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# An even positive integer 'x' has 'y' positive integral facto

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An even positive integer 'x' has 'y' positive integral facto  [#permalink]

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13 Jan 2011, 03:39
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55% (hard)

Question Stats:

60% (02:07) correct 40% (01:37) wrong based on 286 sessions

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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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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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13 Jan 2011, 06:15
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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 cannot 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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An even positive integer 'x' has 'y' positive integral facto  [#permalink]

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16 Nov 2016, 03:42
1
Let’s express x as a product of its prime factors:

$$x = 2^a*3^b*5^c*…$$

Total number of its positive divisors will be

$$(a+1)*(b+1)*(c+1)*… = y$$

Now we’ll do the same for the 4x

$$4x= 2^2*2^a*3^b*5^c*… = 2^{a+2}*3^b*5^c…$$

Total number of positive divisors of 4x will be

$$(a + 3)*(b+1)*(c+1) … = y’$$

For easier calculations let’s take $$(b+1)*(c+1)*… = z$$

We have: $$(a + 3)*z = y’$$

$$(a + 1 + 2)*z = y’$$

$$(a+1)*z + 2*z = y’$$

$$y + 2*z = y’$$

So, as we can see, in order to find out the total # of positive divisors of 4x we need to know # of odd divisors of x, which is not given in the question.

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Re: An even positive integer 'x' has 'y' positive integral facto  [#permalink]

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22 Jul 2017, 11:54
'x' has 'y' positive integral factors including '1' and the number itself.
Thus x should be a prime number.And none of the options fit in except E.
Re: An even positive integer 'x' has 'y' positive integral facto   [#permalink] 22 Jul 2017, 11:54
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