Express 60 as a product of its prime factors. Find the smallest possible integer k such that 60k is a square number.
The correct answer and explanation is:
To express 60 as a product of its prime factors, we start by dividing 60 by the smallest prime numbers.
- 60 is divisible by 2 (since it is even):
60÷2=3060 ÷ 2 = 30 - 30 is divisible by 2 again:
30÷2=1530 ÷ 2 = 15 - 15 is divisible by 3 (since the sum of the digits, 1 + 5 = 6, is divisible by 3):
15÷3=515 ÷ 3 = 5 - 5 is a prime number.
Thus, the prime factorization of 60 is:
60=22×3×560 = 2^2 × 3 × 5
To find the smallest integer kk such that 60k60k is a square number:
A square number is a number that can be expressed as the product of identical factors, meaning each prime factor’s exponent in its prime factorization must be even.
The prime factorization of 60 is 22×31×512^2 × 3^1 × 5^1. To make 60k60k a perfect square, the exponents of all the primes must be even:
- The exponent of 2 is already even (222^2).
- The exponent of 3 is odd (313^1), so we need to multiply by one more 3 to make it even.
- The exponent of 5 is odd (515^1), so we need to multiply by one more 5 to make it even.
Thus, we multiply 60 by 3×5=153 × 5 = 15 to make 60k60k a square number. Therefore, the smallest integer kk is:
k=15k = 15
Now, multiplying 60×1560 \times 15 gives:
60×15=90060 \times 15 = 900
The prime factorization of 900 is:
900=22×32×52900 = 2^2 × 3^2 × 5^2
Since all the exponents are even, 900 is a perfect square. Thus, the smallest kk is 15, and 60k60k is now a square number.