Divisibility problem

$(100^2-99^2)(99^2-98^2)\dots(3^2-2^2)(2^1-1^2)=199\cdot197\cdot\dots\cdot5\cdot3$

Which is equal to $\frac{199!}{2\cdot4\cdot6\dots\cdot198}$

First find the powers of 3 in the numerator;

$\left\lfloor\frac{199}{3}\right\rfloor+\left\lfloor\frac{199}{9}\right\rfloor+\left\lfloor\frac{199}{27}\right\rfloor+\left\lfloor\frac{199}{81}\right\rfloor=66+22+7+2=97$

Now find the powers of 3 in the denominator;

Write the denominator as

$2\cdot4\cdot6\dots\cdot198=2^{99}(1\cdot2\cdot3\dots\cdot99)$

and then, the same procedure

$\left\lfloor\frac{99}{3}\right\rfloor+\left\lfloor\frac{99}{9}\right\rfloor+\left\lfloor\frac{99}{27}\right\rfloor+\left\lfloor\frac{99}{81}\right\rfloor=33+11+3+1=48$

Therefore the difference $97-48=49$ is the largest $n$ .