Is 2^6 * 3^10 * 5^12 * 7^15 * 1^18 a prefect square? Explain how you know.

Answer 1

to be a perfect square, all its prime factors must be squared.

Since 7 occurs an odd number of times, 7^2 cannot divide the number.

So, no, it is not a perfect square.

It is almost a perfect square, though, since

2^6 * 3^10 * 5^12 * 1^18 = (2^3 * 3^5 * 5^6)^2

Answer 2

every pair of same factors must be a perfect square and any integer base raised to an even exponent must be a perfect square, so
all factors are perfect square except 7^15
which is 7^14 x 7
7^14 is a perfect square, but if I multiply that by 7 it no longer is.
we could write the whole thing as
(2^3 * 3^5 * 5^6 * 7^7 * 1)^2*7
the final *7 wrecks it all