Let $x$ be a prime number and $y$ be an integer.If following expression is an integer then find maximum possible value of$y$.
$$\frac{1}{xy}+\frac{2}{xy}+\frac{3}{xy}+\frac{4}{xy} + \cdots + \frac{54}{xy}+\frac{55}{xy}$$
My answer:$$\frac{1}{xy}+\frac{2}{xy}+\frac{3}{xy}+\frac{4}{xy}+\cdots+\frac{54}{xy}+\frac{55}{xy}=\frac{2^2 \times 5 \times7 \times11}{xy}$$
Obviously $x\in\{2,5,7,11\}$ and for $x ≠2$, $y$ can't be a factor of $x$ because then the expression is not an integer number. So $y≠5,7,11$. and for $x=2$ for the same reason $y≠4$. So the maximum of $y$ should be $2$.
Is $y=2$ really the maximum value of $y$ ? or I am missing something?
Thank you!
Best Answer
The correct value should be $y = 770$.
You have almost solved the problem. Once you see that the sum is equal to $(2^2 \times 5 \times 7 \times 11)/xy,$ you can just set $x = 2$ so that $y = 11 \times 7 \times 5 \times 2 = 770$. This satisfies the desired constraints, and it's clear that we can't do any better (we've exhausted all of the prime factors available, and changing the value of $x$ will only decrease the value of $y$).
Note that the expression is still an integer:
$$\frac{2^2 \times 5 \times 7 \times 11}{xy} = \frac{1540}{2 \times 770} = 1.$$