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If n is a positive integer, and the product of all integers

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by lucas211 » Wed May 25, 2016 9:51 am

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If n is a positive integer, and the product of all 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

Thanks in advance :-)
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by Brent@GMATPrepNow » Wed May 25, 2016 10:41 am
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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by [email protected] » Wed May 25, 2016 6:14 pm
Hi lucas211,

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.

Final Answer: B

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by Matt@VeritasPrep » Thu May 26, 2016 2:44 pm
If we've got n! = 9 * 10 * 11, we need n! to contain 9, 10, and 11 somewhere in its prime factorization.

Since 11 can only appear in multiples of 11, we need a multiple of 11. The smallest positive one is 11 itself, so n ≥ 11, and the least possible n is 11 itself.
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by shwetagoyal121 » Sun May 29, 2016 11:27 pm
Brent@GMATPrepNow 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
The answer would be 11.

Look at this question from the prospective options angle as well.

The first probable answer is 10 ; (because that is the smallest value)

Product of all the integers till n= 990 x (Some other number say Y)

Case I:

Now if n=10, whether the product is a multiple of 990 will depend on whether Y has 11 as a factor or not.

case II:

now if n=11 (next smallest value)

the product has all the factors to make it a multiple of 990 irrespective of the value of Y.

Hence 11 is the right answer.
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Prime Factorization

by SampathKp » Mon Dec 23, 2019 1:45 pm
lucas211 wrote:If n is a positive integer, and the product of all 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

Thanks in advance :-)
Prime factorization of 990 we get

990 = 2x3^2x5x11.

The implication of the above is that even though we have 10 (2x5) as factor of 990, 11 must be there for the product of prime factors to count to 990.
Without 11, the product of positive integers wont count to 990.

Every multiple of 990 will have prime factors of 990 and 'may' also have an additional prime factor . Take for e.g 12,870. (990X13) it will have 13 as an additional prime factor but will contain 2, 3^2, 5 , 11 also as factors.

Hence least possible value on n is 11 and not 10 even though 10 is factor of 990 and will be factor all multiples of 990.

Answer is B.
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by Scott@TargetTestPrep » Sat Dec 28, 2019 8:12 pm
lucas211 wrote:If n is a positive integer, and the product of all 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

Thanks in advance :-)
We are given that n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990. Thus:

n! / 990 = integer

To determine the minimum value of n, we need to break 990 into its prime factors.

990 = 10 x 99 = 5 x 2 x 3^2 x 11

Thus:

n! / (5 x 2 x 3^2 x 11) = integer

Since n! must be divisible by 2, 5, 9, and 11, the minimum value of n must be 11. Recall that 11! is divisible by 11 and by any positive integer less than 11. In other words, 11! / 990 = integer.

Answer: B

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