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If n is an integer and n^4 is divisible by 32, which of the

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Director
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If n is an integer and n^4 is divisible by 32, which of the  [#permalink]

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New post 25 Jul 2018, 02:48
thank you Bunuel :) one question :-) why do you assume that least value is 8 ? :? whats your reasoning ? can you please explain :-)
as \(n\) is an integer, the least value of \(n\) is obtained when \(k=8\) -


And how did you figure out these values :? \(8\) for \(k=2^3*2^4\), \(12\) for \(k=2^3*3^4\)

for instance this \(k=2^3*2^4\), how should I understand it ? for example for n to be 4 k= 8 but you didn't break down this into exponents unlike other examples :?


hey pushpitkc, may be you can explain/answer the above questions :-) thanks!
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If n is an integer and n^4 is divisible by 32, which of the  [#permalink]

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New post 26 Jul 2018, 05:39
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dave13 wrote:
thank you Bunuel :) one question :-) why do you assume that least value is 8 ? :? whats your reasoning ? can you please explain :-)
as \(n\) is an integer, the least value of \(n\) is obtained when \(k=8\) -


And how did you figure out these values :? \(8\) for \(k=2^3*2^4\), \(12\) for \(k=2^3*3^4\)

for instance this \(k=2^3*2^4\), how should I understand it ? for example for n to be 4 k= 8 but you didn't break down this into exponents unlike other examples :?


hey pushpitkc, may be you can explain/answer the above questions :-) thanks!


Hey dave13

Let me try and explain this question once again.

If \(n^4\) is divisible by 32, we have been asked to find a remainder(from answer options) when n is divided by 32.

First, we have to prime-factorize 32 which is \(2^5\).

Now, n has to contain a minimum of \(2^2\) in order for \(n^4\) to be divisible by \(32(2^5)\). In order to validate our answer,
we can test a smaller value - If n had only one 2, then \(n^4\) would contain \(2^4\) and that would not be divisible by 32.

Hope this clears your confusion!
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If n is an integer and n^4 is divisible by 32, which of the &nbs [#permalink] 26 Jul 2018, 05:39

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