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

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

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12 Sep 2005, 13:18
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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Senior Manager
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12 Sep 2005, 13:27
EDITED

will go with 4, take 4^4 is divisible by 32 and 4/32 gives reminder 4...

I went about in a wierd way that i cannot explain...but i took less than a minute....
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Last edited by ranga41 on 13 Sep 2005, 06:40, edited 1 time in total.

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12 Sep 2005, 14:44
I agree with ranga's answer, although I have to note that 28/32 gives a remainder of 28. I just started picking numbers and ended up with 4 as the first positive integer, whose 4th power is divisible by 32.

I tried to come up with a more mathematical explanation and this is the best I can offer:

n^4 is divisible by 32, so n^4 = 2^5 * x, where x is some not so random integer
n = 2 * ((2*x)^(1/4))

For n to be an integer, x must be equal to (2^3)*y, where y is some positive number, whose quad root is an integer. If we take the simplest case where y=1, n=2 * ((2^4)^(1/4)) = 4

So 4 is the smallest + integer, whose 4th power is divisible by 32.

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VP
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12 Sep 2005, 16:55
ans is 4. Here is how i got it.

n ^ 4 is diisible by 32 or 2 ^ 5.

now for this to happen each n, when factorised, should have 2^n where n > 1

So the modified question become:
find remainder when 2 ^ n (n > 1) is divided by 32 or
2 ^ n % 32, n > 1

clearly remainders can be 2^2, 2^3, 2^4 or 4, 8, 16

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12 Sep 2005, 18:34
B. 4 should be correct. because n^4 is divided by 32 if n is at least 4 or multiple of 4. 6, 10 doesnot work for n because n^4 is not divided by 32 evenly. so, if n=4, then n divided by 32 leavs 4 as reminder.

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Current Student
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12 Sep 2005, 20:34
you guys make this look soo eazy...how many of you did this in under 2 mins? I must say you guys are great...I think if you can do this in under 1 min...you are math is definetly solid...

OA is 4...

it took me 4 mins...to figure this out...

n^4=32*m where m is an integer...

so we can break 32 down to its prime factors...2^5....

n^4=2^5 * m...so we know that n is at the least has 2^2 as its prime factor...

think of it like, if n^4 was equal to 2^4 then we know that n=2m....right!

anyway back to the problem

so n at a minimum is 4...

so we now get

4x=32m+c (where m is an integer, c is an integer and is meant to be a reminader)

so in order for x to be an integer...

x=8m+c/4 (c has to be 4 or a multiple of it!)

4 it is...

not sure..if what i did makes sense but I get the same ans as you...which so happens to be the OA...will post OE as soon as I get it...

so lets do 4/32 what is the remainder

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12 Sep 2005, 20:34
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