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Bunuel
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A number is called “terminating in base N” if that number can be expre 01 Mar 2017, 03:49
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C
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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
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C
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A number is called “terminating in base N” if that number can be expre 01 Mar 2017, 04:19
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Bunuel
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
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A/N^B=5/24

N^B=(24 * A)/5

24=2^3*3

Now since R.H.S has 3 and 2 L.H.S must also have a 3 and 2. Plugging lowest such no from the options(6) we get;

6^B = (2^3 * 3) * A / 5

Now A can be 3^2 * 5 (Choosing in such a way to make RHS an integral power of LHS)

(3*2)^B = (2^3 * 3) * (3^2 * 5) / 5

(3*2)^B = 3^3 * 2^3 = (3*2)^3

Hence option C.

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A number is called “terminating in base N” if that number can be expre 01 May 2018, 10:00
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Bunuel
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
Show more

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
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A number is called “terminating in base N” if that number can be expre 23 Feb 2019, 11:43
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Bunuel
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
Show more

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

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A number is called “terminating in base N” if that number can be expre 27 Feb 2019, 05:57
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Denominator 24 must be manipulated to N^B format such that N has minimum value.
Factorizing 24 = 2^3 * 3^1

For 24^1: N= 24, B= 1
For 2^4*3^2= 144: N= 12, B=2
For 2^3*3^3= 6^3: N=6, B=3
For 2^4*3^4= 6^4: N=6, B=4
and so on.

So, Max N is 24 and Min N is 6.
Ans C
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A number is called “terminating in base N” if that number can be expre 28 Feb 2019, 01:58
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Shobhit7
Denominator 24 must be manipulated to N^B format such that N has minimum value.
Factorizing 24 = 2^3 * 3^1

For 24^1: N= 24, B= 1
For 2^4*3^2= 144: N= 12, B=2
For 2^3*3^3= 6^3: N=6, B=3
For 2^4*3^4= 6^4: N=6, B=4
and so on.

So, Max N is 24 and Min N is 6.
Ans C
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Careful with the maximum. The fraction can be scaled up as, for instance, \(\frac{10}{48}\) or \(\frac{50}{240}\), and so on, so in fact there is no maximum value of \(N\). Your reasoning for the minimum looks correct, though.
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A number is called “terminating in base N” if that number can be expre 13 Jan 2026, 17:09
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