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# Number Theory

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How many zeros are at the end of (100!)(200!)(300!)(400!) when multiplied out?

May 1, 2022

#1
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(100/5 + 100/25 + 200/5 + 200/25 + 200/125 +300/5 + 300/25 + 300/125 + 400/5 + 400/5 + 400/5)= 387 zeroes.

May 1, 2022
#3
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oops I made a typo on this when calculating, ignore.

HoldMyBeer  May 1, 2022
#2
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Floor{[100/5]+[100/25]+[200/5]+[200/25]+[200/125]+[300/5]+[300/25]+[300/125]+[400/5]+[400/25]+[400/125]}==246 zeros:

144000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000==246 zeros.

May 1, 2022
#4
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We calculate the trailing zeroes of each of 100!, 200!, 300!, and 400!.

Note that we have this nice formula for calculating the trailing zeroes of n! : $$\#\text{ trailing zeros} = \displaystyle\sum_{k = 1}^\infty \left\lfloor\dfrac{n}{5^k}\right\rfloor$$.

Then $$100!$$ has $$\left\lfloor\dfrac{100}5\right\rfloor + \left\lfloor\dfrac{100}{5^2}\right\rfloor = 24$$ trailing zeros.

Similarly, 200! has 49 trailing zeros, 300! has 74 trailing zeros, and 400! has 99 trailing zeros.

Then, (100!)(200!)(300!)(400!) has 24 + 49 + 74 + 99 = 246 trailing zeros.

May 2, 2022