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22 Mar 2004, 18:06
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what is the maximum value of m if 12!/2^m is an integer?
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23 Mar 2004, 15:17
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Just my 2 cents. Let's find how many 2^n we can find in each term of expression 12!
12: 2^2
10: 2^1
8: 2^3
6: 2^1
4: 2^2
2: 2^1
add up the exponents: 2+1+3+1+2+1 = 10.
In order for the expression to be an integer, then numerator has to include denominator terms. Therefore, max. "m" can be is 10
12: 2^2
10: 2^1
8: 2^3
6: 2^1
4: 2^2
2: 2^1
add up the exponents: 2+1+3+1+2+1 = 10.
In order for the expression to be an integer, then numerator has to include denominator terms. Therefore, max. "m" can be is 10
23 Mar 2004, 17:04
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hallelujah1234
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hallelujah1234
m = [12/2]+[12/4]+[12/8] = 10
Hi,
Is this some standard formula?
Thanx.
Yeah, for primes.
Hey Halle can you express the formula in variables. In other words, 12/2+12/4+12/8 is not ten, so one must continue to the end of the sequence? Pauls solution is nice, however, GMAT might ask the same question only say24!/2^n and there simply isnt enought time to write them out. Thanks
