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If the prime factorization of the integer q can be expressed as \(a^{2x} b^{x} c^{3x-1}\), where a, b, c, and x are distinct positive integers, which of the following could be the total number of factors of q?
(A) 3j + 4, where j is a positive integer
(B) 5k + 5, where k is a positive integer
(C) 6l + 2, where l is a positive integer
(D) 9m + 7, where m is a positive integer
(E) 10n + 1, where n is a positive integer
Guys - does this question makes sense to you? I am struggling. OA is not provided either.
(A) 3j + 4, where j is a positive integer
(B) 5k + 5, where k is a positive integer
(C) 6l + 2, where l is a positive integer
(D) 9m + 7, where m is a positive integer
(E) 10n + 1, where n is a positive integer
Guys - does this question makes sense to you? I am struggling. OA is not provided either.
28 Jan 2012, 17:27
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enigma123
Given: \(q=a^{2x}*b^x*c^{3x-1}\). # of distinct factors of q is \((2x+1)*(x+1)*(3x-1+1)=(2x+1)*(x+1)*3x\).
Now, \(3x(x+1)(2x+1)\) is both multiple of 3 and even (as either x or x+1 is even). Let's check the answer choices:
A. 3j+4 --> not a multiple of 3 (1 more than multiple of 3) --> discard;
B. 5k+5 --> can be even and multiple of 3, for example if k=5, then 5k+5=30 --> keep;
C. 6l+2 --> not a multiple of 3 (2 more than multiple of 3) --> discard;
D. 9m+7 --> not a multiple of 3 (1 more than multiple of 3) --> discard;
E. 10n+1 --> not even (10n+1=even+odd=odd) --> discard.
So, only 5k+5 can possibly be the total number of factors of q (in fact the lowest value of k would be 17 for which x=2).
Answer: B.
Hope it's clear.
P.S. In case one doesn't know.
Finding the Number of Factors of an Integer
First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.
The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)
Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
For more on number properties check: math-number-theory-88376.html
31 Jul 2018, 23:32
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This is how i solved this question:
given = a^2x * b ^x * C ^3x-1
Lets assume x = 2
so it becomes a^4*b^2*c^5
Total factors = 5*3*6 = 90
now equate 90 to each of the equation given.
1)3j+4 = 90
j = 86/3 (not divisible)
2) 5k+5 = 90
k = 85/5 (divisible)
3) 6l+2 = 90
l = 88/6 (not divisible)
4) 9m+7 = 90
m = 83/9 (not divisble)
5) 10n+1 = 90
n = 89/10 (not divisible )
So Answer = B
given = a^2x * b ^x * C ^3x-1
Lets assume x = 2
so it becomes a^4*b^2*c^5
Total factors = 5*3*6 = 90
now equate 90 to each of the equation given.
1)3j+4 = 90
j = 86/3 (not divisible)
2) 5k+5 = 90
k = 85/5 (divisible)
3) 6l+2 = 90
l = 88/6 (not divisible)
4) 9m+7 = 90
m = 83/9 (not divisble)
5) 10n+1 = 90
n = 89/10 (not divisible )
So Answer = B
