Final answer:
The smallest integer p that makes 540p a square number is 5. Factoring 540 into its prime factors and making all exponents even, we find that the square root of 540p is a product of its prime factors: 2 × 3 × 3 × 5, which equals 90. Hence, √(540p) as the product of its prime factors is 90.
Step-by-step explanation:
To find the smallest integer p such that 540p is a square number, we first need to factorize 540 into its prime factors. The prime factorization of 540 is 22 × 33 × 5.
For a number to be a square, all exponents in its prime factorization must be even. Here, the exponent of 5 is odd, so we need an additional factor of 5 to make it even. Therefore, p must be at least 5.
Now we have 540p as a perfect square, which is 22 × 33 × 52.
The square root of this number is the product of the primes, each raised to half their exponent in the perfect square, which gives us √(540p) = 21 × 33/2 × 51
= 2 × 3 × 3 × 5
= 90.
Thus, the smallest integer p for which √(540p) is a square number and the product of its prime factors is 90.