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If k is the smallest positive integer such that 2,940k

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Intern
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Joined: 16 Feb 2013
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If k is the smallest positive integer such that 2,940k  [#permalink]

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New post 04 May 2013, 10:17
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A
B
C
D
E

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Question Stats:

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If k is the smallest positive integer such that 2,940k is the square of an integer, then k must be

A. 3
B. 5
C. 6
D. 15
E. 21
Intern
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Joined: 23 Apr 2013
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Re: If k is the smallest positive integer such that 2,940k  [#permalink]

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New post 04 May 2013, 10:52
1
streamingline wrote:
If k is the smallest positive integer such that 2,940k is the square of an integer, then k must be

A)3
B)5
C)6
D)15
E)21


Spoiler: :: OA
15


\(2940 = 2^2 * 3 * 5 * 7^2\)

So the smallest positive integer to be multiplied to make it a perfect square is 3 * 5 = 15

Correct Option is D :)
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Re: If k is the smallest positive integer such that 2,940k  [#permalink]

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New post 04 May 2013, 19:35
Thanks, didn't realise K was multiplied by 2940
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Re: If k is the smallest positive integer such that 2,940k  [#permalink]

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New post 11 Jan 2016, 08:23
streamingline wrote:
If k is the smallest positive integer such that 2,940k is the square of an integer, then k must be

A. 3
B. 5
C. 6
D. 15
E. 21



Hi,

Pls clarify whether K is a unit digit of the given No. or it is in multiplication?
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Re: If k is the smallest positive integer such that 2,940k  [#permalink]

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New post 11 Jan 2016, 08:24
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If k is the smallest positive integer such that 2,940k  [#permalink]

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New post 06 Jul 2018, 10:04
1
streamingline wrote:
If k is the smallest positive integer such that 2,940k is the square of an integer, then k must be

A. 3
B. 5
C. 6
D. 15
E. 21

• Perfect square rule: A perfect square always has an even number of powers of prime factors.
That rule is often easier to remember this way:
all prime factors must come in pairs. Couplets.

There are no single "copies" of a prime factor in a perfect square.
If there is only one "copy" of a prime factor, or an odd number of copies, then the integer is not a perfect square.

• \(2,940 * k\) is a perfect square

Prime factorize: \(2,940= 2 * 2 *3*5*7*7\)
This situation will not make a perfect square.
The 3s and the 5s are not in pairs. We need one more copy of each.
(We have \(3^1\) and \(5^1\), but exponents on prime factors must be even.)

\(k\) is the number that provides the extra \(3\) and the extra \(5\)
\(k = 3*5 = 15\)

ANSWER D

\(2,940 * (3*5) = 44,100\), which is \(210^2\)
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If k is the smallest positive integer such that 2,940k   [#permalink] 06 Jul 2018, 10:04
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