Integers: Questions Involving Fractions

If $$x$$ is a prime number greater than 5, $$y$$ is a positive integer, and $$5y=x^2+x$$, then $$y$$ must be divisible by which of the following? >I. $$5$$ >II. $$2x$$ >III. $$x+1$$

Correct. [[snippet]] I. Since $$y=\frac{x(x+1)}{5}$$ is an integer, $$x(x+1)$$ must be divisible by 5 to cancel out the 5 in the denominator. Since $$x$$ is a prime number greater than 5, it is definitely *not* divisible by 5. It follows that $$(x+1)$$ must be divisible by 5. You're looking for a prime number $$x$$ that is 1 less than a multiple of 5. $$x=19$$ fits the bill since $$x+1=20$$ is divisible by 5. In that case, >$$\displaystyle y = \frac{19\cdot 20}{5} = 19\cdot 4$$. This is not divisible by 5, so $$y$$ doesn't *have* to be divisible by 5. II. In the equation $$y=\frac{x(x+1)}{5}$$, $$y$$ is definitely divisible by $$x$$, but is $$y$$ divisible by 2? Remember that $$x(x+1)$$ is the product of two consecutive integers. We know that $$x$$ is a prime number greater than 5 and that therefore it is odd. Thus $$x+1$$ is *even*. Since $$x+1$$ is even, $$5y$$ is also even, and therefore $$y$$ is even (i.e., divisible by 2). It follows that $$y$$ is divisible by $$2x$$. III. With the same plug-in that was used for option I, you can see that $$y=19\cdot 4$$ is not divisible by $$x+1=20$$. Since you have found a single example where $$y$$ is not divisible by $$x+1$$, it follows that $$y$$ does not *have* to be divisible by $$x+1$$.

Incorrect. [[snippet]]

Incorrect. [[snippet]] You're right about II, but are you sure about III?

Incorrect. [[snippet]] You're right about II, but are you sure about I?

Incorrect. [[snippet]]

I only

II only

III only

I and II only

II and III only