$a=2450$? i did 49*50 is this first grdae?!

Denote $f(x)$ as the LHS. Note that for $x\leq 1$, we have $f(x)=4950-99x$ and for $x\geq 99$ we have $f(x)=99x-4950$ If $x\in[1,99]\cap\mathbb{Z}$, then $$f(x)=\sum_{k=1}^{x-1} k+\sum_{k=1}^{99-x} k$$$$=\frac{x(x-1)+(99-x)(100-x)}{2}$$$$=\frac{x^2-x+x^2-199x+9900}{2}$$$$x^2-100x+4950$$Making a rough sketch of our entire function, we see that there will be a unique solution only when the solution is the closest integer to the $x$ which minimizes $x^2-100x+4950$ This is $x=50$, which gives $a=\frac{(50)(49)+(49)(50)}{2}=2450$

RedFireTruck wrote: $a=2450$? i did 49*50 is this first grdae?! I found this pdf https://imomath.com/othercomp/Slo/SloMO01.pdf and it has problems for "1st grade" up to "4th grade" so I assume they are just question difficulty levels.

How is this problem 1st grade level.

Playing around with small expressions like $y=|x-1|+|x-2|$, $y=|x-1|+|x-2|+|x-3|$, we see that y is linear for all values of x except at a point of symmetry. In the first case, the symmetry point is $x=\frac{3}{2}$ and $x=2$ in the second. Inductively, $x=50$ is the point of symmetry in the given expression, using which we obtain $a=2450$.