Question

What is the largest power of \[12\] that would divide \[49!\] ?

Answer 1

Hint: 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 of \[3\]$ = \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.