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Multiples of numbers

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Multiples of numbers

by adi » Sat Mar 08, 2008 2:19 pm
Hello all,

I really hate these types of problems. They seem easy, yet I always have problems with them. I got this one right but it was mainly thanks to guessing.

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If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?

A) 10
B) 11
C) 12
D) 13
E) 14

----------------------------------

*** Answer should be 11 ***
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Re: Multiples of numbers

by netigen » Sat Mar 08, 2008 2:39 pm
[quote="adi"]Hello all,

I really hate these types of problems. They seem easy, yet I always have problems with them. I got this one right but it was mainly thanks to guessing.

----------------------------------

If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?

A) 10
B) 11
C) 12
D) 13
E) 14

----------------------------------

*** Answer should be 11 ***[/quote]


This one sounds easy if you think this way:

prime factors of 990 = 11 * 3 * 3 * 5 * 2

which means that 11 has to be in the list of number from 1 to n (since 990 is a factor of product of all integers from 1 to n)

hence the least possible value for n = 11

Hope this makes sense.
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by xilef » Tue Mar 11, 2008 11:09 am
It says that 1 to n, INCLUSIVE, is a multiple of 990, so if n=11, then 11! is divisible by 990. If it is exclusive then the answer is 10 and 11, otherwise the answer is 10.
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by [email protected] » Sun Jun 29, 2008 1:05 pm
For this question, why just because 11 is a multiple of 990 does this mean that the product of all integers from 1 to 11 (or 11!) is a multiple of 990.

For an example, 5 is a prime factor of 25, but 5! is not a multiple of 25?

Thanks!

Eric
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by Ian Stewart » Sun Jun 29, 2008 1:48 pm
[email protected] wrote:For this question, why just because 11 is a multiple of 990 does this mean that the product of all integers from 1 to 11 (or 11!) is a multiple of 990.

For an example, 5 is a prime factor of 25, but 5! is not a multiple of 25?

Thanks!

Eric
Again, look at the prime factorization of 990:

990 = 2*3^2*5*11

Since 11! is divisible by 2, 3^2, 5 and 11, 11! is divisible by 990.
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