This-question-is-posted-on-the-request-of-mrW1-See-comments-of-my-answer-to-Q-15543-Find-the-last-last-non-zero-digit-of-the-expansion-of-2000-

Question Number 15742 by RasheedSoomro last updated on 14/Jun/17

This question is posted on the request of mrW 1

You can't use 'macro parameter character #' in math mode

Find the last last non - zero digit of the expansion

of 2000!

Commented by mrW1 last updated on 13/Jun/17

Thank you Mr Rasheed for solving

the question!

Answered by RasheedSoomro last updated on 13/Jun/17

^{∙} 2000! has as many ending zeros

as many times 10 is factor of it .

^{∙} 10 is as many times factor of 2000!

as many times 5 is factor of of it .

^{∙} 5 is

(⌊ \frac{2000}{5} ⌋ + ⌊ \frac{2000}{25} ⌋ + ⌊ \frac{2000}{125} ⌋ + ⌊ \frac{2000}{625} ⌋)

= 400 + 80 + 16 + 3 = 499 times

factor of 2000!

That all means 2000! has 499 ending 0^{'} s

and if p = \frac{2000!}{10^{499}} then p has no ending 0 .

And how is this related to last non - zero digit

of 2000! ?

{It}^{'} s as that

Last non - zero digit of 2000! (say d)

= unit digit of p = 2000! / 10^{499}

Hence p \equiv d (\mod 10)

Now,

2000! = (1 \dots 4) (6 \dots 9) \dots (1996 \dots 1999) (5.10 \dots 2000)

= 5^{400} \overset{- 400 brackets -}{(1 \dots 4) (6 \dots 9) \dots (1996 \dots 1999)} . 400!

400! = (1 \dots 4) (6 \dots 9) \dots (396 \dots 399) (5.10 \dots 400)

= 5^{80} \overset{- 80 brackets -}{(1 \dots 4) (6 \dots 9) \dots (396 \dots 399)} . 80!

80! = (1 \dots 4) (6 \dots 9) \dots (76 \dots 79) (5.10 . . 80)

= 5^{16} \overset{16 brackets}{(1 \dots 4) (6 \dots 9) \dots (76 \dots 79)} . 16!

16! = (1 \dots 4) (6 \dots 9) (11 \dots 14) (16) (5.10.15)

= 5^{3} (1 \dots 4) (6 \dots 9) (11 \dots 14) {(16) . 3!}

= 5^{3} \overset{3 brackets}{(1 \dots 4) (6 \dots 9) (11 \dots 14)} (96)

Total 499 brackets (as many as there are zeros)

2000! = 5^{499} {(1 \dots 4) (6 \dots 9) \dots (1996 \dots 1999)}

\times {(1 \dots 4) (6 \dots 9) \dots (396 \dots 399)}

\times {(1 \dots 4) (6 \dots 9) \dots (76 \dots 79)}

\times {(1 \dots 4) (6 \dots 9) (11 \dots 14)} (96)

p = \frac{2000!}{2^{499} . 5^{499}} = (\frac{1}{2^{499}}) {(1 \dots 4) (6 \dots 9) \dots (1996 \dots 1999)}

\times {(1 \dots 4) (6 \dots 9) \dots (396 \dots 399)}

\times {(1 \dots 4) (6 \dots 9) \dots (76 \dots 79)}

\times {(1 \dots 4) (6 \dots 9) (11 \dots 14)} (96)

p = {(\frac{1 \dots 4}{2}) (\frac{6 \dots 9}{2}) \dots (\frac{1996 \dots 1999}{2})}

\times {(\frac{1 \dots 4}{2}) (\frac{6 \dots 9}{2}) \dots (\frac{396 \dots 399}{2})}

\times {(\frac{1 \dots 4}{2}) (\frac{6 \dots 9}{2}) \dots (\frac{76 \dots 79}{2})}

\times {(\frac{1 \dots 4}{2}) (\frac{6 \dots 9}{2}) (\frac{11 \dots 14}{2})} (96)

Now, we know that or can prove that

\frac{(5 k + 1) (5 k + 2) (5 k + 3) (5 k + 4)}{2} \equiv 2 (\mod 10)

Commented by Tinkutara last updated on 13/Jun/17

We have last non - zero digit of n!

= 2^{A} \times A! \times B!

where A = [\frac{n}{5}] and B = remainder

when n is divided by 5 .

Last non - zero digit of 2000!

= 2^{400} \times 400! \times 0! = 2^{400} \times 400!

Last non - zero digit of 400!

= 2^{80} \times 80!

Last non - zero digit of 80!

= 2^{16} \times 16!

Last non - zero digit of 16!

= 2^{3} \times 3! \times 1! = 48 \equiv 8

Last non - zero digit of 2000!

\equiv 6 \times 400! = 6 \times 6 \times 80! \equiv 6 \times 80!

\equiv 6 \times 6 \times 16! \equiv 6 \times 16! \equiv 6 \times 8 \equiv 8

Similarly last non - zero digit of 1000!

= 2^{200} \times 200!

Last non - zero digit of 200!

= 2^{40} \times 40!

Last non - zero digit of 40!

= 2^{8} \times 8!

Last non - zero digit of 8!

= 2^{1} \times 1! \times 3! = 2 \times 6 \equiv 2

Last non - zero digit of 1000!

= 6 \times 200! = 6 \times 6 \times 6 \times 2 \equiv 2

Commented by RasheedSoomro last updated on 13/Jun/17

I {don}^{'} t know that formula!

Commented by RasheedSoomro last updated on 13/Jun/17

Continue from my above answer

Sorry that there was no space and my few lines

{couldn}^{'} t be included in the above answer .

I write them below

So each bracket is congruentto 2 (\mod 10) :

\frac{* . * . * . *}{2} \equiv 2 (\mod 10) [499 congruences) \dots . . (i)

96 \equiv 6 (\mod 10) \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots (ii)

(i) \times (ii) : p \equiv 2^{499} . 6 (\mod 10)

p \equiv 8.6 (\mod 10) [∵ 2^{4 k + 3} \equiv 8]]

p \equiv 8 (\mod 10)

Hence the last non - zero digit of 2000! = 8

Commented by mrW1 last updated on 13/Jun/17

Great job done!

Commented by mrW1 last updated on 13/Jun/17

To tinkutara :

Your solution is fantastic!

How did you get the solution ? Is it a

method developed by yourself ? Can

you supply more details about the

method ?

Commented by Tinkutara last updated on 14/Jun/17

I checked the method on internet . You

can type “ last non - zero digit of a

{factorial}^{″} on Google to get the formula

but I {don}^{'} t know its proof .

Commented by mrW1 last updated on 14/Jun/17

thanks!

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