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Quants - Problems

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Director
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Quants - Problems [#permalink] New post 12 Sep 2012, 22:54
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
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Re: Quants - Problems [#permalink] New post 13 Sep 2012, 01:13
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getgyan wrote:
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


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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PhD in Applied Mathematics
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Director
Director
User avatar
Affiliations: SAE
Joined: 11 Jul 2012
Posts: 509
Location: India
Concentration: Strategy, Social Entrepreneurship
GMAT 1: 710 Q49 V37
GPA: 3.5
WE: Project Management (Energy and Utilities)
Followers: 20

Kudos [?]: 86 [0], given: 262

Re: Quants - Problems [#permalink] New post 13 Sep 2012, 01:29
getgyan wrote:
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



Hey !

Kudos to you. :)

Thanks!
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First Attempt 710 - first-attempt-141273.html

Re: Quants - Problems   [#permalink] 13 Sep 2012, 01:29
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