Chicago Cubs and modular arithmetic

Representing $x$ in the form of 77 as $77a+15$ for $a \ge 0$ .

Representing $x$ in the form of 286 as $286b+15$ for $b \ge 0$ .

Now equating both we get:

$77a+15=286b+15 \\ 7a=26b$

Now when $a=0$ we get $x=15$ but $x>15$ and $26 | a$ so $a=26$ and $x=2017$ .

We have $286=11.26$ & $77=11.7$ .

LCM of7 & 26 is 182.

So LCM of 286 & 77 is 11.182 = 2002 which both 286 & 77 divides.

So the least number leaving remainder 15 is $\boxed{2002+15=2017}$ .

I used Chinese Remainder Theorem

Consider that in the space of possible solutions we will want the lowest common multiple of 77 and 286 that is less than this number by 15. As we can interpret x congruent to 15 in under modulo 77 as all the multiples of 77 plus 15, the same can be said for the second equation. This leads back to the first statement I made.

$LCM(77,286)+15=2017$

Go Cubs! And to the end of the Billy Goat Curse!

LCM(77,286)=7×11×26=2002. x is equivalent to 15 mod(2002) as well.which implies that x=2002q+15 for some integer q.Therefore the lowest possible value of x such that x>15 is x=2002×1+15=2017. So the final answer is 2017.