Question

Expressed as the product of prime factors ,198=2x3^2 x11 and 90=2x3^2x5 use these results to find

the smallest integer,k,such that 198k is a perfect square

Answer 1

For 198: k = 22

For 90: k = 10

Explanation:

For a number to be a perfect square, its prime factors must have even exponents.

198:

If an exponent is not even, you must raise it to the next even exponent.

198 = 2 × 3² × 11

2 and 11 have exponent 1. They need exponent 2.

The means 198 must be multiplied by 2 × 11 = 22

Then you have 198 × 22 = 2² × 3² × 11², and you have a perfect square.

198 × 22 = 198k

k = 22

90:

If an exponent is not even, you must raise it to the next even exponent.

90 = 2 × 3² × 5

2 and 5 have exponent 1. They need exponent 2.

The means 90 must be multiplied by 2 × 5 = 10

Then you have 90 × 10 = 2² × 3² × 5², and you have a perfect square.

90 × 10 = 90k

k = 10