#1**+1 **

100! ends in a lot of zeroes, way more than 17, so the answer is just simply \(\boxed{0}\).

For a more detailed explanation, the number of zeroes that 100! will end in is equal to the number of 5's the prime factorization of 100! is, because 5*2=10, and there are obviously a lot more 2's than 5's in the prime factorial of 100!.

There are \(\left \lfloor{\frac{100}{5}}\right \rfloor+\left \lfloor{\frac{100}{25}}\right \rfloor = 20+4 = 24\) zeroes at the end of 100! which means that the last 17 digits will all be zero, which will mean that the sum will be 0.

textot Feb 2, 2021

#1**+1 **

Best Answer

100! ends in a lot of zeroes, way more than 17, so the answer is just simply \(\boxed{0}\).

For a more detailed explanation, the number of zeroes that 100! will end in is equal to the number of 5's the prime factorization of 100! is, because 5*2=10, and there are obviously a lot more 2's than 5's in the prime factorial of 100!.

There are \(\left \lfloor{\frac{100}{5}}\right \rfloor+\left \lfloor{\frac{100}{25}}\right \rfloor = 20+4 = 24\) zeroes at the end of 100! which means that the last 17 digits will all be zero, which will mean that the sum will be 0.

textot Feb 2, 2021