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

Question

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

Bunuel · Accepted Answer

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

n^4  being divisible by 32 implies  n^4=32k=2^5k , for some integer k.

Taking the fourth root of this equation gives:  n=2\sqrt {2k} .

Since n is an integer, the least value of n ( n_{min} ) is obtained when k = 8. Hence,  n_{min}=2\sqrt {2*8}=4 .

4 divided by 32 gives a remainder of 4.

Answer: B.

To elaborate more: as n is an integer and  n=2\sqrt {2k} , then  \sqrt {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 , and so on. Basically, n will be a multiple of 4. So  \frac{n}{32}  can give us the remainder of 4, 8, 12, 16, 20, 24, 28, 0 - all multiples of 4. The only multiple of 4 in the answer choices is 4.

Answer: B.

rao · Answer

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.

chix475ntu · Answer

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

gurpreetsingh · Answer

IMO B

Take  n^4 = 2^5 * K  where K can be of form  K = 2^{4m-5}

=>  n = 2^m

now m can be any integer greater than 1 
( m cannot be 1 as K is an integer thus, 4m-5 > 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.

gurpreetsingh · Answer

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^{4z-5} 
=> n is of the form  2^z

for z =1 , 4z-5 <0  thus not possible.
for  z = 2 , n = 4

when 4 is divided by 32 the remainder is 4.

mainhoon · Answer

Gurpreet,
Can you please explain why does 32A be of the form 2^(4z)? Thanks

gurpreetsingh · Answer

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}

TGC · Answer

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

Beat720 · Answer

Following is how I tackle this problem,

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

dave13 · Answer

Bunuel , can you please explain how from here  n^4=2^5k  you got ths  n=2\sqrt {2k}  ?

Bunuel · Answer

By taking the fourth root:

n^4=2^5k ;

n=\sqrt {2^5k}=\sqrt {2^4*2*k}=2\sqrt {2k} .

saubhikbhaumik · Answer

Concept tested: Remainder of product of 2 numbers = product of 2 remainders

That means (remainders of n/32)^4 will be divisible by 32 as n^4 is divisible by 32

Let put the options nad test the concept.

Let the remainder be (a) = 2
2x2x2x2 = 16 -> Not divisible by 32

Let the remainder be (b) = 4
4x4x4x4 = 64 x 4

64 being divisible by 32 64x4 is also divisible by 32.

Hence, B

Adnan123 · Answer

n^4/32 , r = 0
n^4 = 32k
32 = 2^5

minimum value of k where 4V32k is an integer is k = 8

n = 4V32(8) = 4
4/32, remainder = 4

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

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