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

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Joined: 09 Mar 2016
Posts: 1225
If n is an integer and n^4 is divisible by 32, which of the  [#permalink]

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25 Jul 2018, 01:48
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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26 Jul 2018, 04:38
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.

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Joined: 26 Feb 2016
Posts: 3325
Location: India
GPA: 3.12
If n is an integer and n^4 is divisible by 32, which of the  [#permalink]

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26 Jul 2018, 04:39
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.

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Joined: 28 Aug 2018
Posts: 14
Re: If n is an integer and n^4 is divisible by 32, which of the  [#permalink]

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31 Oct 2018, 00:23
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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Joined: 12 Jul 2018
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Schools: ISB '20, NUS '21
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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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06 Dec 2018, 23:50
[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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07 Dec 2018, 01:22
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

For other subjects:
ALL YOU NEED FOR QUANT ! ! !
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Re: If n is an integer and n^4 is divisible by 32, which of the &nbs [#permalink] 07 Dec 2018, 01:22

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