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Singapore Math Olympiad

Functional Equation Problem from SMO, 2018 – Question 35

Try to solve this problem number 35 from Singapore Mathematics Olympiad, SMO, 2018 based on Functional Equation.

Problem – Functional Equation (SMO Entrance)


Consider integers \({1,2, \ldots, 10}\). A particle is initially -at 1 . It moves to an adjacent integer in the next step. What is the expected number of steps it will take to reach 10 for the first time?

  • 82
  • 81
  • 80
  • 79

Key Concepts


Functional Equation

Equation

Check the Answer


Answer : 81

Singapore Mathematical Olympiad

Challenges an Thrills – Pre – College Mathematics

Try with Hints


If you got stuck into this problem we can start taking an expected number of steps to be \(g_{n}\). We need to remember at first the particle was in 1 then it will shift to the next step so for n no of position we can expressed it as n and n -1 where n = 2,3,4,……..,100.

Now try the rest…………..

Now let’s continue after the last hint …………

Then \(g_{n+1} = \frac {1}{2} (1+g_{n} + g_{n+1} )+ \frac {1}{2}\)

which implies , \(g_{n+1} = g_{n} + 2\)

Now we know that,\(g_{2} = 1\). Then \(g_{3} = 3\), \(g_{4}= 5\),………………,\(g_{10}=17\)

\(g = g_{2}+g_{3}+g_{4}+………………..+g_{10} = 1+3+…………………+17 = 81\)[ Answer]

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Singapore Math Olympiad

Functional Equations Problem | SMO, 2012 | Problem 33

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


Answer: 105

Singapore Mathematical Olympiad, 2012

Challenges and Thrills – Pre – College Mathematics

Try with Hints


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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Singapore Math Olympiad

Problem on Functional Equation | SMO, 2010 | Problem 31

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2010 based on functional equation.

Problem – Functional Equation (SMO Entrance)


Consider the identity \(1+2+……+n = \frac {1}{2}n(n+1)\). If we set \(P_{1}(x) = \frac{1}{2}x(x+1)\) , then it is the unique polynomials such that for all positive integer n,\(p_{1}(n) = 1+2+…………..+n\) . In general, for each positive integer k, there is a unique polynomial \(P_{k} (x) \) such that :

\(P_{k} (n) = 1^k + 2^ k+3^k +………………+n^k\) for each n =1,2,3……………

Find the value of \(P_{2010} (-\frac {1}{2})\) .

  • 2
  • 5
  • 6
  • 0

Key Concepts


Polynomials

Functional Equation

Check the Answer


Answer : 0

Singapore Mathematics Olympiad

Challenges and Thrills – Pre College Mathematics

Try with Hints


If you got stuck in this question we definitely can start from here:

In the question given above say k is the positive even number :

so let \(f(x) = P_{k} – P_(x-1)\)

Then \(f(n) = n^k \) for all integer \(n\geq 2\) (when f is polynomials)

Like this then \(f(x) = x^k\)(again for all \( x \geq 2\) .

If you got stuck after first hint try this one

\(P_{k} (-n + 1) – P_{k}(-n) = f(-n +1) = (n-1)^k\)…………………………….(1)

Again, \(P_{k} (-n + 2) – P_{k}(-n+1) = f(-n +2) = (n-2)^k\)…………………………………(2)

Now taking n = 1;The \(eq^n\)(1) becomes, \(P_{k}(0) – P_{k}(-1) = f(0) = 0^{k}\),

And for \(eq^(n)\) (2) ; \(P_{k}(1) – P_{k}(0) = f(1) = 1^{k}\).

Now sum these equation and try to solve the rest………..

Summing this two equation we get , \(P_{k}(1)-P_{k}(-n) = 1^{k} + 0^{k}+1^{k}+…….+(n-1)^k\).

so , \(P_{k}(-n)+P_{k}(n-1)=0\)

Again if \(g(x) = (P_{k}(-x)+P_{k}(x-1)\)

Then g(n) is equal to 0 for all integer \(n\geq2\)

As g is polynomial, g(x) =0;

So , \(P_{k}(-\frac {1}{2}) + P_{k}(-\frac {1}{2}) = 0\)

so \(p_{k}(-\frac {1}{2}) = 0\) …………(Answer)

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AMC 10 Singapore Math Olympiad

Functional Equation Problem from SMO, 2013 – Senior Section

Try this beautiful Functional Equation Problem from SMO, Singapore Mathematics Olympiad, 2013.

Problem – Functional Equation (SMO Test)

Let M be a positive integer .It is known that whenever \(|ax^2 + bx +c|\leq 1\) for all

\(|x|\leq 1\) then \(|2ax + b |\leq M \) for all \(|x|\leq 1\). Find the smallest possible value of M.


  • 4
  • 5
  • 6
  • 10

Key Concepts


Functional Equation

Function

Check the Answer


Answer: 4

Singapore Mathematics Olympiad

Challenges and Thrills – Pre – College Mathematics

Try with Hints


We cant this sum by assuming a,b,c as fixed quantity.

Let \( f(x) = ax^2 + bx + c \).

Then \( f(-1) = a – b + c \) ; \( f(0) = c \) ; \( f(1) = a + b + c\) ;

Try to do the rest of the sum …………………………..

Suppose \( |f(x)|\leq 1\) for all \(|x|\leq 1 \) . Then

\( |2ax + b| = | (x – \frac {1}{2} ) f(-1) – 2 f(0) x + (x+\frac {1}{2} f(1) |\)

\(\leq |x – \frac {1}{2}| + 2 |x| + |x + \frac {1}{2}|\)

\(\leq |x – \frac {1}{2} | + |x+\frac {1}{2}| + 2 \)

\(\leq 4 \)

Now I guess you have already got the answer but if not ………….

From the last step we can conclude ,

\(|2 x^2 – 1|\leq 1 \) whenever \(|x|\leq 4\) and \(|2x| = 4 \)

is achieved at \(x = \pm 1\).

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Math Olympiad Singapore Math Olympiad

Functional Equation Problem | SMO, 2013 – Problem19 (Senior Section)

Try this beautiful problem from Singapore Mathematics Olympiad based on Functional Equation.

Problem – Functional Equation (SMO Exam)


Let f and g be functions such that for all real numbers x and y,

\( g (f (x+y)) = f( x ) + (x+y) g (y)\).

Find the value of \( g(0) + g (1) + ……………………+ g (2013) \)

  • 1
  • 3
  • 2
  • 0

Key Concepts


Functional Equation

Funcion

Arbitrary Numbers

Check the Answer


Answer: 0

Singapore Mathemaics Olympiad

Challenges and thrills

Try with Hints


We can start this problem by considering y = -x.

Then \( g (f (0) ) = f (x) \) for all x. This \(f\) is is a constant function ; namely

\( f (x) = c \) for some c.

Try the rest of the sum ……………………………………………………

For all value of x,y we have

\( (x+y) g(y) = g(f(x+y)) – f(x) = g(c) – c = 0 \)

Since x + y is arbitrary , we must have \( g (y) = 0 \) for all y .Hence

\( g (0) + g ( 1 ) + …………………………..+ g(2013) = 0 \) (Answer).

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Math Olympiad USA Math Olympiad

Polynomial Functional Equation – Random Olympiad Problem

Understand the problem

Find all the real Polynomials P(x) such that it satisfies the functional equation: P(2P(x)) = 2P(P(x)) + P(x)^{2} \forall real x .

Source of the problem

Unknown 

Topic
Functional Equation, Polynomials
Difficulty Level
7/10
Suggested Book
Excursion in Mathematics  Challenges and Thrills in Pre College Mathematics

Start with hints

Do you really need a hint? Try it first!

Well, it is really good that the information polynomial is given! You should use that. What is the first thing that you check in a Polynomial Identity? Degree! Yes, check whether the degree of the Polynomial on both the LHS and RHS are the same or not. Yes, they are both the same n^2 .  But did you observe something fishy?  
Now rewrite the equation as P(2P(x)) - 2P(P(x)) = P(x)^{2}. Do the Degree trick now… You see it right? Yes, on the left it is n^2 and on the RHS it is 2n . So, there are two cases now… Figure them out!

Case 1: 2n = n^2 … i.e. P(x) is either a quadratic or a constant function. Case 2:  P(2P(x)) - 2P(P(x)) has coefficient zero till x^2n. We will study case 1 now. Case 1: 2n = n^2… i.e. P(x) is either a quadratic or a constant function. P(2P(x)) - 2P(P(x)) = P(x)^{2} = P(2y) - 2P(y) = y^{2} where y = P(x) . Now, expand using P(x) = ax^2 + bx +c, it gives 2ay^2 -c = y^2 … Now find out all such polynomials satisfying this property. For e.g. \frac{x^2}{2} is a solution. If P(x) is constant, prove that P(x) = 0 / \frac{-1}{2} .  

Case 2: P(2y) - 2P(y) = y^{2}. Assume a general form of P(x) = $latex $and show that P(x) must be quadratic or lesser degree by comparing coefficients as you have a quadratic on RHS and n degree polynomial of the LHS.  Now, we have already solved it for quadratic or less degree.  
  • Always Compare the Degree of Polynomials in identities like this. It provides a lot of information.
  • Compare the coefficients of Polynomials on both sides to equalize the coefficient on both sides.
  • Find all polynomials P(2P(x)) - 8P(P(x)) = P(x)^{2}.
  • Find all polynomials \( P(cP(x)) – d P(P(x)) = P(x)^{2} \) depending on the values of c and d.

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Math Olympiad

A functional inequation

Understand the problem

Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
\[f(x + y) + y \le f(f(f(x)))\]
holds for all $x, y \in \mathbb{R}$.
Source of the problem
Benelux MO 2013
Topic
Functional Equations
Difficulty Level
Easy
Suggested Book
Functional Equations by BJ Venkatachala

Start with hints

Do you really need a hint? Try it first!

Note that the RHS does not contain y. Thus it should be possible to play with different values of y to get rid of the inconvenient f\circ f\circ f.
Taking y=f(f(f(x)))-f(x), we get f(f(f(x)))\le f(x). Now show that f(x)\le f(f(f(x))). Thus $f(f(f(x)))=f(x)$.
Taking x=0, we get f(y)+y\le f(0). Now try to show that f(y)+y\ge f(0).
Taking y=-x we get the desired inequality from hint 3. Thus f(y)+y=f(0)\implies f(y)=f(0)-y for all y.

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Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year.

Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

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Triangle Problem | PRMO-2018 | Problem No-24

Try this beautiful Problem on Trigonometry from PRMO -2018.You may use sequential hints to solve the problem.

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Value of Sum | PRMO – 2018 | Question 16

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Chessboard Problem | PRMO-2018 | Problem No-26

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Chocolates Problem | PRMO – 2018 | Problem No. – 28

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Functional equation dependent on a constant


Understand the problem

Find all real numbers $a$ for which there exists a non-constant function $f :\Bbb R \to \Bbb R$ satisfying the following two equations for all $x\in \Bbb R:$
i) $f(ax) = a^2f(x)$ and
ii) $f(f(x)) = a f(x).$
Source of the problem
Baltic Way 2016
Topic

Functional Equation

Difficulty Level
Easy
Suggested Book
Functional Equations by BJ Venkatachala

Start with hints

Do you really need a hint? Try it first!

Show that the choices a=0,1 work.

Show that af(f(x))=a^2f(f(x)). As we have already dealt with a=0, this gives af(f(x))=f(f(x))
Hint 3 gives (a-1)f(f(x))=0. As a=1 has already been dealt with, we must consider the option f(f(x))\equiv 0.
Hint 3 gives af(x)\equiv 0. As a\neq 0, we have f(x)\equiv 0. This contradicts the fact that f is non-constant. Hence, a=0,1 are the only options.

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Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year.

Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

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Try this beautiful Problem on Trigonometry from PRMO -2018.You may use sequential hints to solve the problem.

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Measure of Angle | PRMO-2018 | Problem No-29

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Good numbers Problem | PRMO-2018 | Question 22

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Digits Problem | PRMO – 2018 | Question 19

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Chocolates Problem | PRMO – 2018 | Problem No. – 28

Try this beautiful Problem on Combinatorics from PRMO -2018.You may use sequential hints to solve the problem.

Trigonometry | PRMO-2018 | Problem No-14

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Categories
Math Olympiad

Find all arithmetic functions

Understand the problem

Find all functions $f : \mathbb{N}\rightarrow \mathbb{N}$ satisfying following condition:
\[f(n+1)>f(f(n)), \quad \forall n \in \mathbb{N}.\]

Source of the problem

IMO longlist 1977

Topic
Functional equations
Difficulty Level
Medium
Suggested Book
Functional Equations by B J Venkatachala

Start with hints

Do you really need a hint? Try it first!

Prove (by induction on m) that if n\ge m then f(n)\ge m.
Note that hint 1 implies that f(n+1)>f(n).
Note that hint 2 implies that n+1>f(n), i.e. f(n)\le n.
Hint 3, combined with hint 1 gives f(n)=n for all n\in\mathbb{N}.

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Connected Program at Cheenta

Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year.

Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

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Categories
Math Olympiad

Application of Cauchy’s Functional Equation – INMO 2018 Problem 6

An Application of Cauchy’s Functional Equation


Cauchy’s functional equation is a description of a function. Lets look at Indian National Math Olympiad 2018’s Problem 6 which can be solved as an application of Cauchy’s Functional Equation:

$$ f(x + y) = f(x) + f(y) $$

INMO 2018 Problem 6

Let N denote the set of all natural numbers and let (f : N\rightarrow N) be a function such that
(a) (f{(mn)} = f {(m)} f{(n)}) for all m,n in N ;
(b) m+n divides (f {(m)} + f {(n)} ) for all m, n in N
Prove that there exists an odd natural number (k) such that (f {(n)} = n^k) for all n in N.

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