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by magizhan » Mon Apr 12, 2010 6:00 pm
Since m^2 is divisible by 48, possible values of m^2 are 1*48, 2*48, 3*48....

factorising 48 we get 3*2^4, Since m is an integer m^2 should have even powers of prime factors.

First possible value of m^2 is 3*48 = 3^2*2^4, which implies m=12.

Possible values of m are therefore 12, 24, 36..

As m could be any number in the above series, the maximum possible integer that divides m should satisfy any value of m in the above series.

Hence 12. Answer C.
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by this_time_i_will » Mon Apr 12, 2010 6:28 pm
yea solution is correct. How many minutes did u take to solve this. I took close to 3. :(
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by magizhan » Mon Apr 12, 2010 7:02 pm
Close to 2. Though I could arrive at the solution in less than a minute there was a confusion on why not 36. Needed a closer look at the question with more clarity to zero in on 12.

One more pointer is that 'largest number' question will mostly not have the largest number in the answer choices as the answer.
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by dxgamez » Mon Apr 12, 2010 7:46 pm
close to 1 min. from the qn, we should know that m can have multiple values. thus, we shud take the minimum multiple of 48.

option a and b are out since their squares doesnt even reach 48.

48 = 2^4 x 3

we can add a 3 to make 2^4 x 3^2 which is equal to 144. thus m would be 12.

m can then be 24, 36 as shown by magizhan..but the largest integer that divides all of them is 12.
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