Trailing zeros

Number Theory Level 2

( 6 ⋅ 1 ) ! = 72 0 ( 6 ⋅ 2 ) ! = 4790016 00 ( 6 ⋅ 3 ) ! = 6402373705728 000 \begin{aligned} & \,(6\cdot{\color{#3D99F6}1})! = 72{\color{#3D99F6}0} \\& \,(6\cdot{\color{#D61F06}2})! = 4790016{\color{#D61F06}00} \\& \,(6\cdot{\color{#20A900}3})! =6402373705728{\color{#20A900}000} \end{aligned} From the above, it can be observed that ( 6 ⋅ n ) ! \,(6\cdot n)! has exactly n n trailing zeros. Is it true that ( 6 ⋅ 2018 ) ! \,(6\cdot {\color{#E81990}2018})! will have exactly 2018 {\color{#E81990}2018} trailing zeros ?


Notation: ! ! denotes the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .

Yes No

Naren Bhandari by Naren Bhandari , 21, India

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4 solutions

This rule will get broken at (6*5)! as zeroes will then be formed by - 10,20,30 each will directly contribute one zero and 5,15 each will contribute one zero as 5 present in their factors will give a zero after multiplying with 2 and 25 will contribute 2 zeroes as it contains two fives as factors. so 3+2+2=7. And hence it goes on increasing as more and more numbers will further contain more fives as their factors hence a single number will be contributing more zeroes therefore upto 2018 it will be greater than 2018.

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Chew-Seong Cheong
Apr 28, 2018

The number of trailing zeros of n ! n! is given by z ( n ) = ∑ k = 1 ∞ ⌊ n 5 k ⌋ \displaystyle z(n) = \sum_{k=1}^\infty \left \lfloor \frac n{5^k} \right \rfloor . For n = 6 ⋅ 2018 n=6\cdot 2018 , z ( 6 ⋅ 2018 ) > ⌊ 6 ⋅ 2018 5 ⌋ = 2421 ≠ 2018 z(6\cdot 2018) > \left \lfloor \dfrac {6\cdot 2018}5 \right \rfloor = 2421 \ne 2018 .

No , ( 6 ⋅ 2018 ) ! (6\cdot 2018)! does not have the exactly 2018 trailing zeros.

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Md Zuhair
Apr 28, 2018

As my code worked, It may have 3023 zeroes. So that general case is false.

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Rishabh Bhardwaj
Apr 30, 2018

Simple, we need to calculate number of 5 and 2 pairs in the factorial (essentially number of 5 would do!!!). you will have "at least" a pair with an addition of 6 each time inside the factorial. As you move to fact(6x4), such pairs are 4, while for fact(6x5), it's 6 (6-5's and more than 6 two's) . The pattern doesn't continue even for 5-6s. BTW, you can determine the exact number of zeroes in the factorial.

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