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Multiple of 1440

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by Brent@GMATPrepNow » Tue Dec 30, 2008 9:28 am
n is a positive integer, and k is the product of all integers from 1 to n. If k is a multiple of 1440, then what is the least possible value of n?
A) 8
B) 12
C) 16
D) 18
E) 24
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Re: Multiple of 1440

by logitech » Tue Dec 30, 2008 9:40 am
1440 = 2^5 3 ^2 5

So we will need at least one 5 and two 3s and five 2s

1 2 3 4 5 6 = 2^4 3 ^2 5

When n=6 we have our 5 and two 3's but we are missing one 2

n can not be 7, so the next 2 factor is in 8

I will go with A
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by mental » Tue Dec 30, 2008 9:46 am
Lets factorize 1440 = 10*12*12 = 2^5*3^2*5

2 * 3 * (2^2) * 5 * (2*3) * 2

1*2*3*4*5*6*2

after 6 we have 1 more 2 left, that comes from next even integer ie 8

min value of n = 8

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I remembered 6! = 720 , that is 1440/2
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by Brent@GMATPrepNow » Tue Dec 30, 2008 9:47 am
Nice work - the answer is A.
I'll post my solution as well, since I already wrote it:

Begin by finding the prime factorization of 1440, which is 1440 = 2×2×2×2×2×2×3×3×5,
So, our product k must contain at least five 2’s, two 3’s and one 5
Begin at 1 and keep adding additional numbers to the product k, until we meet this criterion.
 1×2×3×(2×2)×5×(2×3)×7×(2×2×2) (STOP here – we have our five 2’s, two 3’s and one 5)
We need only progress from 1 to 8 until we have all of the prime factors of 1440 in the product.
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