When k^4 is divided by 32, the remainder is 0. Which of the following

k is a positive integer. When \(k^4\) is divided by \(32\), the remainder is \(0\). Which of the following could be the remainder when \(k\) is divided by \(32\)?

­Since the remainder of \(k/32\) is \(0\), we know the following:

\(k^4 = x * 32\)

\(k^4 = x * 2^5\)

\(k^4 = x * 2 * 2^4\)

Since \(k\) is an integer, it must also be the case that the fourth root of \(x * 2\) is an integer. Thus, the following must also be true:

\(x * 2 = y^4 * 2^4\)

So,

\(k^4 = y^4 * 2^4 * 2^4\)

\(k = y * 2 * 2\)

\(k = y * 4\)

So, \(k\) is a multiple of \(4\). Thus, since \(32\) is a multiple of \(4\), any remainder of \(k/32\) must be a multiple of \(4\).­­­­

I. \(2\)

II. \(4\)

III. \(6\)

Only \(4\) is a multiple of \(4\).

A. I only

B. II only

C. III only

D. I and II only

E. II and III only

Correct answer: B

32 = 2^5

k^4=2^8 *n^4 (where n is a integer)
k=2^2*n = 4*n

when k will be divided by 32 the remainder will either be a multiple of 4 or zero (as 32 is also a multiple of 4)

So only ii = 4 is possible

Option B

wishmasterdj

\(k^4\) is divisible by 32, so \(k^4\) must contain 32 in it.
Distribute all prime factors of 32 in terms of power of 4 as k is to the power of 4.

\(k^4=32y=2^5y=2^4*2y\)
Now k should have one of each of base on the right side \(2^4*2*y\), so 2, 2 and y.
That is k=2*2*y, where y is any positive integer.

If k=4y, then when ever we divide by 32, k will give us a remainder as a multiple of 4.

For example
k=4*1, remainder =4
k=4*2, remainder =8=4*2
k=4*8, remainder =0=4*0
k=4*9, remainder =36=32+4=4

2 and 6 are not multiples of 4. Hence only II

B
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k^4 divided by 32 leaves a remainder of 0, can express as \(k^4 = 32n = 2^5n\) for some integer n.

Taking the fourth root of this equation, we get \(k = 2\sqrt[4]{2n}\).

Given that k is an integer and \(k = 2\sqrt[4]{2n}\), it follows that \(\sqrt[4]{2n}\) must also be an integer. Therefore, k can assume values like:

4 for \(n = 2^3\);
8 for \(n = 2^3*2^4\);
12 for \(n = 2^3*3^4\);
20 for \(n = 2^3*5^4\);
24 for \(n = 2^3*2^4*3^4\);
28 for \(n = 2^3*7^4\);
32 for \(n = 2^3*2^8\) and so on.

Essentially, k will always be a multiple of 4. Consequently, when dividing k by 32, the possible remainders are 4, 8, 12, 16, 20, 24, 28, or 0 – all multiples of 4.

Answer: B.
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Bunuel chetan2u, please tag this one under GMAT Prep Focus

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Added the tag. Thank you!
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Can someone explain this again?

Two of our top experts already provided their explanations. You should be more specific with what specifically you need help rather than declare “I don’t understand anything”

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Hi chetan2u,

I can't figure out how \(k^4 = 2^4*2*y\) leads to \(k = 2*2*y\). Could you explain this?

Hi MartyMurray,

Could you explain the jump in logic from K^4 = 2a*2^4 onwards?

It is understood that for K to be an integer, 2a must be something in the form of (xy)^4. 

Therefore, 2a = (xy)^4 . How do you proceed from here and determine that it is a multiple of 4?

­

why does it follow that ­\(\sqrt[4]{2n}\) must also be an integer? Couldn't be something like 1.5 or 2.5 that when multiplied by 2 becomes an integer?­

­

Crucial to know: if \(x\) is a positive integer, then \(\sqrt{x}\) is either a positive integer or an irrational number. (It cannot be a reduced fraction such as \(\frac{7}{3}\) or \(\frac{1}{2}\)).

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is there any video for this? thx!