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shubhangi
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ps 22 Mar 2004, 18:06
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what is the maximum value of m if 12!/2^m is an integer?

8
10
12
11



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ps 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
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lvb9th
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ps 23 Mar 2004, 17:04
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hallelujah1234
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m = [12/2]+[12/4]+[12/8] = 10

Hi,

Is this some standard formula?

Thanx.

Yeah, for primes.
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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
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ps 23 Mar 2004, 17:10
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lvb9th
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
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[a] + [b] is not always equal to [a+b]

[a] = integral part of a
[5.3] = 5
[-5.3] = -6
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ps 23 Mar 2004, 18:51
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hallelujah1234
lvb9th
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

[a] + [b] is not always equal to [a+b]

[a] = integral part of a
[5.3] = 5
[-5.3] = -6
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Ok, I GOT THAT PART. But why stop at [12/8] why not go to [12/10], using your solution what is 24!/2^n, or any factorial of your choice over 2^n. I just want to understand the formula correctly. Ok, how about this 16!/2^n, the max n is [16/2]+[16/4]+[16/8]+[16/16]=15? Oh, I think I got it, in the example above stop at 8 since 2^4>12. :shh



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