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.
Answer: D.
THEORY:
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*5^2Total 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.htmlBACK 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.
Answer: E.
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