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If n is an integer and n^4 is divisible by 32, which of the
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Updated on: 23 Oct 2013, 23:44
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If n is an integer and n^4 is divisible by 32, which of the following could be the remainder when n is divided by 32? (A) 2 (B) 4 (C) 5 (D) 6 (E) 10
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Originally posted by mustdoit on 03 Mar 2010, 00:35.
Last edited by Bunuel on 23 Oct 2013, 23:44, edited 2 times in total.
Renamed the topic, edited the question and added the OA.




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Re: n^4 is divisible by 32. Find remainder for n/32
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03 Mar 2010, 12:51
mustdoit wrote: If n is an integer and n4 is divisible by 32, which of the following could be the remainder when n is divided by 32?
(A) 2 (B) 4 (C) 5 (D) 6 (E) 10
Solving this would clarify lots of concepts !!! Solution coming.... 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 \(4\). Answer: B. To elaborate more: as \(n\) is an integer and \(n=2\sqrt[4]{2k}\), then \(\sqrt[4]{2k}\) must be an integer. So \(n\) can take the following values: \(4\) for \(k=8\), \(8\) for \(k=2^3*2^4\), \(12\) for \(k=2^3*3^4\)... Basically \(n\) will be multiple of \(4\). So \(\frac{n}{32}\) can give us the reminder of 4, 8, 12, 16, 20, 24, 28, 0  all multiples of 4. Only multiple of 4 in answer choices is 4.
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Re: n^4 is divisible by 32. Find remainder for n/32
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03 Mar 2010, 06:39
Given: n^4/32 = int n^4/2^5 = int 1/2(n/2)^4 = int
try n= int = 2,4,6,8... becuase n/2 must be an integer
n=2 wont work because 1/2(n/2)^4 needs to be an integer. n=4, result = 1/2 (2)^4 = 8, so the remainder of n/32 = remainder of 8/32 = 8 (not an option) n=8, result = 1/2(8/2)^4 = 1/2(4)^4 = 1/2x16x16= 16x8, so the remainder of n/32 = remainder of 16x8/32 = 4 (option B)
Once you understand that the question is asking 1/2(n/2)^4 must be an integer then you can quickly run the numbers and figure out the remainder of n/32.




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Re: n^4 is divisible by 32. Find remainder for n/32
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03 Mar 2010, 07:10
rao_1857 wrote: Given: n^4/32 = int n^4/2^5 = int 1/2(n/2)^4 = int
try n= int = 2,4,6,8... becuase n/2 must be an integer
n=2 wont work because 1/2(n/2)^4 needs to be an integer. n=4, result = 1/2 (2)^4 = 8, so the remainder of n/32 = remainder of 8/32 = 8 (not an option) n=8, result = 1/2(8/2)^4 = 1/2(4)^4 = 1/2x16x16= 16x8, so the remainder of n/32 = remainder of 16x8/32 = 4 (option B)
Once you understand that the question is asking 1/2(n/2)^4 must be an integer then you can quickly run the numbers and figure out the remainder of n/32. Maybe I'm missing something but why are you calculating the result, and then getting the remainder of the result divided by 32? Shouldn't you be getting the remainder of n divided by 32? For example, using your statement above: n=4, result = 1/2 (2)^4 = 8, so the remainder of n/32 = remainder of 8/32 = 8 (not an option) Result is indeed 8, but you need to divide 4 by 32 (0 remainder 4), not 8 by 32 If its done the way I described above, the answer could be A, D or E



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Re: n^4 is divisible by 32. Find remainder for n/32
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03 Mar 2010, 07:33
2^4=16 so not possible and 2^5 = 32
so v need (2^2)^2
4 satisfies this condition



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Re: n^4 is divisible by 32. Find remainder for n/32
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03 Mar 2010, 08:32
mustdoit wrote: If n is an integer and n4 is divisible by 32, which of the following could be the remainder when n is divided by 32?
(A) 2 (B) 4 (C) 5 (D) 6 (E) 10
Solving this would clarify lots of concepts !!! Solution coming.... 32 = 2^5 n^4/32 is an integer n^4 is even and at the minimum (n^4 and n) should be a multiple of 4. so the min value of n is 4 and remainder is 4 B



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Re: n^4 is divisible by 32. Find remainder for n/32
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03 Mar 2010, 23:15
Bunuel!! Thanks for the explanation.



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Re: n^4 is divisible by 32. Find remainder for n/32
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04 Mar 2010, 06:03
IMO B Take \(n^4 = 2^5 * K\) where K can be of form \(K = 2^{4m5}\) => \(n = 2^m\) now m can be any integer greater than 1 ( m cannot be 1 as K is an integer thus, 4m5 > 0 ) now m=2,3,4....so on gives n =4, 8 , 16........36, 40... so on now 4 can be the remainder when n=36. Since we are dividing by 32, remainder must be multiple of 4 as n is multiple of 4.
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Re: n^4 is divisible by 32. Find remainder for n/32
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27 Aug 2010, 09:50
I have made typo mistakes and that makes it bad solution. I dont rem where I was dreaming that day. Since n^4 is divisible by 32 => \(n^4 = 32*A\) but for n to be an integer \(32*A\) should be reduced to an integer when we take 4th root on both sides thus 32*A must be of the form \(2^{4z}\) => \(A = 2^{4z5}\) => n is of the form \(2^z\) for\(z =1 , 4z5 <0\) thus not possible. for \(z = 2 , n = 4\) when 4 is divided by 32 the remainder is 4.
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Re: n^4 is divisible by 32. Find remainder for n/32
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03 Sep 2010, 20:48
Gurpreet, Can you please explain why does 32A be of the form 2^(4z)? Thanks
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Re: n^4 is divisible by 32. Find remainder for n/32
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03 Sep 2010, 21:58
mainhoon wrote: Gurpreet, Can you please explain why does 32A be of the form 2^(4z)? Thanks 32*A = n^4 and n is integer. take 4th root on both the sides. Both LHS and RHS should have integral values. This is only possible when 32A = 2^{4z}
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Re: n^4 is divisible by 32. Find remainder for n/32
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04 Sep 2010, 23:00
Always use method explained by bunuel above in this type of the problems



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Re: If n is an integer and n^4 is divisible by 32, which of the
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19 Apr 2014, 21:18
Concept Tested :
Remainders
background:
a*b*c/d then
if Rem(a/d) =x, Rem(b/d)=y, Rem(c/d)=z then
x*y*z/d
Sol:
Try numbers using options.
(1). 2
means 32+2 =34
REM(34*34*34*34/32) must be 0
REM(2*2*2*2/32) is 16
(2). 4
means 32+4=36
REM(36*36*36*36/32) must be 0
REM(4*4*4*4/32) is 0
Hence the answer



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Re: If n is an integer and n^4 is divisible by 32, which of the
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08 Jul 2015, 11:54
mustdoit wrote: If n is an integer and n^4 is divisible by 32, which of the following could be the remainder when n is divided by 32?
(A) 2 (B) 4 (C) 5 (D) 6 (E) 10 Following is how I tackle this problem, please give comment!32 = 2^5. n^4 = n*n*n*n > n^4 is divisible by 32 only if n is divisible at least by 4 (because n^4 need one more factor of "2" to be divisible by 2^5) > The smallest value of n is 4 > remainder of n : 32 = 4 > B



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Re: If n is an integer and n^4 is divisible by 32, which of the
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18 Jul 2016, 02:52
option 1
Remainder 2 means 32+2=34 n^4= 34^4= 34*34*34*34/32 is not an integer
option 2
Remainder 4 means 32+4=36 n^4= 36*36*36*36/32 is an integer means n^4 is divisible by 32 if we take n=4
so B is correct



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If n is an integer and n^4 is divisible by 32, which of the
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15 Oct 2016, 06:34
Bunuel wrote: mustdoit wrote: If n is an integer and n4 is divisible by 32, which of the following could be the remainder when n is divided by 32?
(A) 2 (B) 4 (C) 5 (D) 6 (E) 10
Solving this would clarify lots of concepts !!! Solution coming.... 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 \(4\). Answer: B. To elaborate more: as \(n\) is an integer and \(n=2\sqrt[4]{2k}\), then \(\sqrt[4]{2k}\) must be an integer. So \(n\) can take the following values: \(4\) for \(k=8\), \(8\) for \(k=2^3*2^4\), \(12\) for \(k=2^3*3^4\)... Basically \(n\) will be multiple of \(4\). So \(\frac{n}{32}\) can give us the reminder of 4, 8, 12, 16, 20, 24, 28, 0  all multiples of 4. Only multiple of 4 in answer choices is 4. Hey BunuelThanks for your solution....could you please let me know if this is the right way of solving this? n^4 is divisible by 32, implies \(\frac{n^4}{2^5}\) clearly, if n=2, then the above condition does not stand true, as we will have one 2 remaining in the denominator, hence n has to be a multiple of at least 4 (because we need another 2 in the Nr)Chose a value of n which is greater than 32 and multiple of 4. \(\frac{36}{32}\), remainder = 4 OR could we simply say Remainder/Denominator = 4/32?
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Re: If n is an integer and n^4 is divisible by 32, which of the
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15 Oct 2016, 06:56
IMO the number n should be a multiple of 2 greater than 2 itself (2^2, 2^3, 2^4, etc.) ; not necessarily of multiple of 4.
So, we try the first option : 2^2 = 4
4/32 = 0 + remainder 4
So, B is the right answer (4 COULD be a remainder of 4/32).



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If n is an integer and n^4 is divisible by 32, which of the
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15 Jul 2017, 10:53
This is how I did it.
We're given that \(n^4 ≡ 0 (mod 32)\)
So right off the bat, we know \(n^4\) is congruent to 32, 64, 96...
We could solve this, but maybe the fastest way is just to plug in the answer choices.
Let's try "2."
\(2^4\) = \(4^2\) = 16, and \(16 ≢ 0 (mod 32)\) This isn't our answer.
Let's try "4."
\(2^4\) = \(16^2\) = 256, and \(256 ≡ 0 (mod 32)\)
(because 256 is evenly divisible by 32.)
So "n" could be 4.
Answer B.



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Re: If n is an integer and n^4 is divisible by 32, which of the
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25 Jul 2018, 01:24
Bunuel wrote: mustdoit wrote: If n is an integer and n4 is divisible by 32, which of the following could be the remainder when n is divided by 32?
(A) 2 (B) 4 (C) 5 (D) 6 (E) 10
Solving this would clarify lots of concepts !!! Solution coming.... 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 \(4\). Answer: B. To elaborate more: as \(n\) is an integer and \(n=2\sqrt[4]{2k}\), then \(\sqrt[4]{2k}\) must be an integer. So \(n\) can take the following values: \(4\) for \(k=8\), \(8\) for \(k=2^3*2^4\), \(12\) for \(k=2^3*3^4\)... Basically \(n\) will be multiple of \(4\). So \(\frac{n}{32}\) can give us the reminder of 4, 8, 12, 16, 20, 24, 28, 0  all multiples of 4. Only multiple of 4 in answer choices is 4. Bunuel, can you please explain how from here \(n^4=2^5k\) you got ths \(n=2\sqrt[4]{2k}\) ?



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Re: If n is an integer and n^4 is divisible by 32, which of the
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25 Jul 2018, 01:28
dave13 wrote: Bunuel wrote: mustdoit wrote: If n is an integer and n4 is divisible by 32, which of the following could be the remainder when n is divided by 32?
(A) 2 (B) 4 (C) 5 (D) 6 (E) 10
Solving this would clarify lots of concepts !!! Solution coming.... 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 \(4\). Answer: B. To elaborate more: as \(n\) is an integer and \(n=2\sqrt[4]{2k}\), then \(\sqrt[4]{2k}\) must be an integer. So \(n\) can take the following values: \(4\) for \(k=8\), \(8\) for \(k=2^3*2^4\), \(12\) for \(k=2^3*3^4\)... Basically \(n\) will be multiple of \(4\). So \(\frac{n}{32}\) can give us the reminder of 4, 8, 12, 16, 20, 24, 28, 0  all multiples of 4. Only multiple of 4 in answer choices is 4. Bunuel, can you please explain how from here \(n^4=2^5k\) you got ths \(n=2\sqrt[4]{2k}\) ? By taking the fourth root: \(n^4=2^5k\); \(n=\sqrt[4]{2^5k}=\sqrt[4]{2^4*2*k}=2\sqrt[4]{2k}\).
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Re: If n is an integer and n^4 is divisible by 32, which of the &nbs
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