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If the product of the integers from 1 to n is divisible by 490, what

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If the product of the integers from 1 to n is divisible by 490, what [#permalink]

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New post 29 Mar 2011, 07:30
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If the product of the integers from 1 to n is divisible by 490, what is the least possible value of n?

(A) 7
(B) 14
(C) 21
(d) 28
(E) 35
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Aug 2017, 00:45, edited 1 time in total.
Renamed the topic and edited the question.

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Re: If the product of the integers from 1 to n is divisible by 490, what [#permalink]

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New post 29 Mar 2011, 07:45
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490 = 7 x 7 x 5 x 2 = 14 x 7 x 5

Let the number be 'N'

N/490 = integer

Therefore, if the denominator is to be completely cancelled out from the numerator, N must contain 14, 7 and 5.

=> the smallest value of n must be 14.

Sol: 'B'
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Re: If the product of the integers from 1 to n is divisible by 490, what [#permalink]

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New post 29 Mar 2011, 19:29
490 = 7 * 7 * 5 * 2

1 * .. *n = 7 * 7 * 5 * 2 * k

So it needs to have at least two 7s and one 5 and one 2 as factor

So n = 14, as factorial upto 7 has only one 7, and next 7 occurs at 14!, 5 and 2 are automatically included.

Answer - B
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Re: If the product of the integers from 1 to n is divisible by 490, what [#permalink]

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New post 13 Aug 2017, 07:37
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Re: If the product of the integers from 1 to n is divisible by 490, what   [#permalink] 13 Aug 2017, 07:37
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