The number of even factors of 21600 is

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:

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.

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