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[vaivish]

If k^4 is divisible by 32, which of the following could be the remainder when k is divided by 32?
I. 2
II. 4
III. 6
(A) I only
(B) II only
(C) III only
(D) I and II
(E) None


Oa is b.

[anju]

Answer is b
since there are values provided in the option, let's start backsolving
let's assume k = 2 so k^4 = 16 which is not completely divisible by 32
let's assume k = 4 so k^4 = 256 which when divided by 32 gives 8, so when 4 is divided by 32 the remainder is 4 - OPTION b is correct
let's take k = 6 so k^4 = 36*36 which is not completely divisible by 32

So option B

[lali_mew]

Anju, you are right about the answer.
The approach needs a correction though.

The options 2,4 and 6 are remainders.
Thus,
If the remainder = 2, k = 34. 34^4 is not divisible by 32
If the remainder = 4, k = 36. 36^4 IS divisible by 32
If the remainder = 6, k = 38. 38^4 is not divisible by 32

[dalwow]

Is there some quick way you found if 34^4, 36^4 and 38^4 is divisible by 32? Did you just work it out or is there a faster way? Thanks

[Ian Stewart]

Strictly speaking, the above approaches are backwards. The question does not say "If the remainder is 2 when k is divided by 32, is k^4 divisible by 32?", which is the question the above solutions answer. Fortunately it doesn't make a difference to the answer here, but it will on other questions.

If k^4 is divisible by 32, which of the following could be the remainder when k is divided by 32?
I. 2
II. 4
III. 6

If you know k^4 is divisible by 32, then k^4 is divisible by 2^5. What must k be divisible by? If k were divisible by 2 but *not* by 2^2, then k^4 would only be divisible by 2^4 = 16. So k must be divisible by at least 2^2 = 4. Thus, k is a multiple of 4, and since 32 is also a multiple of 4, when k is divided by 32, the remainder will be a multiple of 4. If that is unclear, algebraically, we know k = 4a for some a. When we divide k by 32, we can write, with q as the quotient, and r as the remainder:

k = 32q + r
4*a = 32q + r
r = 4*a - 32q
r = 4(a - 8q)

so r must be a multiple of 4. Only II is possible.

I'd note that if you were going to do the problem 'backwards', anju has chosen the best numbers here; in a remainder problem, if you are going to pick numbers, you will always get the smallest (and therefore easiest) set of numbers by letting the quotient be zero, (and not one). If the remainder is 2 when k is divided by 32, k could certainly be equal to 2.

[kshin78]

Is there some quick way you found if 34^4, 36^4 and 38^4 is divisible by 32? Did you just work it out or is there a faster way? Thanks

The options 2,4 and 6 are remainders.
Thus,
If the remainder = 2, k = 34. 34^4 is not divisible by 32
If the remainder = 4, k = 36. 36^4 IS divisible by 32
If the remainder = 6, k = 38. 38^4 is not divisible by 32



Take 34 and prime factorize = 2 X 17 = has one 2 and one 17. if you ^4, you'll end up w/ four 2s and four 17s. You'll need FIVE 2s to be divisible by 32.

Take the same approach and you'll see that 38 doesn't work but 36 does.