[12/2] + [12/4] + [12/8] = 6 + 3 + 1 = 10
Greatest value of n is 10
number properties
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pemdas
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12!/2^n could be restated as following
(1 *2 *3 *2^2 *5 *(2*3) *7 *2^3 *9 *(2*5) *11 *(3*2^2)/2^n OR 2^10 factored out
2^10(1*3*5...)/2^n which makes n=10
(1 *2 *3 *2^2 *5 *(2*3) *7 *2^3 *9 *(2*5) *11 *(3*2^2)/2^n OR 2^10 factored out
2^10(1*3*5...)/2^n which makes n=10
sud21 wrote:If the factorial of 12 is divisible by 2^n, what is the greatest value of n?
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ArunangsuSahu
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2 is a common factor for the even integers so 2 to 12 has 6
4 has an extra 2 , 8 has two extra 2s, 12 ha one extra 2
So 6+1+2+!=10
4 has an extra 2 , 8 has two extra 2s, 12 ha one extra 2
So 6+1+2+!=10
