Multiples of numbers

[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.

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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

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*** Answer should be 11 ***

[netigen]

[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.

[xilef]

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.

[[email protected]]

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

[Ian Stewart]

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.