What is the probability that the product of the numbers will be a multiple of 6?

Answer 1

Assuming 2 randomly selected positive integers are multiplied together, I came up with a 5 in 12 chance (41 2/3 percent )

Here's how I figured it: take the first number and divide by six and examine the remainder (called modulo 6). You have 6 different possibilities (0,1,2,3,4,5), each with a 1 in six chance of occurring.

If mod 6 = 0, then the first number is a multiple by 6, and it doesn't matter what the 2nd number is; all products are multiples of 6 as well.

If mod 6 = 1 or 5, then the 2nd number must be a multiple of six to get the product to be multiple of 6. The chance of the 2nd number a multiple of six is 1 in 6.

If mod 6 = 2, then these are numbers like 2,8,14,20... They are even numbers, so if the second number is a multiple of 3, then the product is multiple of 6. The chance of the 2nd number a multiple of three is 1 in 3.

Mod 6 = 3, then first number is multiple of 3 (like 3,9,15,21), and the second number must be even (chance is 1 in 2).

Mod 6 = 4, then first number is like 4,10,16,22 It is an even number so the 2nd number must be multiple of 3 (chance is 1 in 3).

So in order how I described, we have (1 + 1/6 + 1/6 + 1/3 + 1/2 + 1/3)*(1/6) which equals 5/12.