The largest exponent of 15 in $100!$ is \[\]
A.48\[\]
B.24\[\]
C.12\[\]
D. None of these \[\]

Hint: We find the highest power on 3 as $r$ such that ${{3}^{r}}$ divides 100! and highest power on 5 as $S$ such that ${{5}^{S}}$ divides $100!$1 from the formula $k=\left[ \dfrac{n}{p} \right]+\left[ \dfrac{n}{{{p}^{2}}} \right]+\left[ \dfrac{n}{{{p}^{3}}} \right]+...$ where ${{p}^{k}}$ exactly divides $100!$ with $p$ being a prime. We use the fact that if two divisor exactly divide a number then their product also divide that number to find $t$ such that ${{15}^{t}}={{\left( 3\times 5 \right)}^{t}}$ divides $100!$1.

Complete step-by-step solution
We know that if there are two numbers $p$ and $q$ and they exactly divide any number $n$ and then their product $pq$ also exactly divides the number $n$. We also know that the largest exponent on any prime $p$ is $k$ such that ${{p}^{k}}$ exactly divides $n!$ then we have,
\[k=\left[ \dfrac{n}{p} \right]+\left[ \dfrac{n}{{{p}^{2}}} \right]+\left[ \dfrac{n}{{{p}^{3}}} \right]+...\]
Here $\left[ x \right]$ for any real $x$ returns the greatest integer less than equal to $x$. If $a,b$ are any two integers such that $ a < b $ then we have,
\[\left[ \dfrac{a}{b} \right]=0\]
We are asked to find in the question the largest exponent of 15 in $100!$. We find two relative primes $p,q$ whose product is 15 from its prime factorization
\[15=3\times 5\]
So we have $p=3,q=5$ and also $n=100$. If 3 and 5 will exactly divide 100! , then their product 15 will divide 100!. Let the largest power on $p=3$ be $r$ in 100!. So we have,
\[\begin{align}
  & r=\left[ \dfrac{n}{p} \right]+\left[ \dfrac{n}{{{p}^{2}}} \right]+\left[ \dfrac{n}{{{p}^{3}}} \right]+... \\
 & \Rightarrow r=\left[ \dfrac{100}{3} \right]+\left[ \dfrac{100}{{{3}^{2}}} \right]+\left[ \dfrac{100}{{{3}^{3}}} \right]+\left[ \dfrac{100}{{{3}^{4}}} \right]+\left[ \dfrac{100}{{{3}^{5}}} \right]...\left( \because 100<{{3}^{m}}\forall m\ge 5 \right) \\
 & \Rightarrow r=33+11+3+1+0+0+0... \\
 & \Rightarrow r=48 \\
\end{align}\]
Let the largest power on $q=5$ be $s$ in 100!. So we have,
\[\begin{align}
  & s=\left[ \dfrac{n}{q} \right]+\left[ \dfrac{n}{{{q}^{2}}} \right]+\left[ \dfrac{n}{{{q}^{3}}} \right]+... \\
 & \Rightarrow s=\left[ \dfrac{100}{5} \right]+\left[ \dfrac{100}{{{5}^{2}}} \right]+\left[ \dfrac{100}{{{5}^{3}}} \right]+\left[ \dfrac{100}{{{5}^{4}}} \right]+... \\
 & \Rightarrow s=20+4+0+0+0...\left( \because 100<{{5}^{m}}\forall m>2 \right) \\
 & \Rightarrow s=24 \\
\end{align}\]
So we can write ${{p}^{r}}{{q}^{s}}={{3}^{48}}{{5}^{24}}$ will exactly divide 100! With $r=48,s=24$ as highest power on 3 and 5 respectively. .We can write in symbols as
\[\begin{align}
  & {{3}^{24}}\cdot {{5}^{48}}|100! \\
 & \Rightarrow \left( {{3}^{24}}\cdot {{5}^{24}} \right)\cdot {{5}^{24}}|100! \\
 & \Rightarrow {{\left( 3\cdot 5 \right)}^{24}}\cdot {{5}^{24}}|100! \\
 & \Rightarrow {{15}^{24}}|100! \\
\end{align}\]
So the highest power on 15 in $100!$ is 24 and the correct option is B.

Note: If $pq$ exactly divides the number $n$ then it is not necessary that $p,q$ divides the number $n$. If both $p,q$ divide $n$ then they have to be relative primes like in this problem. If we want to find the result in a short time then we only need to find the power on the larger relative prime between $p,q$ after converting the given number into $p\times q$ such that $p,q$ are relative primes.