What is the number of Zeros at the end of the product of the nubers from 0 to 100.?

Question

Steps and Method please..........

Answer 1

easy

I think you mean 1 to 100 as if starting is zero then result is 0

number of 0s at the end k means 10^k devides

so 2^k devides and 5^k devides

number of numbers divisible by 5 is much less
20 numbers 5,10,... ( 100.5) count 20 fives
5 numbers (25,50,) each count 2 fives but once is counted above so 4 more
nu number counts 3 fives

so 5^24 devides the numbers
so zeroes is 24
(if you wanted to count 2 50 minumum as 50 even numbes > 24

Answer 2

1, because the product of the numbers from 0 to 100 is equal to 0.

Did you mean 1 to 100?

There are 50 even numbers between 1 and 100 (2,4,...,100).
There are 25 multiples of 4=2^2 (4,8,...,100).
There are 12 multiples of 8=2^3 (8,16,...,96).
There are 6 multiples of 16=2^4 (16,32,...,96).
There are 3 multiples of 32=2^5 (32,64,96).
And 1 multiple of 64=2^6 (64).

So, in the product P, there is 1 factor of 2^6, 3-1=2 factors of 2^5 (because the one multiple of 2^6 has already been counted), 6-3=3 factors of 2^4, 12-6=6 factors of 2^3, 25-12=13 factors of 2^2 and 50-25=25 factors of 2.

Thus, summarizing all factors of the powers of 2, there is a factor of 2^6 * (2^5)^2 * (2^4)^3 * (2^3)^6 * (2^2)^13 * 2^25 = 2^(6+10+12+18+26+25) = 2^97.

Similarly, there are 20 multiples of 5 (5,10,...,100) in P.
4 multiples of 25=5^2 (25,50,75,100).

Thus, in P, there are 4 factors of 5^2 and and 20-4=16 factors of 5, so P contains a factor of (5^2)^4 * 5^16 = 5^(8+16) = 5^24.

And, since 10=5*2, P contains 24=(minimum of 97 and 24) factors of 10, so there are 24 zeros at the end.

Answer 3

there is only one zero because 0*1*2*3*...*100 =0

Answer 4

1,because 0X1X2......X100=0
there is 1 0 in 0