A number is called “terminating in base N” if that number can be expressed

[BTGModeratorVI]

A number is called “terminating in base N” if that number can be expressed as A/N^B for some integers A and B. What is the smallest positive integer N for which 5/24 is terminating in base N?

A. 1
B. 2
C. 6
D. 8
E. 24

Answer: C
Source: Veritas

[swerve]

A number is called “terminating in base N” if that number can be expressed as A/N^B for some integers A and B. What is the smallest positive integer N for which 5/24 is terminating in base N?

A. 1
B. 2
C. 6
D. 8
E. 24

Answer: C
Source: Veritas

Denominator \(24\) must be manipulated to \(N^B\) format such that \(N\) has a minimum value.
Factorizing \(24 = 2^3 \cdot 3^1\)

For \(24^1: N= 24, B= 1\)
For \(2^4\cdot 3^2= 144: N= 12, B=2\)
For \(2^3\cdot 3^3= 6^3: N=6, B=3\)
For \(2^4 \cdot 3^4= 6^4: N=6, B=4\)
and so on.

So, the maximum value of \(N\) is \(24\) and the smallest positive integer \(N\) is \(6\).

Therefore, the correct answer is C

[Brent@GMATPrepNow]

A number is called “terminating in base N” if that number can be expressed as A/N^B for some integers A and B. What is the smallest positive integer N for which 5/24 is terminating in base N?

A. 1
B. 2
C. 6
D. 8
E. 24

Answer: C
Source: Veritas

We must find a fraction that's EQUIVALENT to 5/24 so that the EQUIVALENT fraction can be written in the form A/(N^B)

24 = (2)(2)(2)(3)
So, the denominator of the EQUIVALENT fraction must have at least three 2's
Notice that if we take 24 = (2)(2)(2)(3) and multiply both sides by 9, we get: 24(9) = (2)(2)(2)(3)(9)
Rewrite as: 216 = (2)(2)(2)(3)(3)(3)
Rewrite as: 216 = [(2)(3)][(2)(3)][(2)(3)]
Rewrite as: 216 = [(2)(3)]³
Rewrite as: 216 = 6³

This means we can take: 5/24
And multiply top and bottom by 9 to get the EQUIVALENT fraction: 45/216
Now rewrite the denominator as follows: 45/6³
In other words, 5/24 = 45/216 = 45/6³
Since 45/6³ is written in the form A/(N^B), we can see that N = 6

Answer: C

Cheers,
Brent

[Scott@TargetTestPrep]

A number is called “terminating in base N” if that number can be expressed as A/N^B for some integers A and B. What is the smallest positive integer N for which 5/24 is terminating in base N?

A. 1
B. 2
C. 6
D. 8
E. 24

Answer: C
Source: Veritas

To solve this problem, we need to first find the smallest integer k such that 24k is a power of N.

Since 24 = 2^3 x 3, k must be 3^2 so that 24k = (2^3 x 3) x (3^2) = 2^3 x 3^3 = 6^3.

Thus we see that N = 6. (Note: 5/24 = (5 x 3^2)/(24 x 3^2) = 45/6^3 which is in A/N^B format.)

Answer: C