Expansion-of-1000-has-249-0-s-at-the-end-Find-the-first-non-zero-digit-from-right-1000-d000-00-What-is-the-value-of-d-

Question Number 13724 by prakash jain last updated on 22/May/17

Expansion of 1000! has 249, 0^{'} s at the end

Find the first non - zero digit from

right .

1000! = \dots \dots d 000 \dots 00

What is the value of d ?

Commented by prakash jain last updated on 02/Jun/17

249 5^{'} s come from following

200 multiples of 5

40 multiples of 25 (1 extra)

8 multiples of 125 (1 extra)

1 multiples of 625 (1 extra)

249 = ⌊ \frac{1000}{5} ⌋ + ⌊ \frac{1000}{25} ⌋ + ⌊ \frac{1000}{125} ⌋ + ⌊ \frac{1000}{615} ⌋

The question is asking for nonzero

digits just before the zeros .

Commented by RasheedSoomro last updated on 03/Jun/17

Thanks for clarification . I felt mistake of

my thinking after a while . {That}^{'} s why

I deleted my comment .

Commented by RasheedSoomro last updated on 12/Jun/17

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

Answered by RasheedSoomro last updated on 12/Jun/17

If d is required digit then

\frac{1000!}{10^{249}} \equiv d (\mod 10)

\frac{1000!}{2^{249} \times 5^{249}} \equiv d (\mod 10)

\frac{1000!}{2^{52} \times (5 \times 10 \times 15 \times \dots \times 1000)} \equiv d (\mod 10)

All the 5^{'} s come in 5 \times 10 \times \dots \times 1000 .

Some 2^{'} s are also included in 5 \times \dots \times 1000

One 2 per each multiple of 10^{'} s = 100 2^{'} s

One extra 2 per each multiple of 20^{'} s = 50 2^{'} s

One extra 2 per each multiple of 40^{'} s = 25 2^{'} s

One extra 2 per each multiple of 80^{'} s = 12 2^{'} s

One extra 2 per each multiple of 160^{'} s = 6 2^{'} s

One extra 2 per each multiple of 320^{'} s = 3 2^{'} s

One extra 2 per each multiple of \underline{640^{'} s = 1 2^{'} s}

Total 2^{'} s included in 5 \times 10 \times \dots \times 1000 = 197 2^{'} s

Remaining 2^{'} s = 249 - 197 = 52

\frac{1000!}{2^{52} \times (5 \times 10 \times 15 \times \dots \times 1000)} \times 2^{52} \equiv d \times 2^{52} (\mod 10)

\frac{1000!}{5 \times 10 \times 15 \times \dots \times 1000} \equiv d \times 2^{52} (\mod 10)

\frac{1000!}{5 \times 10 \times 15 \times \dots \times 1000} \equiv 6 (\mod 10)

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

So

d \times 2^{52} \equiv 6 (\mod 10) \dots \dots \dots \dots (i)

We can verify that

2^{4} \equiv 6 (\mod 10)

{(2^{4})}^{3} \equiv 6^{3} \equiv 6 (\mod 10)

2^{12} \times 2 \equiv 6 \times 2 \equiv 2 (\mod 10)

2^{13} \equiv 2 (\mod 10)

{(2^{13})}^{4} \equiv 2^{4} \equiv 6 (\mod 10)

2^{52} \equiv 6 (\mod 10)

1 \times 2^{52} \equiv 1 \times 6 \equiv 6 (\mod 10)] \times 1 \dots \dots (ii)

2 \times 2^{52} \equiv 2 \times 6 \equiv 2 (\mod 10)] \times 2

3 \times 2^{52} \equiv 3 \times 6 \equiv 8 (\mod 10)] \times 3

4 \times 2^{52} \equiv 4 \times 6 \equiv 4 (\mod 10)] \times 4

5 \times 2^{52} \equiv 5 \times 6 \equiv 0 (\mod 10)] \times 5

6 \times 2^{52} \equiv 6 \times 6 \equiv 6 (\mod 10)] \times 6 \dots . . (iii)

7 \times 2^{52} \equiv 7 \times 6 \equiv 2 (\mod 10)] \times 7

8 \times 2^{52} \equiv 8 \times 6 \equiv 8 (\mod 10)] \times 8

9 \times 2^{52} \equiv 9 \times 6 \equiv 4 (\mod 10)] \times 9

Comparing (i) with (ii) & (ii), we see that

d = 1, 6

The answer is 1 or 6

This answer is wrong

For right answer see my answer

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

Commented by RasheedSoomro last updated on 10/Jun/17

Main Steps

\frac{1000!}{10^{249}} \equiv d (\mod 10)

\frac{1000!}{2^{52} \times (5 \times 10 \times 15 \times \dots \times 1000)} \equiv d (\mod 10)

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

d \times 2^{52} \equiv 6 (\mod 10)

1 \times 2^{52} \equiv 1 \times 6 \equiv 6 (\mod 10)] \times 1 \dots . . (ii)

6 \times 2^{52} \equiv 6 \times 6 \equiv 6

d = 1, 6

Commented by mrW1 last updated on 05/Jun/17

t h a n k y o u! i a m v e r y i m p r e s s e d!

i n e e d m o r e t i m e t o u n d e r s t a n d y o u r

w o r k i n g .

p l e a s e c h e c k! t h e l a s t n o n - z e r o d i g i t

s h o u l d b e 2 .

Commented by RasheedSoomro last updated on 05/Jun/17

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

correct it . But the answer {doesn}^{'} t came

2 . Answer after correcting error comes 1

May be there is an other error too .

I^{'} m trying to make my answer error - free .

Commented by RasheedSoomro last updated on 05/Jun/17

I had edited this post and once changes

had been public

But now after few hours, I am seeing an old version of

the post . I mean changes disappeared!!!

! ?!!! ? ? ?!!!

Commented by mrW1 last updated on 05/Jun/17

I s a w y o u r c h a n g e d p o s t i n d e e d . H o w

c o u l d t h e c h a n g e s g o l o s t ?

Commented by RasheedSoomro last updated on 05/Jun/17

Thanks for witness! I have reported

to the developer of the forum also .

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 11/Jun/17

1000! = 2^{α} 3^{β} 5^{γ} \dots \dots . .

n ⩽ x < n + 1 \Rightarrow [x] = n

α = [\frac{1000}{2}] + [\frac{1000}{2^{2}}] + [\frac{1000}{2^{3}}] + [\frac{1000}{2^{4}}] + [\frac{1000}{2^{5}}] +

+ [\frac{1000}{2^{6}}] + [\frac{1000}{2^{7}}] + [\frac{1000}{2^{8}}] + [\frac{1000}{2^{9}}] + [\frac{1000}{2^{10}}] + \dots .

= 500 + 250 + 125 + 62 + 31 + 15 + 7 + 3 + 1 + 0 = 994

γ = [\frac{1000}{5}] + [\frac{1000}{25}] + [\frac{1000}{125}] + [\frac{1000}{625}] + 0 + \dots =

= 200 + 40 + 8 + 1 + 0 = 249

s o : 1000!, i s d i v i d i b l e b y n u m b e r :

{(2 \times 5)}^{249} . i . e : 1000! h a v e 249 {z e r o}^{'} s o n

{i t}^{'} s e n d .