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Question: Find the number of zeros at the end of \(1090!\)?...

Find the number of zeros at the end of 1090!1090!?

Explanation

Solution

We first need to explain the process of how zero appears at the end of a number. We also explain the condition of n!=1×2×3×....×nn!=1\times 2\times 3\times ....\times n. Then we use the formula of ∑r=1∞[n!ar]\sum\limits_{r=1}^{\infty }{\left[ \dfrac{n!}{{{a}^{r}}} \right]} to find the number of certain factors in n!n! where aa denotes the particular factor.

Complete step-by-step solution:
Here the term n!n! defines the notion of multiplication of first n natural numbers.
This means n!=1×2×3×....×nn!=1\times 2\times 3\times ....\times n.
We need to find the number of zeros at the end of 1090!1090!. Term 0 at the end of a number indicates that the term is divisible by 10 which is the multiplication of 2 and 5.
Therefore, we have to find the number of 2s and 5s are present in the multiplication of 1090!1090!
The process or formula of finding the number of certain factors in n!n! is ∑r=1∞[n!ar]\sum\limits_{r=1}^{\infty }{\left[ \dfrac{n!}{{{a}^{r}}} \right]} where aa denotes the particular factor. Here we need to consider the condition that we have to find the sum till ar≤n!{{a}^{r}}\le n!. The function defines the box function which gives the closest integer value less than the given number.
So, number of 2s in 1090!1090! is ∑r=1∞[1090!2r]\sum\limits_{r=1}^{\infty }{\left[ \dfrac{1090!}{{{2}^{r}}} \right]}. Similarly, the number of 5s in 1090!1090! is ∑r=1∞[1090!5r]\sum\limits_{r=1}^{\infty }{\left[ \dfrac{1090!}{{{5}^{r}}} \right]}.
We get ∑r=1∞[1090!2r]+∑r=1∞[1090!5r]=270\sum\limits_{r=1}^{\infty }{\left[ \dfrac{1090!}{{{2}^{r}}} \right]}+\sum\limits_{r=1}^{\infty }{\left[ \dfrac{1090!}{{{5}^{r}}} \right]}=270.
Therefore, the number of zeros at the end of 1090!1090! is 270.

Note: We need to remember that the box functions will only give integer value and it has to be less than or equal to the given value. We also need the closest integer. We also need to remember that 10 is created by the same number of 2s and 5s. If the number of 2s and 5s present in n!n! is different from each other, then we have to take the least one and the remaining ones will remain unused.