Question 1182542: Find the last non-zero digit of 50!
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52786)   (Show Source):
You can put this solution on YOUR website! .
In web-page
https://www.quora.com/What-is-the-first-non-zero-digit-in-50-factorial-50-How-did-you-get-the-answer-without-a-calculator#:~:text=The%20answer%20is%202%2C%20see,compute%20the%20factorization%20of%2050!
you can find different solutions for any taste.
The ANSWER is 2.
Even better form solution, with examples, including the case 50!, is under this link
https://www.justquant.com/numbertheory/how-to-find-last-non-zero-digit-in-a-factorial/
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Also, a good idea is to present 50! as the product of five groups of numbers
50! = (1*2*3*4*. . .*9*10) * (11*12*13*. . .*19*20) * (21*22*23*. . .*29*30) * (31*32*33*. . .*39*40) * (41*42*43*. . .*49*50)
It is clear that the last non-zero digit of 50! is the product of last non-zero digits of all five groups.
Regarding 1st group, 10! = 3628800; the last non-zero digit is 8.
Regarding 2nd group, (11*12*13*. . . *18*19*20) = 670442572800; the last non-zero digit is 8.
Regarding 3rd group, (21*22*23*. . . *28*29*30) = 109027350432000; the last non-zero digit is 2.
Regarding 4th group, (31*32*33*. . . *38*39*40) = 3075990524006400; the last non-zero digit is 4.
Regarding 5th group, (41*42*43*. . . *48*49*50) = 37276043023296000; the last non-zero digit is 6.
Now, the product of the last non-zero digits in these groups is 8*8*2*4*6 = 3072;
so, my DIRECT EMPIRICAL EXPERIMENT C O N F I R M S that the last non-zero digit in the number 50! is  .
Thanks to my MS EXCEL, making these boring calculations for me easy, fast and with no errors ( ! )
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Check out this page
https://math.stackexchange.com/questions/950502/how-to-find-the-last-non-zero-digit-of-50
The solution by Edward Jiang is probably the most straight forward process (that doesn't involve any complicated notation or concepts). That solution is the second provided on that page.
Recall that the exclamation marks indicate factorial
Eg: 5! = 5*4*3*2*1
We can use the rule that
(b!)/(a!) = b*(b-1)*...*(b-a+1)
So for example, if a = 10 and b = 20, then
(b!)/(a!) = b*(b-1)*...*(b-a+1)
(20!)/(10!) = 20*(20-1)*...*(20-10+1)
(20!)/(10!) = 20*19*...12*11
(20!)/(10!) = 11*12*...*19*20
Note how we start with 11 (the value just after 10) counting our way up to 20, multiplying along the way.
We could rephrase things as such
(20!)/(10!) = (20*19*...*12*11*10!)/(10!)
(20!)/(10!) = 20*19*...12*11
(20!)/(10!) = 11*12*...*19*20
In the second step, the "10!" terms cancel out.
We would also say
(30!)/(20!) = 21*22*...*29*30
(40!)/(30!) = 31*32*...*39*40
(50!)/(40!) = 41*42*...*49*50
Hopefully this is enough to get you pointed in the right direction.
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