Is 2^6 * 3^10 * 5^12 * 7^15 * 1^18 a prefect square? Explain how you know.
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
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
To determine if the expression 2^6 * 3^10 * 5^12 * 7^15 * 1^18 is a perfect square, we can analyze the exponents of each prime factor.
First, let's simplify the expression by evaluating any exponents with a base of 1 since any number raised to the power of 0 is equal to 1. Thus, we can ignore the term 1^18, as it doesn't affect the overall result.
So, the expression becomes: 2^6 * 3^10 * 5^12 * 7^15.
To check whether this product is a perfect square, we need to examine the exponents of each prime factor. For a number to be a perfect square, all the exponents must be even. In other words, the exponents must be divisible by 2.
Let's analyze each exponent:
- The exponent of 2 is 6. Since 6 is divisible by 2, the exponent is even.
- The exponent of 3 is 10. Since 10 is not divisible by 2, the exponent is odd.
- The exponent of 5 is 12. Since 12 is divisible by 2, the exponent is even.
- The exponent of 7 is 15. Since 15 is not divisible by 2, the exponent is odd.
Since not all of the exponents are even, the expression 2^6 * 3^10 * 5^12 * 7^15 is not a perfect square.