Klenex wrote:

The number of even factors of 21600 is

A. 32

B. 42

C. 60

D. 25

E. 52

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: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

Back to the original question:Make a prime factorization of the number: \(21,600 =2^5*3^3*5^2\).

According to the above the number of factors is \((5+1)(3+1)(2+1)=72\).

Now, get rid of powers of 2 as they give even factors --> you'll have \(3^3*5^2\) which has (3+1)(2+1)=12 factors. All the remaining factors will be odd, therefore 21,600 has 72-12=60 even factors.

Answer: C.

P.S. Please do not reword or shorten questions when posting. Thank you.

Thank you for the answer. But I am not able to figure out the 60 even factors

As Even*Even=Even & Even*Odd=Even , I considered 2^5*3^3 which has (5+1)(3+1)=24 factors and 2^5*5^2 which has (5+1)(2+1)=18 factors.

So, total 24+18=42 factors. What am I missing here?