Question

The largest non-negative integer $k$ such that ${56^k}$ divides $25!$ is:
A. $2$
B. $3$
C. $4$
D. $5$

Answer 1

Hint:We know the formula that the largest power of the prime $p$ which divides $n!$ is given by:
$\left[ {\dfrac{n}{p}} \right] + \left[ {\dfrac{n}{{{p^2}}}} \right] + \left[ {\dfrac{n}{{{p^3}}}} \right] + ............$ where $\left[ x \right]$is the greatest integer function of $x$.Using this we try to solve the question.

Complete step-by-step answer:
We know that any problem related to the divisibility of a factor is solved by the greatest integer function.
Now you must be wondering what is the greatest integer factor?
So the greatest integer factor or GIF is represented by the above bracket and when any number lies between $n,n + 1$, then GIF of that number will be integer just smaller than that number.
For example: $3.6{\text{ in [3}}{\text{.6]}} = 3$
${\text{GIF of [ - 1}}{\text{.25] = - 2}}$
GIF for any integer will be the same as that integer
$[3] = 3,[4] = 4$
Now we are given to find the largest non-negative integer $k$ such that ${56^k}$divides $25!$.
So as we know that the formula for the largest power of the prime $p$ which divides $n!$ is given by:
$\left[ {\dfrac{n}{p}} \right] + \left[ {\dfrac{n}{{{p^2}}}} \right] + \left[ {\dfrac{n}{{{p^3}}}} \right] + ............$ where $\left[ x \right]$is the greatest integer function of $x$ and $p$ is the prime.
So here we are asked to find $k$ such that ${56^k}$divides $25!$.
$n = 25$
$56 = 2 \times 2 \times 2 \times 7$
Therefore the largest power of $2$ divides $25!$ is
$\left[ {\dfrac{{25}}{2}} \right] + \left[ {\dfrac{{25}}{{{2^2}}}} \right] + \left[ {\dfrac{{25}}{{{2^3}}}} \right] + \left[ {\dfrac{{25}}{{{2^4}}}} \right] + \left[ {\dfrac{{25}}{{{2^5}}}} \right] + ............$
$\left[ {12.5} \right] + \left[ {6.25} \right] + \left[ {3.125} \right] + \left[ {1.56} \right] + \left[ {0.76} \right]$
$12 + 6 + 3 + 1 + 0 = 22$
But we know that $56 = {2^3}(7)$
$56 = {2^k}(7)$
We got for $2$, the maximum power=$22$
Now for ${2^3}$, it would be $\left[ {\dfrac{{22}}{3}} \right] = 7$
Hence the largest power of $8$ which divides $25!$$ = 7$
Now the largest power of $7$ which divides $25!$ is
$\left[ {\dfrac{{25}}{7}} \right] + \left[ {\dfrac{{25}}{{{7^2}}}} \right] + \left[ {\dfrac{{25}}{{{7^3}}}} \right]$
$3 + 0 + 0 = 3$
As we know that the largest power of $8$ which divides $25!$$ = 7$
And the largest power of $7$ which divides $25!$ is $3$
So, combining the largest power of $(8 \times 7) = 56$ which divides $25!$ would be $3$
So $k = 3$

So, the correct answer is “Option B”.

Note:We can write:
$25! = (25)(24)(23)(22),,,,,,,,,,,,,,(1)$
$25! = (7 \times 8) \times (14 \times 4) \times 21 \times 6 \times 2......$
$25! = (7 \times 8) \times (14 \times 4) \times (7 \times 2 \times 2 \times 2) \times ......$
$25! = (56) \times (56) \times (56) \times p$ where other products is taken as $p$
So $25! = {56^3}(p)$
Hence $k = 3$