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

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

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?­

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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 \(\frac{1}{2}\)).

Given: k^4 is integer and
Implies: k needs to be an integer too.

Given: k^4 / 32 or k^4 / 2^5 leaves no remainder
Ok, so

If k = 1, then (1^4 / 2^5) has remainder
If k = 2, then (2^4 / 2^5) has remainder
If k = 3, then (3^4 / 2^5) has remainder
If k = 4 (or 2^2), then (2^2)^4 / 2^5) has no remainder.

AHA! then k = 4a (where a is any number, since (multiples of 4)^4 will also get divided fully by 32, leaving no remainder)

Now, what the question stem asks is, what will be the remainder of k/32 which we re-write as 4a/32, i.e remainder will be in multiple of 4's

(Remember For example, remainder of 3/8 is 3, then similarly remainder of 4a/32 is 4a) 

Thus the only option is II.­

 
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We can solve this quickly if we understand the concept of division very well. 

k is a positive integer. Hence exponent of every prime factor of k^4 must be a multiple of 4. Why? Discussed here:
https://anaprep.com/number-properties-f ... ct-square/

k^4 is divisible by 2^5 which means that 2^8 is a factor. Hence k must have 4 as a factor i.e. k = 4a

Using the concept of division, if we have k balls, we will be able to make groups of 4 each. If 8 of those groups are joined together to make groups of 32, we will still have groups of 4 remaining. Hence whatever the remainder, it will always be divisible by 4.
Hence 2 and 6 are not possible. 

Answer B

Concept of Division: 
https://youtu.be/A5abKfUBFSc

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­32 = 2^5

i.e. k^4 has 2^5
which is possible oy if k has atleast 2^2

CONCEPT: The remainder is always the multiple of GCD of the numerator and Denominator

therefore, K when divided by 32 should leave the remainder 4 (if the remainder is not zero)

Answer: Option B

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­Watch this solution to see how breaking down an apparently simple question into clear, logical steps can significantly speed up your problem-solving process.



This approach not only ensures accuracy but also helps you avoid common pitfalls that might waste precious exam time.
By following this methodical strategy, you'll learn how to tackle similar questions with confidence and efficiency.­

Rule: k^x divided by y gives you the same reminder of r^x divided by y where r is the remainder you obtain dividing k^1 by y.

In this case x=4, y=32 and r are the answer choices.
So, if k^4 divided 32 gives you a remainder of 0, also (answer choices)^4 should give you a remainder of 0.
Check all three and see that only II. is divisible by 32, thus gives you a remainder of 0 when divided by 32.

What I don't understand is, if we are left with the expression 4x/2^5=2^2x/2^5 and we assume x to be 1 2^2/2^5 why can't we further reduce our numerator and denominator to 2/2^4, which would make a remainder of 2 also possible?­

PS: never mind, remainders are only applicable to the number by which we devide, by reducing the numrerator and denominator by 2, the remainder will also be reduced by 2, leading to a remainder of 2. With my question, I changed the initial question, hence I was not divding by 32 any longer but by 16, hence a remainder of 2 was possible. I leave it here, maybe someone has an equal misconception regarding remainders as I did.

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You cannot reduce like that. For example, 8/6 yields the remainder of 2. However, if you reduce by 2 and get 4/3, youd get the remainder of 1, which is not correct

Thank you, I just realised it myself ­

PPS: My inference about the reduction of the remainder is only partially true. It is only possible if AND ONLY if both numerator and denominator can be reduced by the same number, then the remainder will also be reduced by the same number. In our case 4/32=2/16 leads to a remainder change from 4 to 2. However, 5/32=R:5 and if we reduce 5/2, we will not get an integer, hence this rule does not apply here, as remainders are used instead of decimals but never together. Also, if numerator is larger than denominator, this also leads to equal scenarios. 36/32=R:4 ->18/16=R:2. However 35/32=R:35 but (35/2)/(32/2) not possible as we would have a decimal and not an integer any longer and hence we would nee to further reduce the number (where possible) or bring it back to the initial equation.

You're asking an excellent question. Indeed, there is a way to use properties of the binomial theorem to simplify our approach, especially when dealing with higher powers. Let's explore this:

1. Binomial Theorem for (a + b)^n:
When we have (a + b)^n, the last term is always b^n.

2. In our case, we're looking at (32m + r)^4, where r is the remainder.
The last term here will be r^4.

3. Key insight: When dealing with remainders, we only care about the last term.
All other terms will be multiples of 32 (our divisor) and won't affect the remainder.

4. So, we only need to consider r^4 mod 32.

Now, let's apply this to our options:

I. 2: 2^4 = 16 (not divisible by 32)
II. 4: 4^4 = 256 = 32 * 8 (divisible by 32)
III. 6: 6^4 = 1296 = 32 * 40 + 16 (not divisible by 32)

This method allows us to focus only on the remainder raised to the 4th power, rather than expanding the entire (32m + r)^4.

For higher powers, this becomes even more useful:

- For k^6 divided by 32, we'd only need to check r^6 mod 32.
- For k^8 divided by 32, we'd check r^8 mod 32.

This approach leverages the binomial theorem without requiring us to expand the entire expression, making it more manageable in exam conditions.

To develop this intuition:
1. Practice calculating small numbers raised to various powers.
2. Focus on the patterns of the last digits when raising numbers to powers.
3. Remember that for questions about divisibility or remainders, often only the last term of the binomial expansion matters.

From Claude

Try to make k^4 equal to 32, and you realize it needs to actually equal 2^8, then solve for k, and then you're basically done:

Given: k^4/32 = nq + 0, k/32 = xq + r, r=?

K^4/32 = n
K^4 = 32n = 2^5 * n
K = 4th root of (2^5 * n) = 2 * 4th root(2n) = 2*2, where n=8
>> k = 4

4 = 32(0) +r
r = 4