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In the series of all consecutive integers from 1 to n inclusive, the

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In the series of all consecutive integers from 1 to n inclusive, the [#permalink] New post   12 Aug 2024, 02:21
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In the series of all consecutive integers from 1 to n inclusive, the product of all the numbers in the series is divisible by 990.


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Re: In the series of all consecutive integers from 1 to n inclusive, the [#permalink] New post   20 Sep 2024, 03:27
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990 = 2*3^2*5*11, because 11 is a prime number, the series of all consecutive integers from 1 to n must include 11, then n > 9
=> A
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Re: In the series of all consecutive integers from 1 to n inclusive, the [#permalink]
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