q - 3q) = 13.
It follows that D must be a factor of 13, and since D > 7, then D must be equal to 13.
From the second equality, multiplying by 2, we obtain 2N = 6 * 13q + 40 = (3 * 13)*(2q) + 3 * 13 + 1, which means that the remainder when 2N is divided by 3D = 3 * 13 is 1.
Answer A.
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