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If n is a positive integer and the product of all integers

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Intern
Joined: 18 Feb 2018
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Re: If n is a positive integer and the product of all integers  [#permalink]

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18 Mar 2018, 13:04
Can someone please confirm why the answer is not 10? (i.e. 5*2). I understand how to do prime factorization just want to confirm my understanding of why the answer is not 2*5. Thanks for your help.
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Re: If n is a positive integer and the product of all integers  [#permalink]

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18 Mar 2018, 13:15
rsciretta14 wrote:
Can someone please confirm why the answer is not 10? (i.e. 5*2). I understand how to do prime factorization just want to confirm my understanding of why the answer is not 2*5. Thanks for your help.

rsciretta14 because prime factors of 990 are 2 * 5 * 3 * 3* 11. the answer is not 10 because 11 prime factor is missing
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Joined: 03 Jul 2018
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Re: If n is a positive integer and the product of all integers  [#permalink]

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22 Sep 2018, 01:19
SunnyDelight wrote:
just entered this in google and you're right! (1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11) / 990 = 40 320

meanwhile (1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10) / 990 = 3 665.45455 wouldn't work

It makes sense because the highest prime factor of 990 must also be a factor of any multiple of 990.

I guess I need to do more practice problems involving multiples and number properties.

Thank you man! This helped me understand the problem.
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Re: If n is a positive integer and the product of all integers  [#permalink]

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11 Nov 2018, 21:24
prime factorization of 990=3*3*11*2*5

As 11 is required so 11 is the answer
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Re: If n is a positive integer and the product of all integers  [#permalink]

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21 Jan 2019, 06:56
rsciretta14 wrote:
Can someone please confirm why the answer is not 10? (i.e. 5*2). I understand how to do prime factorization just want to confirm my understanding of why the answer is not 2*5. Thanks for your help.

Let me try to explain for the sake of self-edification.

n is a +int (because negative factorial is undefined), and saying the product of all ints 1 to n is the same as saying n!

So we are looking for a n! that will equal 990*k (k is some number which has all the other factors of n!)
In other words, 1*2*3*...*n = 990*k

Prime factorization of 990 = 9*10*11 = (3^2)*(2*5)*11
So we have 1*2*3*...*n = 2*9*5*11 * k

The least possible n has to be a number that makes our equation true.
The reason it can't be 10 is because 10! won't have the prime number 11, which we need to make 990 on LHS of equation.

Writing it out:
1*2*3*4*5*6*7*8*9*10*11 = 1*2*3*_*5*_*_*_*_*_*11 *k
So for the equation to be true the other factors of n! (4, 6 [2*other 3 here], 7, 8, 9, 10) have to be in k.

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Re: If n is a positive integer and the product of all integers &nbs [#permalink] 21 Jan 2019, 06:56

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