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by crackgmat007 » Sun Oct 11, 2009 12:38 pm
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

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by NikolayZ » Mon Oct 12, 2009 12:25 am
Hello !

I think it is A.

Since K=n!, and k is a multiple of 1440. Then we need to find the smallest value of n! that is divisible by 1440.
6!=720
7!=720*7=5040 ( not divisible by 1440)
8!=5040*8=40320
8!/1440=28. Hence, 8.
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by bullzeye » Mon Oct 12, 2009 4:23 am
I agree with Nikolay Z.

Other solution, with just basic math is this:
you know that the number you're looking for is a multiple of 1440. So the smallest number you are actually looking for is 1440. So let's break down 1440 (which is really easy even tough it's a big number). We know 1440 is divisible by 10 and 144 is a 12 squared. Which leave us with 12x12x10 = 3x4x3x4x2x5. So let's think that through - 10 is a multiple of 5 so we know that n>=5. We need to find the integers less than 5 that are multiples of 1440. we Have 2, 3 and 4, so we are left with 3 and 4(3x4x3x4x2x5). So the smallest multiple of 3 (remember n>=5) is 6. 8 is the smallest for 4 (that leaves us with an integer).

So the answer is 8.
Excuse me if my idea is not presented clear enough.
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by mehravikas » Mon Oct 12, 2009 7:16 pm
Assuming that k = 1440, smallest value of 'n' can only be 8
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