Hi Everyone.

Can anyone please explain the logic behind this problem?

If n is the product of 2, 3, and a two-digit prime number, how many of its factors are greater than 6?

Hi Everyone.

Can anyone please explain the logic behind this problem?

If n is the product of 2, 3, and a two-digit prime number, how many of its factors are greater than 6?

Consider a two digit prime number as 11.

then n=2x3x11=66

Then its factor are : 1,2,3,6,11,66 i.e. there are two factors above 6.

This would apply for any two digit prime number, so 2 is the answer.

Hey, I think you might have missed out on 22(2 x 11) and 33(3 x 11) being factors of 66.

So, the total number of factors greater than 6 should be 4.

wow, yes. I thought I was going wrong somewhere, thanks for the correction.

Word to Algebra

n = 2 x 3 x P and P > 10

Since 2 x 3 is not greater than 6 and P alone is greater than 6 this problem is simplified to find numbers of positive factors of ānā that have P in it.

We can use the +1 trick but keep P at 1:

2 x 2 x 1 = 4 which is the answer