why is there 24 zeros at the end of 100!.......and 200!?

Question

Thanks a lot

Answer 1

02, 05, 10 = 2 zeros
12, 15, 20 = 2 zeros
22, 25, 30 = 2 zeros
32, 35, 40 = 2 zeros
42, 45, 50 = 2 zeros
52, 55, 60 = 2 zeros
62, 65, 70 = 2 zeros
72, 75, 80 = 3 zeros
82, 85, 90 = 2 zeros
92, 95, 100 = 3 zeros

100! gives you 24 zeros

102, 105, 110 = 2 zeros
112, 115, 120 = 2 zeros
122, 125, 130 = 2 zeros
132, 135, 140 = 2 zeros
142, 145, 150 = 2 zeros
152, 155, 160 = 2 zeros
162, 165, 170 = 2 zeros
172, 175, 180 = 3 zeros
182, 185, 190 = 2 zeros
192, 195, 200 = 3 zeros

Gives you a total of 24 zeros

100! = 24 zeros
200! = 48 zeros

mainly because 72 = 2 * 2 * 18 and 75 = 5 * 5 * 3 which gives you 2 zeros alone.

Answer 2

Think about 100 factorial... it is 1 x 2 x 3 x ... x 99 x 100.

Now certainly 100 will contribute 2 zeroes. And 2 and 5 will make 10, for another zero. and 4 and 15 makes 60 for another zero... basically if you factored all the digits from 1 to 100 into their prime factors, you'd have enough 2s and 5s to make lots of 10s, and each time you multiply by 10 you add a zero at the end.

Counting up 2s, you have:
2 = 2
4 = 2 x 2
6 = 2 x 3
8 = 2 x 2 x 2
etc.

More than enough 2s to cover the number of 5s found.

For 5s, you have:
5 = 5
10 = 2 x 5
15 = 3 x 5
20 = 4 x 5
25 = 5 x 5
...
50 = 2 x 5 x 5
...
75 = 3 x 5 x 5
...
100 = 4 x 5 x 5

So I see 20 fives for all the powers of 5 and then 4 more for all the powers of 25. That's 24 zeroes for 100!

And 200! will have those plus more...
5 --> 105, 110, 115, 120, 125, ..., 195, 200
5 x 5 --> 125, 150, 175, 200
5 x 5 x 5 --> 125

So 200! should have 49 zeroes at the end.

In general, if you had a number n! you would do the following.
Count the multiples of 5
Count the multiples of 5^2 (25)
Count the multiples of 5^3 (125)
Count the multiples of 5^4 (625)
Count the multiples of 5^5 (3125)
etc.


For example:
3125! would have:
625 multiples of 5
125 multiples of 25
25 multiples of 125
5 multiples of 625
1 multiple of 3125

So 3125! would have 781 zeroes at the end!

Answer 3

The multiples of 5 basically determine how many zeroes there are after n! For 100!, there are 20 integers divisible by 5 (5, 10, 15, etc.). In addition, 4 integers are divisible by 25, for an extra multiple of 5. So, the total number of zeroes in 100! is 20 + 4 = 24.

Now, for 200!, you have 40 multiples of 5, 8 multiples of 25, and 1 multpile of 125, giving 40 + 8 + 1 = 49 zeroes.

Answer 4

For 100! it happens that there are 20 multiples of 5, plus
25 = 5 x 5 one more, 21 (the other was already took into account)
50 = 5 x 5 x 2 one more, 22
75 = 5 x 5 x 3 = one more, 23
100 = 5 x 5 x 4 = one more 24.
The other factor to produce the 0 is the number 2, and there is plenty of multiples of 2 in the firs 100 numbers.

The same reasoning applies to 200!.

I apologize for my bad english.

Answer 5

In order for a number to have a zero at the end it must be divisible by 10 (or 5 and 2).

Find out how many of the numbers in 100! have 2 and 5 as factors, arrange them and do some counting.

If a number ends in 00 then it has 100 as a factor
If a number ends in 000 ... 1000 ...
etc

Answer 6

100 should have 21 zeros. nine from 10, 20, ..., 90, two from 100, and ten more from numbers ending in 5 being multiplied by even number.

Answer 7

(100 x 99 x 98 x 97 x.... ....x 3 x 2 x 1)

(200 x199 x 198 x 197 x.... ....x 3 x 2 x 1)

read this: http://puzzles.nigelcoldwell.co.uk/nineteen.htm

Answer 8

There would only be that many zeros if the number was huge.

Answer 9

0,10,20,30,40,50,60,70,80,90,100,110,120,130,140,150,160,170,180,190,200