How many factors of \(2^3∗3^4∗5^5\) are even numbers?
A. 20
B. 30
C. 90
D. 100
E. 120
ALTERNATIVE approach first:
The factors of \(2^3∗3^4∗5^5\) can be written as
\(2^1, 2^2\) and \(2^3\) (only 2 OR its multiples) = 3
\(3^1, 3^2, 3^3\) and \(3^4\) (only 3 OR its multiples) = 4
\(5^1, 5^2, 5^3, 5^4\) and \(5^5\) (only 5 OR its multiples) = 5
Also, possible factors are the multiples of 3 and 5 = 4*5 = 20
So, total even factors are = Only 2's + Multiples of 2 and 3 + Multiples of 2 and 5 + Multiples of 2 and Multiples of 4 and 5
= 3 + 3*4 + 3*5 + 3*20
= 90
PS: The powers 2, 3 and 5 can take various values as follows:
Power of 2 = 3 ways (1 to 3; Can't take 0 as power 2 as \(2^0\) = 1 which is odd)
Power of 3 = 5 ways (0, 1 .. to .. 4)
Power of 5 = 6 ways (0, 1, .. to .. 6)
Hence total even factors = 3*5*6 = 90-----------------------------------------------------------------------------------------------------
Number of factors for a number denoted as P^m * Q^n * R^o * ... = (m+1)*(n+1)*(o+1)* ..
Here we have only three prime factors, so
P = 2, Q = 3 and R = 5 and m = 3, n = 4 and o = 5
Hence total factors = (3+1)*(4+1)*(5+1) = 120
Out of these number of odd factors are = (5)*(6) = 30
Hence even factors are 120 - 30 = 90
Answer C.
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