Question

How many factors of \[{2^5} \times {3^6} \times {5^2}\] are perfect squares?
A. 24
B. 12
C. 16
D. 22

Answer 1

Hint: First of all, find the possible number of ways in which perfect square factors of \[{2^5},{3^6},{5^2}\] can be arranged individually. Then use the multiplicative principle of permutations to get the required answer. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
For a perfect square, the power of each should be even.
The possible factors of \[{2^5}\] are \[{2^0},{2^1},{2^2},{2^3},{2^4},{2^5}\]
So, the possible perfect square factors of \[{2^5}\] are \[{2^0},{2^2},{2^4}\].
Therefore, possible number of ways of arranging the perfect square factors of \[{2^5}\] = 3
The possible factors of \[{3^6}\] are \[{3^0},{3^1},{3^2},{3^3},{3^4},{3^5},{3^6}\]
So, the possible perfect square factors of \[{3^6}\] are \[{3^0},{3^2},{3^4},{3^6}\].
Therefore, possible number of ways of arranging the perfect square factors of \[{3^6}\] = 4
The possible factors of \[{5^2}\] are \[{5^0},{5^1},{5^2}\]
So, the possible perfect square factors of \[{5^2}\] are \[{5^0},{5^2}\].
Therefore, possible number of ways of arranging the perfect square factors of \[{5^2}\] = 2
By using multiplicative principle of permutations, we have
The total number of ways of arranging the perfect square factors of \[{2^5} \times {3^6} \times {5^2}\] are \[3 \times 4 \times 2 = 24\]
Hence there are 24 factors of \[{2^5} \times {3^6} \times {5^2}\] which are perfect squares.
Thus, the correct option is A. 24

Note: In this problem we have used multiplicative principle permutations i.e., if there are \[x\] number of ways of arranging one thing and\[y\] number of ways of arranging another, then the total number of ways of arranging both the things is given in \[xy\] number of ways.