Question

The largest non-negative number $k$ such that ${{24}^{k}}$ divides $12!$ is \[\]
A. 2\[\]
B. 3\[\]
C. 4\[\]
D. 5\[\]

Answer 1

Hint: We design the equation ${{24}^{k}}={{2}^{m}}{{3}^{l}}\lambda $ where $\lambda $ is any positive integer, $m$ is highest power on 2 and $l$ is the highest power on 3. We find $m,l$ using the highest power formula for prime that divides $n!$ with highest power $r$is given by $r=\left[ \dfrac{n}{p} \right]+\left[ \dfrac{n}{{{p}^{2}}} \right]+\left[ \dfrac{n}{{{p}^{3}}} \right]+...$ where $\left[ x \right]$ is greatest integer function. We put $m,l$ in the equation and use the divisibility rule of 24 and find the power on ${{2}^{3}}\times 3=24$. We compare power both side and find $k$.\[\]

Complete step by step answer:
We know that if there are two numbers $p$ and $q$ which are relative prime (it means greatest common factor (GCD) between $p$ and $q$ is 1) and they exactly divide any number $n$ if and only if their product $pq$ also exactly divides the number $n$. It is one of the divisibility rules. \[\]
We know that the largest power on any prime $p$ is $r$ such that ${{p}^{r}}$ divides $n!$ then we have,
\[r=\left[ \dfrac{n}{p} \right]+\left[ \dfrac{n}{{{p}^{2}}} \right]+\left[ \dfrac{n}{{{p}^{3}}} \right]+...\]
Here $\left[ x \right]$ is the greatest integer function which returns greatest integer less than equal to $x$. We are given the question to find the largest non-negative number $k$ such that ${{24}^{k}}$ divides $12!$. Let us consider the prime factorization of 24. We have
\[24=2\times 2\times 2\times 3={{2}^{3}}\times 3=8\times 3\]
If a number is divisible by the relative primes 8 and 3 then the number is also divisible by 24. Let us consider some natural numbers $\lambda $
\[{{24}^{k}}={{2}^{m}}{{3}^{l}}\lambda ......\left( 1 \right)\]
Here $m$ and $l$ are the largest power on 2 and 3 respectively. If ${{24}^{k}}$ divides $12!$ then ${{2}^{m}}$ and ${{3}^{l}}$ also divide$12!$ .We use largest power formula for $n=12$ and $p=2$ to get $m$ as,
\[\begin{align}
  & m=\left[ \dfrac{12}{2} \right]+\left[ \dfrac{12}{{{2}^{2}}} \right]+\left[ \dfrac{12}{{{2}^{3}}} \right]+\left[ \dfrac{12}{{{2}^{4}}} \right]+... \\
 & \Rightarrow m=\left[ 6 \right]+\left[ 3 \right]+\left[ 1.5 \right]+\left[ 0.75 \right]+... \\
 & \Rightarrow m=6+3+1+0+0... \\
 & \Rightarrow m=10 \\
\end{align}\]
We use largest power formula for $n=12$ and $p=3$ to get $l$ as,
\[\begin{align}
  & l=\left[ \dfrac{12}{3} \right]+\left[ \dfrac{12}{{{3}^{2}}} \right]+\left[ \dfrac{12}{{{3}^{3}}} \right]+... \\
 & \Rightarrow l=\left[ 4 \right]+\left[ 1.\overline{3} \right]+\left[ 0.\overline{4} \right]+... \\
 & \Rightarrow l=4+1+0... \\
 & \Rightarrow l=5 \\
\end{align}\]
We put the value of $m,l$ in equation (1) to have,
\[\begin{align}
  & {{24}^{k}}={{2}^{10}}{{3}^{5}}\lambda \\
 & \Rightarrow {{24}^{k}}=\left( 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3\times 3\times 3 \right)\lambda \\
\end{align}\]
We consider the divisibility rule of 24 and arrange the product $24=8\times 3={{2}^{3}}\times 3$will be close to each other. We have,
\[\begin{align}
  & \Rightarrow {{24}^{k}}={{2}^{3}}\times 3\times {{2}^{3}}\times 3\times {{2}^{3}}\times 3\times 2\times {{3}^{2}}\times \lambda \\
 & \Rightarrow {{24}^{k}}={{\left( {{2}^{3}}\times 3 \right)}^{3}}\times 18\lambda \\
 & \Rightarrow {{24}^{k}}={{24}^{3}}\times 18\lambda \\
\end{align}\]
We compare power on 24 on both sides of the equation and find the largest power on $k$ such that ${{24}^{k}}$ divides $12!$ as 3.

So, the correct answer is “Option B”.

Note: We must be careful of the confusion between greatest and smallest integer function. The smallest integer function $\text{SIF}\left( x \right)$ returns smallest integers greater than $x$. The largest power formula is also called Legendre’s formula which can be reformulated to find $r=\dfrac{n-{{s}_{p}}\left( n \right)}{p-1}$ where ${{s}_{p}}$ is the sum of digits base-$p$ expansion of $n$. We can alternatively solve with prime factorization of $12!$.