Question

the number of zeros at the end of 2020!(factorial) ​

Answer 1

Step-by-step explanation:

Find out the number of zeroes at the end of N

Looking at the expression, we can say that the power of 5 will be the limiting factor.

All we need to do is to figure out the number of 5s in the expression.

11, 22, 33, 1717, 8989,… will not give us any 5s.

55 will give us five 5s.

1010 will give us ten 5s.

1515 will give us fifteen 5s.

And so on.

So, the total number of 5s that I have is

TRAIL-4

Answer 2

The number of zeros at the end of 2020! is 605 zeros.

Step-by-step explanation:

• According to the given information, we need to find the number of zeros that are present at the end of the factorial of 2020.
• Now, we know that, when n is a number and we are familiar with the representation of n in decimal form, the number of zeros at the end of the factorial of n, that is, n!, can be found out by the following formula that is used to find out the factorial of a number n, where n is a positive integer, that is n is negative,

$\frac{n}{5}+ \frac{n}{5^{2} }+ \frac{n}{5^{3} }+\frac{n}{5^{4} }+....+ \frac{n}{5^{n} }$

Here k should be selected in such a way, so that, n < $5^{k+1}$.

Now, here, we have n = 2020.

Now, when we take, k as 4, we will get, $5^{4+1} = 5^{5} = 3125$

Now, 3125 > 2020.

Thus, by the condition, the formula representing the factorial of 2020 will be

$\frac{2020}{5}+ \frac{2020}{5^{2} }+ \frac{2020}{5^{3} }+\frac{2020}{5^{4} }\\=505+80.8+16.16+3.232\\=605.192$

This is approximately equal to 605 zeros.

Thus, the number of zeros at the end of 2020! is 605 zeros.

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