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If n is a positive integer and the product of all integers from 1 to n

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Re: If n is a positive integer and the product of all integers from 1 to n  [#permalink]

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New post 14 Aug 2016, 07:04
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Lets breakdown the question:
n is a positive integer
Therefore n>0

the product of all the integers from 1 to n, inclusive, is a multiple of 990

Therefore (1 x 2 x 3 ..... x n) / 990 gives an integer
Therefore n!/990 gives an integer
Therefore n!/(9 x 10 x 11)gives an integer

Question: What is the least possible value of n!
This means what is the lowest value that n! can be so that when it is divided by 9 x 10 x 11, the outcome gives an integer

The lowest value of n! should have all the factors that are in the demoninator so that when divided an integer is formed

11! / (9 x 10 x 11) = (11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / (9 x 10 x 11)
= 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

10! / (9 x 10 x 11) = (10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / (9 x 10 x 11)
= (9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / 11 = mixed number

Any number lower than 10! will give a mixed number.

Any number higher that 11! will give an integer, however 11! is the lowest value that will produce an integer when 990 is the denominator.
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Re: If n is a positive integer and the product of all integers from 1 to n  [#permalink]

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New post 15 Jan 2018, 13:26
Hi All,

For a number to be a multiple of 990, that number must have the exact same prime factors (including duplicates) as 990 (but may have "extra" prime factors as well).

This prompt adds the extra stipulation that N has to be as SMALL as possible, so after factoring 990, we need to find the smallest product that "holds" the 2, 5, 11, and two 3s that make up 990.

1(2)(3)(4)....(11) is the smallest product that does that, so N = 11.

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Re: If n is a positive integer and the product of all integers from 1 to n  [#permalink]

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New post 18 Mar 2018, 14:04
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 from 1 to n  [#permalink]

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New post 21 Jan 2019, 07:56
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.

Please let me know if I made an error.
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If n is a positive integer and the product of all integers from 1 to n  [#permalink]

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New post 17 Jul 2019, 11:06
One other way to put it is to pick answers for 990:

A): 10x99=990
B): 11x90=990 we can use 90 as a value of reference since it's the smallest multiplicator

Then any number bigger than 11 that multiplies 90 will be bigger, don't look further, B is the Answer!

I don't really understand the need to put such complicated answers that just confuse people like me who absolutely suck at maths, I really hope that one day some answers will be more straightforward rather than putting factorials and time consuming operations such as LCM calculation...
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If n is a positive integer and the product of all integers from 1 to n   [#permalink] 17 Jul 2019, 11:06

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