Question

How many zeros will be there at the end of the product \[2{!^{2!}} \times 4{!^{4!}} \times 6{!^{6!}} \times 8{!^{8!}} \times 10{!^{10!}}\] ?
A) $10! + 6!$
B) $2\left( {10!} \right)$
C) $10! + 8! + 6!$
D) $6! + 8! + 2\left( {10!} \right)$

Answer 1

Hint: The multiplication of all positive integers that says “n”, which will be smaller than or equivalent to n is known as the factorial. The factorial of a positive integer is represented by the symbol “n!”.
The formula to find the factorial of a number is,
$n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times \left( {n - 3} \right) \times ...... \times 3 \times 2 \times 1$
The number of trailing zeros in a factorial (n!)= Number of times n! is divisible by 10= Highest power of 10 which divides n! = Highest power of 5 in n!

Step-By-Step answer:
To find the trailing zeros we have to divide by 5 recursively.
Let us consider one small example.
Suppose we have to find the number of trailing zeroes in 123!
We should divide n by 5
$\left[ {\dfrac{{123}}{5}} \right] = 24$
24 are greater than 5. So, we can again divide 24 by 5.
$\left[ {\dfrac{{24}}{5}} \right] = 4$
4 is less than 5. So, we stop here.
Final answer $ = 24 + 4 = 28$
So, in 123! there are 28 trailing zeroes. And we can also count it by; instead of 123 we consider 120 and 120 is 28 times
5.
Similarly, we can solve our given example
\[2{!^{2!}} \times 4{!^{4!}} \times 6{!^{6!}} \times 8{!^{8!}} \times 10{!^{10!}}\]
Here, consider 2! but 2 is less than 5. So, we will stop here.
Now, consider 4! but 4 is also less than 5. So, we will stop here.
Now, 6!
We will divide 6 by 5.
$\left[ {\dfrac{6}{5}} \right] = 1$
1 is less than 5. So we will stop.
Hence, we have 1 zero in 6!
And in 6 there is only one 5.
Any power of 0 is 0.
Now, 8!
$\left[ {\dfrac{8}{5}} \right] = 1$
1 is less than 5. So, we will stop here.
And any power of 0 is always 0.
There is 1 zero in 8!
Now, 10!
We will divide 10 by 5.
$\left[ {\dfrac{{10}}{5}} \right] = 2$
2 is less than 5. So, we will stop here.
Hence here are 2 zeroes.
And the power of 0 is 0.
So, we get 1 zero in 6! + 1 zero in 8! +2 zeroes in 10!
Hence, by the given options we can say that, $6! + 8! + 2\left( {10!} \right)$ is the final answer.

So, option D) is the answer.

Note: There are n! different ways of arranging n distinct objects into a sequence, the permutations of those objects.
Factorials are also used extensively in probability theory and number theory.