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MGMAT factors problem

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MGMAT factors problem

by HPengineer » Thu Aug 12, 2010 11:29 am
This is from the MGMAT flash card set... Im having trouble understanding. Any help would be most appreciated.

x is divisible by 144. If cube root of X is an integer then which of the following is cube root of x definitely divisible by? Chose all that apply.

A.) 4
B.) 8
C.) 9
D.) 12
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by kvcpk » Thu Aug 12, 2010 11:36 am
HPengineer wrote:This is from the MGMAT flash card set... Im having trouble understanding. Any help would be most appreciated.

x is divisible by 144. If cube root of X is an integer then which of the following is cube root of x definitely divisible by? Chose all that apply.

A.) 4
B.) 8
C.) 9
D.) 12
x is divisible by 144 means
x is divisible by 2^4 * 3^2
cube root of x is an integer.
Hence another factor of X should be atleast 2^2 * 3 So that X will be divisible by 2^6*3^3
This is to make the cube root of x as integer.

So X is definitely divisible by 2^2 * 3 = 4*3 = 12

Hope this helps!!
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don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
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by HPengineer » Thu Aug 12, 2010 11:52 am
ok so i think i got it... for cube root of x to be an integer we know that we have to add additional factors to X.

So we can say that X = 2^6 * 3 ^3 now taking the cube root of that our new x = 2^2 *3 i understand how 4 and 12 work now thanks...

I think what tripped me up was taking the cube root of exponents. We just simply divide the exponent by the root we are taking... is that correct?

for example sqrt of 2*4 would be 2^2..
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by kvcpk » Thu Aug 12, 2010 12:25 pm
HPengineer wrote:ok so i think i got it... for cube root of x to be an integer we know that we have to add additional factors to X.

So we can say that X = 2^6 * 3 ^3 now taking the cube root of that our new x = 2^2 *3 i understand how 4 and 12 work now thanks...

I think what tripped me up was taking the cube root of exponents. We just simply divide the exponent by the root we are taking... is that correct?

for example sqrt of 2*4 would be 2^2..
You are Right on Target!!
"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
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by clock60 » Thu Aug 12, 2010 12:28 pm
kvcpk wrote:
HPengineer wrote:This is from the MGMAT flash card set... Im having trouble understanding. Any help would be most appreciated.

x is divisible by 144. If cube root of X is an integer then which of the following is cube root of x definitely divisible by? Chose all that apply.

A.) 4
B.) 8
C.) 9
D.) 12
x is divisible by 144 means
x is divisible by 2^4 * 3^2
cube root of x is an integer.
Hence another factor of X should be atleast 2^2 * 3 So that X will be divisible by 2^6*3^3
This is to make the cube root of x as integer.

So X is definitely divisible by 2^2 * 3 = 4*3 = 12


Hope this helps!!
hi kvcpk
i like your solution , but slightly puzled , why the answer is not 4 , to me it is definitely divisible by 4 can you please explain a litle more
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by HPengineer » Thu Aug 12, 2010 1:37 pm
I think its just a typo or he means that since 12 is good 4 is also good...

THe correct answer is 4 and 12.... Thanks kvcpk u da man or women not sure :)
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by kvcpk » Thu Aug 12, 2010 3:12 pm
HPengineer wrote:I think its just a typo or he means that since 12 is good 4 is also good...

THe correct answer is 4 and 12.... Thanks kvcpk u da man or women not sure :)
Ofcourse Man :)

If the number is divisible by 12, it should be divisible by 4. However, because there were 2 correct options, I chose the larger one.
"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)
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