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multiple of 990

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multiple of 990

by galinaphillips » Mon Jun 04, 2007 4:29 pm

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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?

10
11
12
13
14

The answer of this mba.com practice problem is 11, but don't understand why it's not 10 or how to solve.
Thanks!
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Re: multiple of 990

by jayhawk2001 » Mon Jun 04, 2007 7:56 pm
galinaphillips wrote: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?

10
11
12
13
14

The answer of this mba.com practice problem is 11, but don't understand why it's not 10 or how to solve.
Thanks!
If you get the prime factors of 990, you will see that it includes 11.

Since the positive integer is formed as 1*2*3*...*n, you need 11
somewhere in that product.

990 = 11 * 10 * 9.

Since the product involves all numbers from 1...n, n=11 will
be sufficient. You don't need n to be greater than 11.
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by plodder_81 » Mon Jun 04, 2007 8:05 pm
n is a positive integer and the product of all the integers from 1 to n, inclusive, means we are looking n! which is n * n-1 * n-2 * ...1

It is stated that n! is a multiple of 990 so n! should be divisible by the factors of 990 . 990 = 11 * 10* 9 as well.

11 ! would be the smallest number which contains 11*10*9 . 10 ! wouldn't be divisible by 11 and so it cannot be chosen here
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Thanks

by galinaphillips » Tue Jun 05, 2007 6:19 am
Thanks a lot! It was very helpful! :)
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by lqc172 » Sun Jun 28, 2015 4:36 pm
I still don't quite get this question...

Does "least possible" integer here means an integer that is possible but the smallest?
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by theCEO » Sun Jun 28, 2015 5:24 pm
lqc172 wrote:I still don't quite get this question...

Does "least possible" integer here means an integer that is possible but the smallest?
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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by Brent@GMATPrepNow » Mon Jun 29, 2015 8:06 am
If n is a positive integer and 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
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N
Similarly, we can say:
If N is is a multiple of k, then k is "hiding" within the prime factorization of N

Examples:
24 is divisible by 3 <--> 24 = (2)(2)(2)(3)
70 is divisible by 5 <--> 70 = (2)(5)(7)
330 is divisible by 6 <--> 330 = (2)(3)(5)(11)
56 is divisible by 8 <--> 56 = (2)(2)(2)(7)

So, if if some number is a multiple of 990, then 990 is hiding in the prime factorization of that number.

Since 990 = (2)(3)(3)(5)(11), we know that one 2, two 3s, one 5 and one 11 must be hiding in the prime factorization of our number.

For 11 to appear in the product of all the integers from 1 to n, n must equal 11 or more.
So, the answer is B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Matt@VeritasPrep » Mon Jun 29, 2015 3:17 pm
lqc172 wrote:I still don't quite get this question...

Does "least possible" integer here means an integer that is possible but the smallest?
Yup, the least POSITIVE integer that has both of these properties:

(1) can be expressed as n!, where n is a positive integer
(2) is divisible by 990

Since 990 = 11 * 10 * 9, we need those factors. 11! is the smallest factorial with all three, so n = 11.
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