The least integer whose multiplication with 588 leads to a perfect square is
First, find the prime factorization of 588.
588 ÷ 2 = 294
294 ÷ 2 = 147
147 ÷ 3 = 49
49 ÷ 7 = 7
7 ÷ 7 = 1
So, the prime factorization of 588 is 2² × 3¹ × 7².
To make 588 a perfect square by multiplication, we need to multiply it by an integer such that all the exponents in the resulting prime factorization are even.
The exponents in 588 are: 2 (for prime 2), 1 (for prime 3), and 2 (for prime 7).
The exponents of 2 and 7 are already even.
The exponent of 3 is 1, which is odd. To make it even, we need to multiply by 3 raised to an odd power. The least such power is 1. So, we must multiply by at least 3¹.
Multiplying 588 by 3:
588 × 3 = (2² × 3¹ × 7²) × 3¹ = 2² × 3¹⁺¹ × 7² = 2² × 3² × 7².
The exponents are now 2, 2, and 2, which are all even.
The resulting number is (2 × 3 × 7)² = 42², which is a perfect square.
The least integer whose multiplication with 588 leads to a perfect square is 3.