SOLUTION: Find some value R, {{{R!/((R-3)!(R-4)!)=1}}}. Find the sum of the digits of R. R! over (R-3)!(R-4)! = 1

Question 1152304: Find some value R, . Find the sum of the digits of R.
R! over (R-3)!(R-4)! = 1
Found 2 solutions by math_helper, Edwin McCravy:
Answer by math_helper(2461) (Show Source): You can put this solution on YOUR website!
R!/((R-3)!(R-4)!) = 1

Using trial and error...
8!/(5!*4!) = 8*7*6/(4!) = 336/24 > 1
9!/(6!*5!) = 9*8*7/(5!) = 504/120 > 1
10!/(7!*6!) = 10*9*8/(6!) = 720 / 720 = 1

So R = 10 satisifies the equation.
The sum of the digits of R is equal to

Answer by Edwin McCravy(20060) (Show Source): You can put this solution on YOUR website!

R has to be 4 or more, so we start trying to find R, so we will get 1 Finally found it! R=10 The sum of the digits of 10 is 1+0 = 1 Edwin

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