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

Problem – Functional equations (SMO Test)


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 \(\lfloor 10L \rfloor \).

  • 105
  • 150
  • 102
  • 200

Key Concepts


Functional Equation

Max and Min Value

Check the Answer


But try the problem first…

Answer: 105

Source
Suggested Reading

Singapore Mathematical Olympiad, 2012

Challenges and Thrills – Pre – College Mathematics

Try with Hints


First hint

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……………

Second Hint

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………………

Final Step

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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