what is the first interger to the left of the string of zeroes in 1000! ?

Answer 1

AARGHHH!! I withdraw my earlier attempt. It must be dementia, I'm afraid. It's still asking about 1000! (1000 factorial), of course, but a careful accounting is needed, not only of the effect of all numbers ending in:

A. 1, 2, 3, ... 9 (I did this by a variety of ways, always coming up with a final non-zero digit of 6 for the product of all such sets),

but also of the following two other types of number, which I regrettably neglected the first time around:

B. 10, 20, 30 ...90 (as often as sets of numbers like that occur), and finally, of types:

C. 100, 200, 300, ... 900 (this set occurs but once). The final, unique 4-digit number (1000) only adds three zeros at the end and doesn't affect the final non-zero digit.

O.K., here goes:

Numbers of Type A.

1 x 2 x 3 x ... x 9 = 362880; final non-zero digit is 8.

There are 100 such "basic sets" of type A altogether.
Multiplying them together, set after set, we get the
following repeated cycle of non-zero ending digits
(top row is # of basic sets used so far; bottom row is the corresponding final digit.)

1 2 3 4 5 6 7 8 9 ...
8 4 2 6 8 4 2 6 8 ...

Notice that after every four basic sets have been used, we end up with 6 as the final non-zero digit. Since there are exactly 100 such sets (a multiple of 4), we end up with that same last non-zero digit for the product of ALL numbers of type A. (But we're not done yet.)

Numbers of Type B.

There are exactly 10 sets of numbers of Type B, i.e. ending with exactly one (but not two or three) zeros.

Multiplying just these sets together, we'll have the following analogous results:

1 2 3 4 5 6 7 8 9 10
8 4 2 6 8 4 2 6 8 4

So, the last non-zero digit in these "10's products" is 4 for this lot.

Numbers of Type C.

We have just one set of "100's". For this, the last non-zero digit is 8 (it's just our "basic set" with a large number of zeros following it); only the last non-zero digit is of interest.

So, to summarize, the different sets we've divided our numbers into, when multiplied together, give us final non-zero digits as follows:

Type A: 6
Type B: 4
Type C: 8

Finally, then, the last non-zero digit of the product of everything is the last digit of:

6 x 4 x 8, or 2. (PHEW!!)

Well, it's now the value of 2 (as reported by zanti3 from the 100 factorial website at:

http://www.justinwhite.com/big-calc/1000.html)

(Finish off this URL with 1000.html, instead of 1000... The Yahoo site would not reproduce the full URL from my draft. Alternatively, just go to Yahoo or Google with 1000 factorial 1000 - this worked for me.)

Boy; what a problem! Evidently the jury is still out on the question of my impending dementia. (See zanti3's kind words, below about my earlier fiasco.)

Thanks again to zanti3 for bringing the existence of a (1000 factorial) web site to my attention.

Live long and prosper, everyone.

Answer 2

Hmm... a Web site that actually has 1000! calculated out gives "2" as the first non-zero digit. Don't worry though, Doc, it's a tough problem - I don't think the wrong answer is due to dementia. :)

* * * * * *

Doc, you deserve the points just for the effort. :) I'll certainly vote for you if this goes to the voters. As for the problem, it's a bear - I'm still trying to come up with a nice easy way to get the answer.

------

Wait, I think I have it... the easy solution:

As you count from 1 to 999, the last NON-ZERO digit has to occur the same number of times for each digit 1, 2, 3, up to 9. Since you have 9 possible non-zero digits and 999 numbers, that means the last non-zero digit is 1 (one) 111 times, 2 (two) 111 times, etc.

The last non-zero digit of 9! is 8. Now we just have to repeat this 111 times, i.e. find the last digit of 8^111. That's easy enough, since the last digit of 8^n goes through the cycle 8, 4, 2, 6. Since the remainder of 111/4 is 3, the last digit is 8^111 is the same as 8^3.

So there is your answer: 2

Answer 3

1

Answer 4

pay attention of addictive habit in your self. Your personality form is amazingly in many situations seen contained in the rooms of AA and NA conferences . you're being bombed with painful self-knowledge and that is effortless to locate convenience in drugs and/or booze and/or convenience nutrition. i do not have a short answer for your difficulty. one element you may want to do is to take an honest seem at your habit and why you've consistently tried to thrill human beings. In very nearly each case that is a egocentric motivation. you opt for human beings to love you. So, in a fashion, you've not quite been giving of your self to others except at the same time as it delivers what you opt for in go back. this can be a tricky element to manage. i'm dealing with this procedure myself as i look at how i have acted in the direction of others my entire existence. with the aid of how, the international needs reliable docs. no longer crappy ones like you stated, yet reliable ones. are you able to concentration on being a reliable regular practitioner? are you able to positioned aside your self-detrimental urges and make the finest of the path you're on? If no longer, then do you imagine you may come to a decision to pursuing your dream of being a pilot? nicely, this isn't powerful... I choose I knew a magic way for human beings like you and me to stop being human beings-pleasers.