Find the smallest positive integer k such that 96k is a square number

There are essentially two ways that I know to solve this.

One is to guess.

The other is to find the prime factors of 96. For an integer to be a perfect square, each prime factor has to occur an even number of time. Therefore k would need to be the product of whatever primes occur in 96 an odd number of times.

Given that 9+6 is divisible by 3, we know that 96 is divisible by 3. Doing the division, we find that 96 = 3 * 32.

The prime factor 3 only occurs once. Therefore k must be a multiple of 3.

32 is 2 raised to the 5th power. This is an odd number of 2’s. We need one more. Therefore k must be a multiple of 2.

Therefore the smaller possible value of k is 6.

96 * 6 is a perfect square. It is the same as 64 * 9, or 24 * 24.

Hope it helps!

(23*2)-12=x     x=34

step-by-step explanation: