How many zeros are there at the end of 80!?

Question

I can't use a calculator or any device to just get a numerican answer. I thought of doing the prime factorization and then multiplying, but that process seems so time consuming and hence wrong.

Answer 1

80 has 1 zero at the right side. It can be written as 80.0 if we are sure of the measurement to 3 significant places

Answer 2

For any integer in general, the number of zeros at the end is the number of times it is divisible by ten. Since 10 = 2*5, that would be the number of times it's divisible by 2 or the number of times it's divisible by 5, whichever is less.

80! is the product of all the numbers from 1 to 80. The number of times it's divisible by 5 is 80/5 = 16 (because of the factors 5, 10, 15, 20, ... 80), plus 80/25 = 3 (rounded down from 3.2 - because of the factors 25, 50, 75, each of which has an extra factor of 5 in addition to the one we already counted). That's a total of 19 factors of 5. There are obviously more than 40 factors of 2, for all the even numbers up to 80.

So 5 is a factor of 80! nineteen times, and 2 is a factor more than nineteen times; therefore 10 is a factor nineteen times as well. But 10 is not a factor of 80! twenty times, because if it were, 5 would also be a factor twenty times. So the number ends with 19 zeros.

Answer 3

5, 2x5, 15, 2x2x5 , 5x5, 2x3x5, 35, 2x2x2x5, 45, 2x5x5, 55, 2x2x3x5, 65, 2x5x7, 3x5x5, 2x2x2x2x5 so we have from this 19 5's and 15 2's
There are 40 - 8 = 32 even numbers, not including the number of 10's. The 2's overwhelm the 5's so we don't need to count them any further. The number of zero's are then the number of 5's so there are 19 zero's in 80!

Answer 4

there are 19 zeroes in the end of 80!, just like the poster above said.

The reason for that is this: to have a zero, you have to multiply 2 by 5.

So the question becomes: how many times is 80! divisible by 2, and how many times is it divisible by 5?

it's clear that it's divisible by 2 many more times than by 5, so 5 is the limiting factor.

Now let's count how many times it's divisible by 5. I'm listing all factors fo 5 going into 80!, alongside with "how many 5's they introduce?"

5 - 1
10 - 1
15 - 1
20 - 1
25 - 2
30 - 1
35 - 1
40 - 1
45 - 1
50 - 2
55 - 1
60 - 1
65 - 1
70 - 1
75 - 2
80 - 1

add these up, and you get 19

Answer 5

There's 19 zeroes at the end of 80! This can be figured out from counting all the multiples of 5 from 1 to 80, including doubles at 25, 50, 75.

Answer 6

Just to be sure, I calculated the LSD's of 80!, discarding illrevelent leading digits as I went, and collected the powers of ten every so often as well. I got 19 zeroes at the end, and the first non-zero digit from the right was 4.

Answer 7

If we expand 80!, we get that the number of 5 factors are
80/5 + 80/25 = 19 rounded down to integer.

Then, the number of zeroes (which is obtained by multiplying each of these 5s with a 2) is 19.

Answer 8

19

Answer 9

I'm assuming you are talking about 80 factorial or 80! There would be a lot. 80! can be written in scientific notation as 7.16 x 10^118 (That's 118 zeros at the end!!!)

Answer 10

I only saw one zero to the right of 80. If you have blurred vision you will come up with more eights and zero's or if vision is worse NO eight and NO zero.