Question

What is the highest power of $12$ that divides $54!$

• A. $24$
• B. $25$
• C. $34$
• D. $27$

Answer 1

The correct option is B

$25$

$2^2\times 3=12.$, so we need to count the highest power of $2$ and the highest power of $3$ that will divide$54!$, and then we can use this to find the highest power of $12$.
The method to find the highest powers of $2$ and $3$ are similar to the one outlined in the previous question.
The highest power of $2$ that divides $54!=\left[\frac{54}{2}\right]+\left[\frac{54}{4}\right]+\left[\frac{54}{8}\right]+\left[\frac{54}{16}\right]+\left[\frac{54}{32}\right]=27+13+6+3+1=50$

The highest power of $3$ that divides $54!=\left[\frac{54}{3}\right]+\left[\frac{54}{9}\right]+\left[\frac{54}{27}\right]=18+6+2=26$
Or

$54!$ is a multiple of $2^{50}\times 3^{26}$. Importantly, these are the highest powers of $2$ and $3$ that divide$54!$.
$2^2\times 3=12.$ We need to see what is the highest power of $22\times 3$ that we can accommodate within$54!$
In other words, what is the highest n such that ${(2^2\times 3)}^n$ can be accommodated within $2^{50}\times 3^{26}$.
Let us try some numbers, say, $10,20,30$
${(2^2\times 3)}^{10}=2^{20}\times 3^{10}$, this is within $2^{50}\times 3^{26}$
${(2^2\times 3)}^{20}=2^{40}\times 3^{20}$, this is within $2^{50}\times 3^{26}$
${(2^2\times 3)}^{30}=2^{60}\times 3^{20}$, this is not within $2^{50}\times 3^{26}$
The highest number possible for n is $25$.
${(2^2\times 3)}^{25}=2^{50}\times 3^{25}$, this is with in $2^{50}\times 3^{26}$, but ${(2^2\times 3)}^{26}=2^{52}\times 3^{26}$, this is not within $2^{50}\times 3^{26}$.
So, $54!$ can be said to be a multiple of ${(2^2\times 3)}^{25}$.

Hence, the highest power of $12$ that divides$54!$ is $“25”$.