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If the prime factorization of the integer q can be expressed
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Updated on: 18 Apr 2013, 21:38
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If the prime factorization of the integer q can be expressed as \(a^{2x} b^{x} c^{3x1}\), 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.
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Originally posted by enigma123 on 28 Jan 2012, 17:12.
Last edited by walker on 18 Apr 2013, 21:38, edited 2 times in total.
Added the OA




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Re: Consecutive +ve integers
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28 Jan 2012, 17:27
enigma123 wrote: If the prime factorization of the integer q can be expressed as a^2x b^x c^3x1, 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. Given: \(q=a^{2x}*b^x*c^{3x1}\). # of distinct factors of q is \((2x+1)*(x+1)*(3x1+1)=(2x+1)*(x+1)*3x\). 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 IntegerFirst 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: mathnumbertheory88376.html
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Re: Consecutive +ve integers
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28 Jan 2012, 17:28
Are you a Maths Avatar or something? You are indeed a GMATCLUB LEGEND. Thanks buddy.
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Re: If the prime factorization of the integer q can be expressed
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11 Jul 2015, 13:38
enigma123 wrote: If the prime factorization of the integer q can be expressed as \(a^{2x} b^{x} c^{3x1}\), 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. Straight forward question if you know the formula for number of factors of an integer \(q= a^x * b^y * c^z\) , where \(x,y,z \geq{0}\) and are integers = (x+1)(y+1)(z+1) including 1 and integer q. Now given, q= \(a^{2x} b^{x} c^{3x1}\), thus based on the above formula, the number of factors of q including 1 and q = (2x+1)(x+1)(3x1+1) = 3x(2x+1)(x+1). So we know that the number will be of the form 3x(2x+1)(x+1) , will be an integer and will be true for ALL X. Thus substitute x=1, we get number of factors = 3*3*2 =18 (this eliminates A, as 3 is a factor of 18 and thus can not leave 4 or (43=1) as the remainder. Similarly, eliminate C and D). Now substitute x=2, we get number of factors = 6*5*3 = 90 and this eliminates E as 10 is a factor of 90 and will leave a remainder of 0. Thus B is the only remaining answer choice and is the OA.



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Re: If the prime factorization of the integer q can be expressed
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20 Jun 2016, 17:57
IMO (B)The answer can be thought of in a simple manner. Here's my take.. First, we know that calculating the number of factors by formula will transform into something like \((2x+1)(x+1)(3x)\) when this is simplified..it becomes \(6x^3 + 9x^2 + 3x\) This means two things 1. The number is a multiple of 3. 2. The value of this number depends on the value of x(as x can be taken out as common).
Looking at the options we see that option B is \(5k + 5\) or \(5(k+1)\) or simply a multiple of 5. The question asks "could be"..so the value of x could be 5..right? and there you have it
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If the prime factorization of the integer q can be expressed
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30 Jun 2018, 18:41
Bunuel wrote: enigma123 wrote: If the prime factorization of the integer q can be expressed as a^2x b^x c^3x1, 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. Given: \(q=a^{2x}*b^x*c^{3x1}\). # of distinct factors of q is \((2x+1)*(x+1)*(3x1+1)=(2x+1)*(x+1)*3x\). 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 IntegerFirst 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: http://gmatclub.com/forum/mathnumbertheory88376.htmlHey Bunuel, solving for total number of factors, \(q=6*x^3+9*x^2+3*x\) so q will be always even. But in choice b, if k is even, then q will be odd. How can this be resolved ?



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Re: If the prime factorization of the integer q can be expressed
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30 Jun 2018, 23:47
Dineshrambalaji wrote: Bunuel wrote: enigma123 wrote: If the prime factorization of the integer q can be expressed as a^2x b^x c^3x1, 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. Given: \(q=a^{2x}*b^x*c^{3x1}\). # of distinct factors of q is \((2x+1)*(x+1)*(3x1+1)=(2x+1)*(x+1)*3x\). 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 IntegerFirst 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: http://gmatclub.com/forum/mathnumbertheory88376.htmlHey Bunuel, solving for total number of factors, \(q=6*x^3+9*x^2+3*x\) so q will be always even. But in choice b, if k is even, then q will be odd. How can this be resolved ? The question asks: which of the following COULD be the total number of factors of q? Options A, C, D and E, CANNOT be the total number of factors of q, regardless of the values of j, l, m, and n. Option B, 5k + 5, COULD be the total number of factors of q, for specific values of k. For example if k = 17.
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Re: If the prime factorization of the integer q can be expressed
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31 Jul 2018, 23:32
This is how i solved this question: given = a^2x * b ^x * C ^3x1 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




Re: If the prime factorization of the integer q can be expressed &nbs
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