Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

n is a positive integer, and k is the product of all integer [#permalink]
28 Sep 2009, 18:49

00:00

A

B

C

D

E

Difficulty:

5% (low)

Question Stats:

80% (01:43) correct
20% (01:36) wrong based on 10 sessions

n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is

Re: smallest possible value [#permalink]
28 Sep 2009, 19:52

tejal777 wrote:

n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is

Using mathematical brute force I got to 8 although I'm sure there is a much more elegant solution. I figured 1440 is not too big a number for that type of formula so trying a few numbers

If n = 6: k = 1 x 2 x 3 x 4 x 5 x 6 = 720

K is a multiple of 1440 meaning k/1440 must not have remainder.

Using the above if n = 7 then k wont be divisible by 1440 evenly, since it's odd. (since 720 is 1440/2 and hence 7x1440/2 = 7x1440/2) But if n = 8 then it will be divisible by 1440 since it's even.

That took me under 2 mins. It's probably not the best solution but made sense to me

Re: smallest possible value [#permalink]
28 Sep 2009, 21:06

Mikko wrote:

1440=2^5*3^2*5

k=n!=> n! must has at least 5 factors of 2 (2, 4=2*2, 6=2*3...) => n=8 to have at least five 2s n=8 => n! is divided by 5, 3^2 remains 0

n=8

Sorry I'm little confused. In this thread: what-is-the-least-value-for-n-82903.html We have the same question but for 990. Factoring 990 we get = 2*3^2*5*11 So the minimum value for n in that question is 11 and the explanation given was its the largest prime factor.

However, in this question 5 is the largest prime factor. But it seems as thought we're not looking for that but instead, the largest prime factor with a power. i.e. 2^5.

Re: smallest possible value [#permalink]
28 Sep 2009, 22:50

yangsta8 wrote:

Mikko wrote:

1440=2^5*3^2*5

k=n!=> n! must has at least 5 factors of 2 (2, 4=2*2, 6=2*3...) => n=8 to have at least five 2s n=8 => n! is divided by 5, 3^2 remains 0

n=8

Sorry I'm little confused. In this thread: what-is-the-least-value-for-n-82903.html We have the same question but for 990. Factoring 990 we get = 2*3^2*5*11 So the minimum value for n in that question is 11 and the explanation given was its the largest prime factor.

However, in this question 5 is the largest prime factor. But it seems as thought we're not looking for that but instead, the largest prime factor with a power. i.e. 2^5. (I don't think that there is a correlation between Prime 5-the factor and the 5 in 2^5)

Is that correct?

It's coincident that you got 11 is the largest prime and n! has enough factors of 2, 3 or 5 in 2*3^2*5*11 but in your topic, 5 is the largest Prime factor but not enough for the power 2^5 if stop at n=5 (only 2=2^1, 4=2^2=> max is 2^3).

U can approach anyway as long as n! is divisible by all the power-factors of the given number

n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is

A. 8 B. 12 C. 16 D. 18 E. 24

To me Ans is 2.

LCM of 1440=2^5*3^2*5 K=n! so, Ans. 2 Am I right? Please help. _________________

n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is

To me Ans is 2.

LCM of 1440=2^5*3^2*5 K=n! so, Ans. 2 Am I right? Please help.

k is the product of all integers from 1 to n inclusive --> k=n!;

k is a multiple of 1440 --> n!=1440*p=2^5*3^2*5*p, for some integer p --> n! must have at least 2 in power 5, 3 in power 2 and 5 as its factors. Now, 7!=2*3*(2^2)*5*(2*3)*7=2^4*3^2*5*7 not enough power of 2 --> next #: 8!=2*3*(2^2)*5*(2*3)*7*(2^3)=2^7*3^2*5*7=1440*(2^2*7) --> so lowest value of n is 8.

n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is

To me Ans is 2.

LCM of 1440=2^5*3^2*5 K=n! so, Ans. 2 Am I right? Please help.

1440=2^5*3^2*5 => n should have at least one 5, two 3's and 5 2's.

if you take n=6 => k = 6! = 1*2*3*4*5*6 = 2^4 * 3^2 * 5 => does not satisfy we need one more 2.

so the next even number 8 is the answer. _________________

n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is

A. 8 B. 12 C. 16 D. 18 E. 24

To me Ans is 2.

LCM of 1440=2^5*3^2*5 K=n! so, Ans. 2 Am I right? Please help.

k=n! And 1440 divides k 1440 = 2^5 * 3^2 * 5

Counting the powers of 2 in n! Upto 2 (1) Upto 4 (3) Upto 6 (4) Upto 8 (7) So n>=8

Counting powers of 3 in n! Upto 3 (1) Upto 6 (2) So n>=6