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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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Updated on: 14 Aug 2017, 00:45
6
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Difficulty:

25% (medium)

Question Stats:

73% (01:08) correct 27% (01:21) wrong based on 73 sessions

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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

Originally posted by banksy on 29 Mar 2011, 07:30.
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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29 Mar 2011, 07:45
1
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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29 Mar 2011, 19:29
1
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.

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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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18 Mar 2019, 14:56
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Re: If the product of the integers from 1 to n is divisible by 490, what   [#permalink] 18 Mar 2019, 14:56
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