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Least possible value

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by theCodeToGMAT » Thu Oct 03, 2013 1:51 am
n! = 990 x _

990 = 9 x 10 x 11 = 3^2 x 2 x 5 x 11

Now, "n!" must contain "11" so the least possible answer is 11!

if you take "n" as 10 then "11" is missing..

and if you take "n" as 12, then even though it also contains 11 but it's not the least possible value.

Answer [spoiler]{B}[/spoiler]
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by mevicks » Thu Oct 03, 2013 1:52 am
saidov.mikhail 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 least possible value of n?
a) 10
b) 11
c) 12
d) 13
e) 14
The fancy way saying n! is "product of all the integers from 1 to n"

Given: n! is divisible by 990 (Factorize: 11*3*3*2*5)
So the division of n!/990 should be an integer

The least value of n should cancel out all the factors of 990
Here we see that if we take n to be 11
11! will contain all the factors of 990

11! = 11*10*9(this will cancel out two 3s)*8*7*6*5*4*3*2*1

Hence answer is [spoiler]b) 11[/spoiler]

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by GMATGuruNY » Thu Oct 03, 2013 3:12 am
saidov.mikhail 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 least possible value of n?
a) 10
b) 11
c) 12
d) 13
e) 14
990 = 9*10*11.

The product of all the integers from 1 to n, inclusive = n!.
We can plug in the answers for n.
Since the question stem asks for the LEAST POSSIBLE VALUE of n that is divisible by 9*10*11, start with the SMALLEST answer choice.

Answer choice A: n=10
10*9*8*7*6*5*4*3*2*1)/(9*10*11).
Only the values in blue cancel out.
11 cannot divide into 10!.
Eliminate A.

Answer choice B: n=11
(11*10*9*8*7*6*5*4*3*2*1)/(9*10*11).
Success!
All of the values in blue can divide into 11!.

The correct answer is B.
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by Brent@GMATPrepNow » Thu Oct 03, 2013 5:46 am
saidov.mikhail 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 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 = 2x2x2x3
70 is divisible by 5 <--> 70 = 2x5x7
330 is divisible by 6 <--> 330 = 2x3x5x11
56 is divisible by 8 <--> 56 = 2x2x2x7

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
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