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

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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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Re: 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, 05:38
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 which of the answer
options can be the remainder when n is divided by 32.

The first step is to prime-factorize 32 which is \(2^5\). n has to contain a
minimum of \(2^2\) in order for \(n^4\) to be divisible by \(32(2^5)\).
If n had only one 2, then \(n^4\) would contain \(2^4\) and 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  [#permalink]

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New post 26 Jul 2018, 05:39
1
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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Re: If n is an integer and n^4 is divisible by 32, which of the  [#permalink]

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New post 31 Oct 2018, 01:23
Simple observation works -
For n^4 to be divisible by 32 , n^4 must be in the form of 2^5*x. Where x is any other number. Out of all the choices only B) that has n = 4 suffices the condition of 2^5x. Rest all the answer choices will not yield 2^5x if their square are squared.
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Re: If n is an integer and n^4 is divisible by 32, which of the  [#permalink]

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New post 07 Dec 2018, 00:50
[quote=
Given: \(n^4=32k=2^5k\) --> \(n=2\sqrt[4]{2k}\) --> as \(n\) is an integer, the least value of \(n\) is obtained when \(k=8\) --> \(n_{min}=2\sqrt[4]{2*8}=4\) --> \(\frac{n_{min}}{32}=\frac{4}{32}\) gives remainder of \(4\).

quote]

how come 4/32 gives remainder as 4?
Shouldn't it be 8?
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Re: If n is an integer and n^4 is divisible by 32, which of the  [#permalink]

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New post 07 Dec 2018, 02:22
deddex wrote:
[quote=
Given: \(n^4=32k=2^5k\) --> \(n=2\sqrt[4]{2k}\) --> as \(n\) is an integer, the least value of \(n\) is obtained when \(k=8\) --> \(n_{min}=2\sqrt[4]{2*8}=4\) --> \(\frac{n_{min}}{32}=\frac{4}{32}\) gives remainder of \(4\).

Quote:

how come 4/32 gives remainder as 4?
Shouldn't it be 8?



Let me ask you a question: how many leftover apples would you have if you had 4 apples and wanted to distribute in 32 baskets evenly? Each basket would get 0 apples and 4 apples would be leftover (remainder).

When a divisor is more than dividend, then the remainder equals to the dividend, for example:
4 divided by 32 yields the reminder of 3: \(4=32*0+4\);
3 divided by 4 yields the reminder of 3: \(3=4*0+3\);
9 divided by 14 yields the reminder of 9: \(9=14*0+9\);
1 divided by 9 yields the reminder of 1: \(1=9*0+1\).

For more on this check:

5. Divisibility/Multiples/Factors




6. Remainders



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Re: If n is an integer and n^4 is divisible by 32, which of the   [#permalink] 07 Dec 2018, 02:22

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