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17 Nov 2021, 09:45
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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:
85%
(hard)
Question Stats:
50% (02:04) correct
50%
(01:59)
wrong
based on 121
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If k is a positive integer less than 215, and \(\frac{26k}{84}\) is an integer, how many different positive prime factors k can have?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Archit3110
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17 Nov 2021, 09:52
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84=2^2*3*7
26=13*2
We get 13*k/ 2*3*7
Value of k has to be factor of 42
Multiple of 42 <215 with prime factors possible upto 5
We will have 2,3,5,7 prime factors of K
Option C
Posted from my mobile device
26=13*2
We get 13*k/ 2*3*7
Value of k has to be factor of 42
Multiple of 42 <215 with prime factors possible upto 5
We will have 2,3,5,7 prime factors of K
Option C
Bunuel
Posted from my mobile device
17 Nov 2021, 10:36
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Bunuel
Let's find the prime factorization of the numerator and denominator and then simplify.
We get: \(\frac{26k}{84}=\frac{(2)(13)k}{(2)(2)(3)(7)}=\frac{(13)k}{(2)(3)(7)}\)
In order for \(\frac{(13)k}{(2)(3)(7)}\) to be an INTEGER, k must be divisible by 2, 3 and 7.
(2)(3)(7) = 42, which means k must be divisible by 42.
Put another way, k must be a multiple of 42 (as long as k is less than 215, as stated in the question)
If k = 42, then the prime factors of k are 2, 3 and 7
If k = (2)(42) (a multiple of 42), then the prime factors of k are 2, 3 and 7
If k = (3)(42) (a multiple of 42), then the prime factors of k are 2, 3 and 7
If k = (4)(42) (a multiple of 42), then the prime factors of k are 2, 3 and 7
If k = (5)(42) (a multiple of 42), then the prime factors of k are 2, 3, 5 and 7
So, k can have, AT MOST, four different prime factors
Answer: C
