Question

What is the largest power of $12$ that would divide $49!$ ?

Answer 1

To answer this type of question first we need to know the highest power of $3$ in $49!$ and highest power of $4$ in $49!$. Then, we will see the common power between $3$ and $4$ that will be the highest power of $12$ in $49!$.

Complete step by step solution:
So, to find the highest power of $12$ in $49!$, we need to check the highest powers of $4$ and $3$ in it.
We know that $n! = n\left( {n - 1} \right)\left( {n - 2} \right).....\left( 2 \right)\left( 1 \right)$.
Therefore, $49! = 49 \times 48 \times 47 \times ....3 \times 2 \times 1$
We know that all the multiple of $2$ would be one power of $2$.
So, number of multiples of 2$$$ = \left[ {\dfrac{{49}}{2}} \right] = 24$ But, multiples of 4$would yield$2$multiples of$2$. So, number of multiples of$4 = \left[ {\dfrac{{49}}{4}} \right] = 12$ Similarly, number of multiples of $$8 = \left[ {\dfrac{{49}}{8}} \right] = 6 Number of multiples of $$16$$$ = \left[ {\dfrac{{49}}{{16}}} \right] = 3
Number of multiples of 16$$$ = \left[ {\dfrac{{49}}{{32}}} \right] = 1$ So, highest power of 2$in$49!$ = 24 + 12 + 6 + 3 + 1 = 46$ So, number of multiples of3 = \left[ {\dfrac{{49}}{3}} \right] = 16$ Similarly, number of multiples of $$9 = \left[ {\dfrac{{49}}{9}} \right] = 5 Number of multiples of $$27$$$ = \left[ {\dfrac{{49}}{{27}}} \right] = 1
So, highest power of $3$ in $49!$ $= 16 + 5 + 1 = 22$
Hence, the highest power of $$12$$$= 22$.

So, the largest power of $12$ that divides $49!$ is $22$.

Note: The factorial, symbolized by an exclamation mark (!), is a quantity defined for all integers greater than or equal to $0$. For an integer n greater than or equal to $1$, the factorial is the product of all integers less than or equal to n but greater than or equal to $1$. The factorial value of $0$ is defined as equal to $1$. The factorial values for negative integers are not defined. We will apply the same approach for all such types of questions.