tmmyc wrote:The question states the product of 1 to n inclusive is a multiple of 990. This means this product will be 990 multiplied by "other stuff".
First, find the prime factors of 990: 2 3 3 5 11
From here, we know that n must be at least 11.
Let's assume n is 11.
Product of all integers from 1 to n where n = 11: 1*2*3*4*5*6*7*8*9*10*11
Here we see that 990's prime factors are all covered in the product above: 2, 3, 5, and 11 are present. The extra 3 can be taken from the 6. The rest of the product is the "other stuff".
Hence 11 is the least possible value of n.
GMAT PREP QUEST?
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Here was my explanation from the Manhattan GMAT forums.
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the product of 1 to n inclusive is basically n!. It has to be a multiple of 990, in other words, divisible by 990.
990 = 2*3*3*5*11
so:
n! / 2*3*3*5*11 must be an integer, meaning that 2, 3, 3, 5, 11 must be present in the numerator as well.
10! is not divisible by 2*3*3*5*11
but 11!, 12!, 13!, 14! are divisible.
Since we need the least one we choose 11
990 = 2*3*3*5*11
so:
n! / 2*3*3*5*11 must be an integer, meaning that 2, 3, 3, 5, 11 must be present in the numerator as well.
10! is not divisible by 2*3*3*5*11
but 11!, 12!, 13!, 14! are divisible.
Since we need the least one we choose 11
