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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
55% (01:45) correct
45%
(01:42)
wrong
based on 225
sessions
History
Date
Time
Result
Not Attempted Yet
If N is an integer and lies between 20 < N < 300, then for how many values of N is\( \frac{N}{3} \) a perfect cube ?
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
10 Dec 2021, 22:34
Kudos
Bookmarks
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.
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.
11 Dec 2021, 01:44
Kudos
Bookmarks
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 ?
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 ?
