Functional Equations Problem | SMO, 2012 | Problem 33

Try this beautiful Problem from Singapore Mathematics Olympiad, 2012 based on Functional Equations.

Let L denote the minimum value of the quotient of a 3- digit number formed by three distinct divided by the sum of its digits.Determine \(\lfloor 10L \rfloor \).

If you got stuck at first only here is the hint to begin with :

Anyway a three digit number we can be expressed as 100x + 10 y +z depending on the place values. and if we do minimize it :

F(x y z) = \(\frac {100x + 10y + z}{x + y + z}\)

Lets consider that for distinct x , y , z, F(x , y , Z) has the minimum value when x<y<z.

Again we can assume,

\( 0 < a < b < c \leq 9\)

Note ,

F(x,y,z) = \(\frac {100 x + 10 y + z }{x +y + z}\) = 1 + \(\frac {99 x + 9 y }{x+y+z}\)

Try the rest of the sum...............

From the last hint we can say

F(x y z ) is minimum when c = 9 (say)

F(x y 9) = 1+ \(\frac {99x +9y }{x+y+9} = 1 + \frac {9(x + y + 9) + 90 x - 81}{x+y +9} = 10 + \frac {9(10x -9)}{x+y +9}\)

Try to do for the next case for minimum value when b = 8..................

In the last hint we can do the next step which is b= 8:

F(x 8 9) = 10 + \(\frac {9(10x -9)}{x+17}= 10 + \frac {90(a+17)- 1611}{x + 17} = 100 - \frac {1611}{x +17} \)

which has the minimum value of x = 1 and so 10 L = 105.(answer)

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