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

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Difficulty:

45% (medium)

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