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BongBideshini
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When k^4 is divided by 32, the remainder is 0. Which of the following 09 May 2026, 00:33
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Given:
k^4 is divisible by 32 = 2^5.
Suppose k has only one factor of 2:
k = 2 × n (n ≥ 1, as k is positive)
Then:
k^4 = 2^4 × n^4
This contains only 2^4, not enough for divisibility by 2^5.
So k must contain at least 2^2.
Hence:
k = 4n

Take different values of k:
k = 4, 8, 12, 16, 20, 24, 28, 32, ...
Possible remainders when divided by 32:
4, 8, 12, 16, 20, 24, 28, 0, ...
Only multiples of 4 are possible.

Answer: B. II only

Bunuel
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?

I. 2
II. 4
III. 6

A. I only
B. II only
C. III only
D. I and II only
E. II and III only
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When k^4 is divided by 32, the remainder is 0. Which of the following 09 May 2026, 02:24
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Faster ways to crack this one have already been discussed , but it’s instructive to understand the following :

if the remainder when k is divided by 32 were 2, k=32x + 2=2(16x +1) for some integer x

k^4 would be equal to 2^4(16x +1)^4
As 16x + 1 is an odd number , k^4 would not be a multiple of 32, contrary to what we are told . Thus the remainder cannot be 2
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When k^4 is divided by 32, the remainder is 0. Which of the following 11 Jul 2026, 07:18
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Having a hard time understanding this? went through all the replies still dont get it?
any advice tips? any module i can go through?

i get that k^4 will have 2^5 int (or could be any multiple of 2^5) so 2^5a

now k = 2*(2a)^(1/4)

for k to be int, 2a^1/4 has to be integer

got that too

2a should be a even perfect 4th - 8,256,1296

these can be broken down into power of 2s and 3s

how are we reaching 4 as remainder from this? to me it seems all possibilities 2,4, 2*3 are possible as remainder
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When k^4 is divided by 32, the remainder is 0. Which of the following 11 Jul 2026, 08:20
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NetOrb
Having a hard time understanding this? went through all the replies still dont get it?
any advice tips? any module i can go through?

i get that k^4 will have 2^5 int (or could be any multiple of 2^5) so 2^5a

now k = 2*(2a)^(1/4)

for k to be int, 2a^1/4 has to be integer

got that too

2a should be a even perfect 4th - 8,256,1296

these can be broken down into power of 2s and 3s

how are we reaching 4 as remainder from this? to me it seems all possibilities 2,4, 2*3 are possible as remainder
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The key point is that k must have at least two factors of 2.

If k were only even but not a multiple of 4, such as k = 2 or k = 6, then k^4 would contain only 2^4, which is not enough to be divisible by 32 = 2^5.
So k must be a multiple of 4. Therefore, when k is divided by 32, the possible remainders are multiples of 4 only: 0, 4, 8, 12, etc.

Among 2, 4, and 6, only 4 is possible.
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When k^4 is divided by 32, the remainder is 0. Which of the following 24 Jul 2026, 16:26
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This one gave me a lot of trouble but I think I got it now.
k^4 / 2^5 is an integer because the remainder is 0. This means that k^4 must have enough 2's in it to divide evenly
This means k must have at least (2) 2's in it.
If we take that to mean k is a multiple of 2^2
If we want to see if when k is divided by 32, we get the remainders shown, all we have to do is try to construct k so that it produces that remainder and still has (2) 2's in it.

Test remainder 2
k = 32m + 2 = 2(16m+1)
Note: this only has a single 2 in it! So this is wrong.

Test remainder 4
k = 32m + 4 = 4(8m+1)
Note: 4=2^2 so it works! This k has (2) 2's in it.

Test remainder 6
k = 32m + 6 = 2(16m+3)
Note: this only has a single 2 in it! So this is wrong.

4 is the only possible remainder in the options.
Bunuel
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?

I. 2
II. 4
III. 6

A. I only
B. II only
C. III only
D. I and II only
E. II and III only
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When k^4 is divided by 32, the remainder is 0. Which of the following 31 Jul 2026, 18:28
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32 = 2^5
For k^4 to be divisible by 2^5, it must have an equal or greater number of 2's in it's prime factorization.
The minimum number of 2's then must be 2.
2^2 = 4, so we know k is a multiple of 4.
k can be any multiple of 4, including 32, so we imediatley know II works.

For options I and III, we can assess them by thinking about how we get a remainder of 2 and 6 from k, or if it is even possible.

I think it is easies to just list a few multiples and notice the pattern.
k= 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, ....
Notice, that if you divide any of these by 32, you get a repeating pattern of
0, 4, 8, 12, 16, 20, 24, 28
You can never get remainders 2 or 6 because the multiples go in steps of 4.
This is a roundabout, probably inefficient way of thinking about it, but it's a way to do it.
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When k^4 is divided by 32, the remainder is 0. Which of the following 02 Aug 2026, 04:33
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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?

i found the following way helpful:
k^4 / 32 = k*k*k*k/2^5 yields remainder 0
this can happen if k itself is divisible by 2^5 or when the product of remainders in k*k*k*k is divisible by 2^5 when each of k is divided by 2^5

i) if 2 is remainder when k is divided by 2^5 then (2*2*2*2)/2^5 = 2^4/2^5, rem=16. here the product of remainders in k*k*k*k is not divisible by 2^5 so this option is incorrect
ii) if 4 is remainder when k is divided by 2^5 then (4*4*4*4)/2^5 = 2^8/2^5, rem=0. here the product of remainders in k*k*k*k is divisible by 2^5 so this option is correct
iii) if 6 is remainder when k is divided by 2^5 then (6*6*6*6)/2^5 = (2^4 * 3^4)/2^5, rem=16. here the product of remainders in k*k*k*k is not divisible by 2^5 so this option is incorrect
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When k^4 is divided by 32, the remainder is 0. Which of the following 11 Aug 2026, 06:59
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THIS QUESTION IS CONCEPTUALLY VERY IMPORTANT

Let's break it down Easy and conceptually consistent:

1. K^4 is divisible by 32 which is 2^5. since five 2s are present implies K^4 has to have minimum three more 2s
2. Implication: K^4 is minimum 2^8
3. Implies: K is minimum 2^2 i.e. 4
4. Thus values of K can be multiples of 4: 4,8,12,16,...,32,36....
5. When divided by 32, they all will give remainders that are multiples of 4

Thus only one of the options possible: Option C
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When k^4 is divided by 32, the remainder is 0. Which of the following 26 Aug 2026, 10:27
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Since k^4 is divisible by 32, and r is the remainder for k/32, therefore r^4 has to be divisible by 32.
- 2^4 = 16 and 16/32 leaves remainder 16
- 4^4 = 256 which is divisible by 32 (You can directly cancel out as well for ease of calculation)
- 6^4 = 36*36; 36*36/32 will leave a 2 in the denominator.
Hence (ii) only.
Bunuel
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?

I. 2
II. 4
III. 6

A. I only
B. II only
C. III only
D. I and II only
E. II and III only
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