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# If the prime factorization of the integer q can be expressed

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If the prime factorization of the integer q can be expressed  [#permalink]

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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 $$a^{2x} b^{x} c^{3x-1}$$, 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.

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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.
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28 Jan 2012, 17:27
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enigma123 wrote:
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.

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).

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
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28 Jan 2012, 17:28
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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  [#permalink]

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11 Jul 2015, 13:38
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enigma123 wrote:
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.

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= $$a^{2x} b^{x} c^{3x-1}$$, thus based on the above formula, the number of factors of q including 1 and q = (2x+1)(x+1)(3x-1+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 (4-3=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  [#permalink]

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20 Jun 2016, 17:57
1
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  [#permalink]

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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^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.

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).

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: http://gmatclub.com/forum/math-number-theory-88376.html

Hey 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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Posts: 60778
Re: If the prime factorization of the integer q can be expressed  [#permalink]

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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^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.

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).

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: http://gmatclub.com/forum/math-number-theory-88376.html

Hey 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  [#permalink]

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31 Jul 2018, 23:32
3
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 )

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Re: If the prime factorization of the integer q can be expressed  [#permalink]

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08 Nov 2018, 06:16
2
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

Given: Prime factorization , distinct integers , Could be type question
therefore No of factors = (2x+1) (x+1)(3x)
Let x = 1 => No of factors = 3*2*3 = 18
None of the option can give 18 so...
Let x =2 => 5*3*6 =90 factors
check option B (k is +ve integer) = 5k+5 = 90 => k=17
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Re: If the prime factorization of the integer q can be expressed  [#permalink]

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09 Nov 2018, 00:45
Sum of all factors = (2x+1)(x+1)(3x-1+1) = 3[x(x+1)(2x+1)] = 6*3[x(x+1)(2x+1)/6] = 18*[sum of squares of n consecutive integers starting from 1] = 18*integer.
Now check which option is divisible by 18.
Only (B) is the suitable answer.
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Re: If the prime factorization of the integer q can be expressed  [#permalink]

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Re: If the prime factorization of the integer q can be expressed   [#permalink] 27 Dec 2019, 09:08
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