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# factorial

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Find the sum of the last 17 digits of 100!.

Feb 2, 2021

### Best Answer

#1
+319
+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.

Feb 2, 2021

### 1+0 Answers

#1
+319
+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