If x Ω y is defined as the smallest integer of which both x and y are factors, then 10 Ω 32 is how much greater than 6 Ω 20?

A. 2
B. 70
C. 100
D. 160
E. 200

I think it's C.

Can you explain?

"The smallest integer divisible by 10 and 32 is 160, and the smallest integer divisible by 6 and 20 is 60. The difference between them is 100." - Princeton Review

To find the value of the expression "x Ω y," we need to determine the smallest integer that both x and y are factors of.

Let's find the smallest integer that is a factor of both 10 and 32.

The prime factorization of 10 is 2 * 5, and the prime factorization of 32 is 2^5.

To find the smallest integer that both 10 and 32 are factors of, we look for the highest power of each prime factor that appears in either number.

In this case, the highest power of 2 that appears in either number is 2^5, and the highest power of 5 that appears in either number is 5^1.

Therefore, the smallest integer that is a factor of both 10 and 32 is 2^5 * 5^1 = 320.

So, 10 Ω 32 = 320.

Next, let's find the smallest integer that is a factor of both 6 and 20.

The prime factorization of 6 is 2 * 3, and the prime factorization of 20 is 2^2 * 5.

The highest power of 2 that appears in either number is 2^2, and the highest power of 5 that appears in either number is 5^1.

Therefore, the smallest integer that is a factor of both 6 and 20 is 2^2 * 3^1 * 5^1 = 60.

So, 6 Ω 20 = 60.

Now, let's compare the two values: 10 Ω 32 and 6 Ω 20.

10 Ω 32 = 320
6 Ω 20 = 60

The difference between 10 Ω 32 and 6 Ω 20 is 320 - 60 = 260.

Therefore, the answer is not one of the given options (A, B, C, D, or E).