The Largest Power Of 3.......

The given number can be written as:

$\begin{aligned} N & = \prod_{k=1}^{99} \left((k+1)^2-k^2\right) = \prod_{k=1}^{99} \left((k+1)-k\right) \left((k+1)+k\right) \\ & = \prod_{k=1}^{99} \left(2k+1\right) = 3 \times 5 \times \cdots \times 197 \times 199 = 199!! \end{aligned}$

The highest power of 3 in $n!!$ can be found with the following formula:

$\begin{aligned} p_{!!}(n) & = \sum_{k=1}^\infty \left \lfloor \frac {\left \lfloor \frac n{3^k} \right \rfloor + 1}2 \right \rfloor & \small \color{#3D99F6} \text{Putting }n=199 \\ p_{!!}(199) & = \sum_{k=1}^\infty \left \lfloor \frac {\left \lfloor \frac {199}{3^k} \right \rfloor + 1}2 \right \rfloor \\ & = \left \lfloor \frac {\left \lfloor \frac {199}3 \right \rfloor + 1}2 \right \rfloor + \left \lfloor \frac {\left \lfloor \frac {199}9 \right \rfloor + 1}2 \right \rfloor + \left \lfloor \frac {\left \lfloor \frac {199}{27} \right \rfloor + 1}2 \right \rfloor + \left \lfloor \frac {\left \lfloor \frac {199}{81} \right \rfloor + 1}2 \right \rfloor \\ & = \left \lfloor \frac {66 + 1}2 \right \rfloor + \left \lfloor \frac {22 + 1}2 \right \rfloor + \left \lfloor \frac {7 + 1}2 \right \rfloor + \left \lfloor \frac {1+ 1}2 \right \rfloor \\ & = 33 + 11 + 4 + 1 = \boxed{49} \end{aligned}$

Notations:

• $!!$ denotes the double factorial notation .
• $\lfloor \cdot \rfloor$ denotes the floor function .

Note that $\begin{aligned} N &:= \left(100^2-99^2\right)\left(99^2-98^2\right)\cdots\left(3^2-2^2\right)\left(2^2-1^2\right) \\ &= \prod_{k=2}^{100} \left(k^2 - (k-1)^2\right) \\ &= \prod_{k=2}^{100} (k - (k-1)) \cdot (k + (k-1)) \\ &= \prod_{k=2}^{100} (2k-1) \\ &= 199!! \\ &= \frac{199!}{2^{99}99!} \end{aligned}$

Now, the highest power of $3$ that divides $n!$ is given by the formula $\sum_{k=1}^\infty \left\lfloor\frac{n}{3^k}\right\rfloor$ so that the highest power of $3$ that divides $N = \dfrac{199!}{2^{99}99!}$ is $\sum_{k=1}^\infty \left\lfloor\frac{199}{3^k}\right\rfloor - \sum_{k=1}^\infty \left\lfloor\frac{99}{3^k}\right\rfloor = (66 + 22 + 7 + 2) - (33 + 11 + 3 + 1) = \boxed{49}$

Let N = * ( 100^2 - 99^2)(99^2 - 98^2)..................(3^2 - 2^2)(2^2 - 1^2) * i.e. N = (100+99)(99+98)(98+97).....................(3+2)(2+1) * i.e. = * 1x3x5x7x9x................197x199. :. n= { $\frac{195-3}{6}$ +1} + { $\frac{189-9}{18}$ +1} + { $\frac{189-27}{54}$ +1} +1 so by solving we will get n=49