Problem Solving

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

First I determined which integers when raised to the 4th are divisible by 32.
32 = 2^5 so n^4 must have 2^5 as a factor

1^4 no
2^4 no
3^3 no
4^4 = 2^8 yes
5^4 no
6^4 no
7^4 no
8^4 = 2^12 yes

following the same pattern n is a multiple of 4

Trying out n=4: 4 divided by 32 is zero remainder 4
So 4 is an option. (Shortcut I used)

To check: Since n is a multiple of 4 try out all multiples of 4 until the pattern is revealed.

8 divided by 32 is zero remainder 8
12 divided by 32 is zero remainder 12
16 divided by 32 is zero remainder 16
20 divided by 32 is zero remainder 20
24 divided by 32 is zero remainder 24
28 divided by 32 is zero remainder 28
32 divided by 32 is 1 remainder 0
36 divided by 32 is 1 remainder 4
40 divided by 32 is 1 remainder 8

and the cycle continues.

The remainders (8,12,16,20,24,28,0,4,8 ) are possible options but not all of them are answer choices. The only value that is an answer choice is 4. So B is the answer.

Although the explanation may seem long, the computations are really short and the check can be done in your head.

32 = 2^7 so n^4 must have 2^7 as a factor

jbrown2, just to nitpick, 32 = 2^5. However, in this case, the answer does not change due to that!




n^4 is divisible by 32.
n^4 is divisble by 2*2^4.

n^4/(2*2^4) = (1/2)*(n/2)^4. This can be an integer only if n is even and also, something needs to factor out that extra 2. That extra 2 has to come from 'n'. So n has to be divisible by 4.

So, now, since n is divisble by 4, the only remainders when it's divided by 32 can be 4,8,12,16,20,24,28

Answer is B.

whoops! There goes one of those careless mistakes. Good thing the answer didnt change

... a different approach would be to use the binomial theorem ....

.. n is of the form n = 32t+r .. where r is the remainder .....

now, n^4 is divisible by 32 ...so that means (32t+r)^4 is divisible by 32 ... now if the given function is expanded then it will become clear that for the function to be divisible by 32 the remainder has to be a multiple of 4 ...

now, n^4 is divisible by 32 ...so that means (32t+r)^4 is divisible by 32 ... now if the given function is expanded then it will become clear that for the function to be divisible by 32 the remainder has to be a multiple of 4

You're correct. Just to clarify what you are saying.

(a+b)^4 = a^4 + a^3*b + a^2*b^2 + a*b^3 +b^4
so
(32t + r)^4 = (32t)^4 + (32t)^3*r + (32t)^2*r^2 + 32t*r^3 + r^4

On the right side of the equation, all of the terms except for the last term is a definite multiple of 32. For the last term to be a multiple of 32 r has to be a multiple of 4.