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# Number of factors

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Number of factors [#permalink]  13 Jan 2011, 03:39
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33% (02:59) correct 66% (01:24) wrong based on 0 sessions
An even positive integer 'x' has 'y' positive integral factors including '1' and the number itself. How many positive integral factors does the number 4x have?

A. 4y

B. 3y

C. 16y

D. 5y

E. Cannot be determined
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Re: Number of factors [#permalink]  13 Jan 2011, 05:04
I tried for x=2,4 and 6. The integral factors for x and 4x did not have any specific relation.
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Re: Number of factors [#permalink]  13 Jan 2011, 06:15
gmatpapa wrote:
An even positive integer 'x' has 'y' positive integral factors including '1' and the number itself. How many positive integral factors does the number 4x have?

A. 4y
B. 3y
C. 16y
D. 5y
E. Cannot be determined

Probably the easiest way would be to try different numbers:
If x=2 then y=2 --> 4x=8=2^3 and # \ of \ factors=4=2y;
If x=2^2 then y=3 --> 4x=16=2^4 and # \ of \ factors=5=\frac{5y}{3};
Two different answers for two values of x, hence we can not determine the # of factors of 4x.

THEORY:

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

BACK TO THE ORIGINAL QUESTION:

Given: x is even -->so x=2^p*b^q, where b is some other prime factor of x (other than 2) and q is its power (note that x may or may not have other primes, this is just an example). The number of all factors of x is y=(p+1)(q+1) so:
if p=1 then y=2(q+1);
if p=2 then y=3(q+1);
if p=3 then y=4(q+1);
....

Now, 4x=2^2*x=2^{p+2}*b^q and 4x will have (p+2+1)(q+1)=(p+3)(q+1), so:
if p=1 then # \ of \ factors=4(q+1)=2y;
if p=2 then # \ of \ factors=5(q+1)=\frac{5y}{3};
if p=3 then # \ of \ factors=6(q+1)=\frac{6y}{4};
....

So # of factors of 4x depends on the initial power of 2 in x.

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Re: Number of factors   [#permalink] 13 Jan 2011, 06:15
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