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ktzsikka
As per the question statement - then for how many values of N is \(\frac{ N}{3}\)
How did you deduce it is asking - how many N is \(\frac{N}{3}\) a perfect cube ?

some portion of the question got deleted/skipped. Now corrected. Thanks!
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ktzsikka
Here is the OE

Solution:

Step 1: Understand Question Statement
We know that 20 < N < 300 and N is integer.
We need to find for how many Ns is \(\frac{N}{3} \) a perfect cube.

Step 2: Define Methodology
As \(\frac{N}{3} \) is an integer and a perfect cube, we can infer that N is a multiple of 3.

We can assume \( \frac{N}{3} = K\)
Then we need to find all possible K such that K is a perfect cube.

Step 3: Calculate the final answer
\(20 < N < 300\)

Dividing by 3 all sides, we get

6.67<\(\frac{N}{3}\)<100

In other words, \(6.67 < K < 100\)

Now, K is a perfect cube, and the perfect cubes between 6.67 and 100 are 8, 27, and 64.

Thus, 3 values are possible, and the correct answer is Option C.
This is wrong. We are not suppose to take 8, as N should be in between 20 to 300
Bunuel can you please check this one !!

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Vibhatu
ktzsikka
Here is the OE

Solution:

Step 1: Understand Question Statement
We know that 20 < N < 300 and N is integer.
We need to find for how many Ns is \(\frac{N}{3} \) a perfect cube.

Step 2: Define Methodology
As \(\frac{N}{3} \) is an integer and a perfect cube, we can infer that N is a multiple of 3.

We can assume \( \frac{N}{3} = K\)
Then we need to find all possible K such that K is a perfect cube.

Step 3: Calculate the final answer
\(20 < N < 300\)

Dividing by 3 all sides, we get

6.67<\(\frac{N}{3}\)<100

In other words, \(6.67 < K < 100\)

Now, K is a perfect cube, and the perfect cubes between 6.67 and 100 are 8, 27, and 64.

Thus, 3 values are possible, and the correct answer is Option C.
This is wrong. We are not suppose to take 8, as N should be in between 20 to 300
Bunuel can you please check this one !!

Posted from my mobile device

N is not taken to be 8, N/3 is. N/3 can be 8, 27, and 64, hence N can be 24, 81, and 192.
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