I assume you are asking about trailing zeros Every multiple of 5 contributes a zero. Every multiple of 25 contributes a second zero Every multiple of 125 contributes a third zero Every multiple of 625 contributes a fourth zero etc. floor(1000000/5) + floor(1000000/25) + ... floor(1000000/5^n) There are: 200,000 multiples of 5 40,000 multiples of 25 8,000 multiples of 125 (5^3) 1,600 multiples of 625 (5^4) 320 multiples of 3,125 (5^5) 64 multiples of 15,625 (5^6) 12 multiples of 78,125 (5^7) 2 multiples of 390,625 (5^8) = 249,998 trailing zeros To help visualize this, let's try a smaller number like 10! This is 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 10 will give you a trailing zero. 5 x 2 will give you a trailing zero. All the rest won't give you any more zeroes. So for 10! the answer is 2 trailing zeroes. Now try 25! 1 x 2 x 3 x 4 x 5 x 6 x ... x 23 x 24 x 25 Again 2n x 5 --> a multiple of 10 --> one zero added n x 10 --> a multiple of 10 --> one zero added 2n x 15 --> a multiple of 10 --> one zero added n x 20 --> a multiple of 10 --> one zero added 4n x 25 --> a multiple of *100* --> two zeroes added The number of multiples of 5 --> 5 The number of multiples of 25 --> 1 Total trailing zeroes in 25! is 6. So following the logic forward to 1 million factorial the answer is 249,998 trailing zeroes.
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Verified answer
10!
= 3628800
100!
= (10^2)!
= 9.3326215 443944 152681 699238 856267*10^157
1000!
= (10^3)!
= 4.02387 260077 093773 543702 433923*10^2567
10000!
= (10^4)!
= 2.8462596 809170 545189 064132 121199*10^35659
100000!
= (10^5)!
= 2.8242294 079603 478742 934215 780245*10^456573
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I assume you are asking about trailing zeros Every multiple of 5 contributes a zero. Every multiple of 25 contributes a second zero Every multiple of 125 contributes a third zero Every multiple of 625 contributes a fourth zero etc. floor(1000000/5) + floor(1000000/25) + ... floor(1000000/5^n) There are: 200,000 multiples of 5 40,000 multiples of 25 8,000 multiples of 125 (5^3) 1,600 multiples of 625 (5^4) 320 multiples of 3,125 (5^5) 64 multiples of 15,625 (5^6) 12 multiples of 78,125 (5^7) 2 multiples of 390,625 (5^8) = 249,998 trailing zeros To help visualize this, let's try a smaller number like 10! This is 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 10 will give you a trailing zero. 5 x 2 will give you a trailing zero. All the rest won't give you any more zeroes. So for 10! the answer is 2 trailing zeroes. Now try 25! 1 x 2 x 3 x 4 x 5 x 6 x ... x 23 x 24 x 25 Again 2n x 5 --> a multiple of 10 --> one zero added n x 10 --> a multiple of 10 --> one zero added 2n x 15 --> a multiple of 10 --> one zero added n x 20 --> a multiple of 10 --> one zero added 4n x 25 --> a multiple of *100* --> two zeroes added The number of multiples of 5 --> 5 The number of multiples of 25 --> 1 Total trailing zeroes in 25! is 6. So following the logic forward to 1 million factorial the answer is 249,998 trailing zeroes.
1000000 Factorial
The Stirling approximation and Gosper formula help provide an estimate:
n! ~ (n/e)ⁿ √((6n + 1)*π/3)
100000/e ~= 36787.944 ~= 10^4.5657055
(100000/e)^100000 ~= (10^4.5657055)^100000
=10^4565705.5
√(600001*π/3) ~= 793 ~= 10^2.899
10^4565705.5 * 10^2.9 ~= 10^4565708.4172
Thus, 100000! is approximately
26133....(another 4,565,700 digits) ....000
invalid
it is very big
fairly speaking.....its no less than infinity....