My calculation gave me 2963
hopefully that is right...
* * * * *
I suggest 501.
122 zeros.
To calculate the number of zeros in a factorial number, we need to determine the number of factors of 5 in the factorial. In this case, we are looking at 10 to the power of 10 factorial. The number of factors of 5 in 10! is 2 (from 5 and 10). Therefore, the number of zeros in 10 to the power of 10 factorial would be 2.
Two!
20! is 2,432,902,008,176,640,000, so there are four consecutive zeroes at the end of 20!
if you counted one digit for every particle in the universe it would take over a googol universes.
18 factorial is equal to 6402373705728000 - with three consecutive zeroes at the end.
There are 18 zeros.
122 zeros.
242 zeros.
To calculate the number of zeros in a factorial number, we need to determine the number of factors of 5 in the factorial. In this case, we are looking at 10 to the power of 10 factorial. The number of factors of 5 in 10! is 2 (from 5 and 10). Therefore, the number of zeros in 10 to the power of 10 factorial would be 2.
Two!
20! is 2,432,902,008,176,640,000, so there are four consecutive zeroes at the end of 20!
if you counted one digit for every particle in the universe it would take over a googol universes.
To determine the number of trailing zeros in 85 factorial (85!), you count how many times 5 is a factor in the numbers from 1 to 85, as there are always more factors of 2 than 5. This is calculated using the formula: [ \text{Number of trailing zeros} = \left\lfloor \frac{85}{5} \right\rfloor + \left\lfloor \frac{85}{25} \right\rfloor = 17 + 3 = 20. ] Thus, 85! has 20 trailing zeros.
The number 12345678910 contains no zeros. It is a sequence of consecutive digits from 1 to 10, with each digit appearing once and in order. Therefore, the total count of zeros in this number is zero.
Two: addition and factorial - the latter being repeat multiplication.
To find the number of trailing zeros in 200 factorial (200!), you can use the formula that counts the factors of 5 in the numbers from 1 to 200, as there are always more factors of 2 than 5. The number of trailing zeros is given by the sum of the integer divisions: ( \lfloor \frac{200}{5} \rfloor + \lfloor \frac{200}{25} \rfloor + \lfloor \frac{200}{125} \rfloor ). This calculates to ( 40 + 8 + 1 = 49 ). Therefore, 200! has 49 trailing zeros.