Question 1182542
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Check out this page
<a href = "https://math.stackexchange.com/questions/950502/how-to-find-the-last-non-zero-digit-of-50">https://math.stackexchange.com/questions/950502/how-to-find-the-last-non-zero-digit-of-50</a>
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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