Question

If \[N = {12^3} \times {3^4} \times {5^2}\], find the total number of even factors of \[N\].

Answer 1

Hint:
Here, we need to find the total number of even factors of \[N\]. All factors of \[N\]can be written as a product of the prime factors 2, 3, and 5, in the form \[N = {2^x} \times {3^y} \times {5^z}\]. Since an odd number multiplied by an even number is even, the even factors will be all the factors formed by multiplying 2, where the power of 2 is not zero.

Complete step by step solution:
A prime number is a number divisible by only 1 and by itself.
We need to write \[N\]as a product of its primes.
We know that 12 is the product of the prime numbers 2, 2, and 3.
Rewriting the value of \[N\], we get
\[N = {\left( {2 \times 2 \times 3} \right)^3} \times {3^4} \times {5^2}\]
Writing \[N\] as a product of its prime factors, we get
\[N = {\left( {{2^2} \times 3} \right)^3} \times {3^4} \times {5^2}\]
Simplifying the expression using the rules of exponents, we get
  $N = {2^6} \times {3^3} \times {3^4} \times {5^2} \\
  N = {2^6} \times {3^7} \times {5^2} \\ $
Now, we know that all factors of \[N\]can be written as a product of the prime factors 2, 3, and 5 in the form \[N = {2^x} \times {3^y} \times {5^z}\], where \[x = 0,1,2, \ldots ,6\], \[y = 0,1,2, \ldots ,7\], and \[z = 0,1,2\].
\[x\] can take 7 values, \[y\] can take 8 values, and \[z\] can take 3 values.
Since an odd number multiplied by an even number is even, the even factors will be all the factors formed by multiplying 2, where the power of 2 is not zero.
Therefore, in case of all even factors of \[N\], \[x\] can take 6 values (from 1 to 6), \[y\] can take 8 values, and \[z\] can take 3 values.
The number of even factors can be found by multiplying the number of possible values of the powers of the prime factors 2, 3, and 5.

Therefore, the number of even factors of \[N\] \[ = 6 \times 8 \times 3 = 144\].

Note:
Alternate method:
You can also solve this problem by finding the total number of factors of \[N\], then finding the total number of odd factors of \[N\], and subtracting the results.
In \[N = {2^6} \times {3^7} \times {5^2}\], \[x\] can take 7 values, \[y\] can take 8 values, and \[z\] can take 3 values.
Thus, we get the total number of factors of \[N\]as \[\left( {6 + 1} \right)\left( {7 + 1} \right)\left( {2 + 1} \right) = \left( 7 \right)\left( 8 \right)\left( 3 \right) = 168\]
Every odd factor of \[N\]will have a combination of the powers of 3 and 5.
In odd factors of \[N = {2^6} \times {3^7} \times {5^2}\], \[x\] can take 1 value (only 0), \[y\] can take 8 values, and \[z\] can take 3 values.
Thus, we get the total number of odd factors of \[N\]as \[1\left( {7 + 1} \right)\left( {2 + 1} \right) = 1\left( 8 \right)\left( 3 \right) = 24\]
Now, we can get the number of even factors of \[N\] by subtracting the number of odd factors from the total number of factors.
Therefore, the number of even factors of \[N\]\[ = 168 - 24 = 144\].