How many consecutive zeros are there at the end of the expansion of 1,000 factorial?

Finding the number of trailing zeroes in the factorial of a number has a sequence of 3-4 steps which applies for all number. I will list it down for reference

step 1: Divide the given number by 5, the answer will give you the number of multiples. If it gives you a decimal answer, you can just take the whole number

Step 2: Divide the given number with the next power of 5, that is 25( 5 times 5) , find the number of multiples

Step 3: Divide the given number with the next power of 5, that is 125 (5 times 5 times 5) , find the number of multiples

Step 4:Keep doing this with increasing powers of 5 till the given number becomes lesser than the power of 5

Step 5: Now add all the number of multiples you got at each step

This will give you the number of trailing zeroes for the given number

Let's do this process for the given number 1000

Divide by 5

1000÷5 = 200 multiples

1000 ÷25 = 40 multiples (5² = 25)

1000÷125 = 8 multiples (5³=125)

1000÷ 625 = 1.69 we take only the whole number so 1 multiple

now the next power is 3125 which cannot divide 1000 so we can stop this process

now add all the multiples

200+40+8+1 = 249

So the factorial of 1000 has 249 trailing zeroes

46.2+112.3

step-by-step explanation:

step-by-step explanation:

the true statements are:

r = set of real numbers

n = set of natural numbers, 1,2,

w = n u{0} = {0,1,}