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Smallest value of N

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by sitsmegain » Mon May 17, 2010 10:33 am
If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3), what is the smallest possible positive value of n?

Also, please explain me the solution
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by indiantiger » Mon May 17, 2010 10:42 am
lets expand n x (2^5) x (6^2) x (7^3) and this can written as
= nx (2^5)x(7^3)x(2^2x3^2)
now as we can see that 5^2 is completely missing from the expression above
so n must atleast have 5^2
and we can also infer from the expression that in order for 3^3 to be factor of the expression
we would need one more 3 as 3^2 is already present
this means

n = 5*5*3 = 75
Is this the answer?
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by sitsmegain » Mon May 17, 2010 10:51 am
indiantiger wrote:lets expand n x (2^5) x (6^2) x (7^3) and this can written as
= nx (2^5)x(7^3)x(2^2x3^2)
now as we can see that 5^2 is completely missing from the expression above
so n must atleast have 5^2
and we can also infer from the expression that in order for 3^3 to be factor of the expression
we would need one more 3 as 3^2 is already present
this means

n = 5*5*3 = 75
Is this the answer?
Yes you are right, answer is 75
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