I'm happy to help with this. This is a very hard question.

Let's look at the equation 5y=x^2+x
The right side is a multiple of 5, because y is an integer.
The left side can be factored into x*(x+1) ------ what's interesting is that x is a prime number, so it's not going to be divisible by 5 or anything other than 1 and itself. The only way that x*(x+1) could possibly be a multiple of 5 is if x+1 is divisible by 5. Numbers divisible by 5 must end in 5 or 0, so numbers that are one less than a multiple of 5 must end in 4 or 9. No prime numbers end in four, because all primes larger than 2 are even, but a prime number can end with 9 --- the first few primes that end in 9 are: 19, 29, 59, 79, 89, 109, ..... Those are the only possible values of x up to 120. Notice, in every case, x+1 will be divisible by not only 5 but also 10

Let's take x = 19 as an example. x^2+x = x*(x+1) = 19*20 = 380

In general, we can say x^2+x = x*(x+1) = x*k*10 (where k is some other number, the value of which doesn't matter --- we just know that k*10 is some multiple of 10.)

5y = x*k*10

divide by 5

y = x*k*2 = 2x*k

Thus, y is not always divisible by 5, not always divisible by x+1, but it MUST be divisible by 2x. I & III are not always true, but II is always true.

Does all this make sense?

Mike
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Mike McGarry
Magoosh Test Prep