Bunuel wrote:
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
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
You can solve this question even not analyzing option II at all.
Since 5y=x(x+1), then x+1 must be a multiple of 5, thus the units digit of x must be 9 (it cannot be 4, because in this case x would be even and thus not a prime number).
Now, if x=19, then neither I nor III is true. The only option that does not contains I or III is B. So, B must be correct.
Answer: B.
If you interested why II must be true consider this: since x is a prime number greater than 5, then it's odd --> 5y=x(x+1)=odd(odd+1)=odd*even=even --> 5y=even --> y=even.
Also, \frac{y}{x}=\frac{x+1}{5}. we know that x+1 is a multiple of 5, thus \frac{y}{x}=\frac{x+1}{5}=integer --> y is a multiple of x. And since y is even, then y is a multiple of 2x.
Hope it helps.
I. 5
II. 2x
III. x+1
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
You can solve this question even not analyzing option II at all.
Since 5y=x(x+1), then x+1 must be a multiple of 5, thus the units digit of x must be 9 (it cannot be 4, because in this case x would be even and thus not a prime number).
Now, if x=19, then neither I nor III is true. The only option that does not contains I or III is B. So, B must be correct.
Answer: B.
If you interested why II must be true consider this: since x is a prime number greater than 5, then it's odd --> 5y=x(x+1)=odd(odd+1)=odd*even=even --> 5y=even --> y=even.
Also, \frac{y}{x}=\frac{x+1}{5}. we know that x+1 is a multiple of 5, thus \frac{y}{x}=\frac{x+1}{5}=integer --> y is a multiple of x. And since y is even, then y is a multiple of 2x.
Hope it helps.
As per the above highlighted part-how can we say that y is multiple of 2x ?
Can you please explain.