If \(g(x)=f(x)+1-x\), find the value of \(g(2020)\).

It is given that \(f(x)\) is a function defined on \(\mathbb{R}\), satisfying \(f(1)=1\), \(f(x+5)\geq f(x)+5\), and \(f(x+1)\leq f(x)+1\). If \(g(x)=f(x)+1-x\), find the value of \(g(2020)\).

Solution

From the given conditions, we have \(f(x+5)\geq f(x)+5\) and \(f(x+1)\leq f(x)+1\).

Since \(f(x+5)\leq f(x+4)+1\) and \(f(x+4)\leq f(x+3)+1\leq f(x+2)+2\leq f(x+1)+3\), we get

$$ f(x+5)\leq f(x+1)+4. $$

Therefore,

$$ f(x)+5\leq f(x+5)\leq f(x+1)+4\leq f(x)+5. $$

Hence,

$$ f(x+1)+4=f(x)+5, $$

so

$$ f(x+1)-f(x)=1. $$

For \(x=1,2,\ldots,2019\), we have

$$ f(2)-f(1)=1,\quad f(3)-f(2)=1,\quad\cdots,\quad f(2020)-f(2019)=1. $$

Adding all equations,

$$ f(2020)-f(1)=2019. $$

Since \(f(1)=1\),

$$ f(2020)=2020. $$

Now,

$$ g(x)=f(x)+1-x. $$

For \(x=2020\),

$$ g(2020)=f(2020)+1-2020=2020+1-2020=1. $$

Therefore,

$$ \boxed{g(2020)=1}. $$

Solution by: Thin Sokkean

If \((1+2x+3x^2)^{10}=a_0+a_1x+a_2x^2+\cdots+a_{20}x^{20}\), find the coefficients \(a_1,\ a_2,\ a_3,\) and \(a_{20}\).

Solution

From \((1+2x+3x^2)^{10}=a_0+a_1x+a_2x^2+\cdots+a_{20}x^{20}\), we need to find the coefficients of \(x,x^2,x^3,\) and \(x^{20}\).

Rewrite: \((1+2x+3x^2)^{10}=[1+x(2+3x)]^{10}\).

Using Newton's formula \((a+b)^n=\sum_{i=0}^{n}C(n,i)a^{n-i}b^i\), we get \([1+x(2+3x)]^{10}=C(10,0)+C(10,1)x(2+3x)+C(10,2)x^2(2+3x)^2+\cdots+C(10,10)x^{10}(2+3x)^{10}\).

Coefficient of \(x\): \(a_1=2C(10,1)=20\).

Coefficient of \(x^2\): \(a_2=2^2C(10,2)=4\times45=180\).

Coefficient of \(x^3\): \(a_3=12C(10,2)+8C(10,3)=12\times45+8\times120=1500\).

Coefficient of \(x^{20}\): \(a_{20}=C(10,10)3^{10}=3^{10}\).

Therefore, \(a_1=20,\ a_2=180,\ a_3=1500,\ a_{20}=3^{10}\).

Part 2

Given \((1+x+2x^2)^{20}=a_0+a_1x+a_2x^2+\cdots+a_{40}x^{40}\).

When \(x=1\), \(a_0+a_1+a_2+\cdots+a_{40}=4^{20}\).

When \(x=-1\), \(a_0-a_1+a_2-a_3+\cdots-a_{39}+a_{40}=2^{20}\).

Adding the two equations, \(2a_0+2a_2+2a_4+\cdots+2a_{40}=4^{20}+2^{20}\).

Hence, \(a_0+a_2+a_4+\cdots+a_{40}=\frac{4^{20}+2^{20}}{2}\).

Since \(a_{40}=2^{20}\),

\[ a_0+a_2+a_4+\cdots+a_{38} =\frac{4^{20}+2^{20}}{2}-2^{20} =\frac{2^{40}+2^{20}}{2}-2^{20} =\frac{2^{40}-2^{20}}{2}. \]

Therefore, \(\boxed{a_0+a_2+a_4+\cdots+a_{38}=\frac{2^{40}-2^{20}}{2}}\).

Prove that \(\frac{21n+4}{14n+3}\) is irreducible for every natural number \(n\).

Solution

Method 01: Let \((a,b)\) denote the greatest common divisor of \(a\) and \(b\). Using the Euclidean Algorithm,

\[ (21n+4,14n+3)=(7n+1,14n+3)=(7n+1,1)=1. \]

Therefore, \(\frac{21n+4}{14n+3}\) is irreducible. Q.E.D.

Method 02: We prove by contradiction. Assume that \(\frac{21n+4}{14n+3}\) is reducible. Then there exists a prime number \(p\) such that \(21n+4\equiv0\pmod p\) and \(14n+3\equiv0\pmod p\).

Multiplying the first equation by \(2\), we get \(42n+8\equiv0\pmod p\). Subtracting \(3(14n+3)\equiv0\pmod p\),

\[ (42n+8)-(42n+9)\equiv0\pmod p \]

so \(-1\equiv0\pmod p\), which is impossible. Hence, \(\frac{21n+4}{14n+3}\) is irreducible. Q.E.D.

Solution by: Thin Sokkean

Vietnam Math Out Standing Student 2012-13

Vietnam Math Outstanding Student 2012-2013

Problem 01: Solve the equation \(x^{4n}+\sqrt{x^{2n}+2012}=2012\), for all \(n\in\mathbb{N}\).

Problem 02: It is given the sequence \((U_n)\) determined by \(U_1=3\) and \(U_{n+1}=\frac13\left(2U_n+\frac3{U_n^2}\right)\), for all \(n\in\mathbb{N}\). Find the limit \(\lim_{n\rightarrow\infty}U_n\).

Problem 03: Given three non-negative real numbers \(x,y,z\), prove that \(\frac1x+\frac1y+\frac1z\geq\frac{36}{9+x^2y^2+y^2z^2+z^2x^2}\).

Problem 04: Find the positive roots of the equation \(\sqrt{x+2\sqrt3}=\sqrt y+\sqrt z\).

Vietnam Math Outstanding Student 2012-2013

Problem 01: Solve the equation \(x^{4n}+\sqrt{x^{2n}+2012}=2012\), for all \(n\in\mathbb{N}\).

Solution: Let \(t=x^{2n}\). Then the equation becomes \(t^2+\sqrt{t+2012}=2012\).

We have \(t^2=2012-\sqrt{t+2012}\), so adding \(t+\frac14\) to both sides gives \((t+\frac12)^2=t+2012-\sqrt{t+2012}+\frac14=(\sqrt{t+2012}-\frac12)^2\).

Since \(t=x^{2n}\geq0\), we get \(t+\frac12=\sqrt{t+2012}-\frac12\), hence \(t+1=\sqrt{t+2012}\).

Squaring both sides, \(t^2+t+1=t+2012\), so \(t^2-2011=0\).

Therefore, \(t=\sqrt{2011}\), and since \(t=x^{2n}\), we have \(x^{2n}=\sqrt{2011}\).

 

Find all functions \(f:\mathbb{R}\rightarrow\mathbb{R}\) such that \(f(x)f(y)=f(xy-1)+xf(y)+yf(x)\), \(\forall x,y\in\mathbb{R}\).

Solution

Replace \(y=0\), we get \(f(x)f(0)=f(-1)+xf(0)\).

If \(f(0)\neq0\), then \(f(x)=x+c\), where \(c\) is a constant. Substituting into the original equation gives a contradiction. Therefore, \(f(0)=0\), and hence \(f(-1)=0\).

Putting \(x=y=1\), we have \(f(1)^2=2f(1)\), so \(f(1)=0\) or \(f(1)=2\).

Putting \(y=-1\), we get \(f(-x-1)=f(x)\).

Replacing \(y\) by \(-y-1\) in the original equation, using \(f(-y-1)=f(y)\) and \(f(xy+x)=f(x(-y-1)-1)\), we obtain

\[ f(xy-1)+yf(x)=f(xy+x)+xf(y)-(y+1)f(x). \]

For \(x\neq0\), let \(y=\frac1x\), then

\[ f(x+1)=\frac{x+2}{x}f(x), \]

or equivalently,

\[ f(x-1)=\frac{x-1}{x+1}f(x). \]

Now replace \(y=1\) in the original equation:

\[ f(x)f(1)=f(x-1)+xf(1)+f(x) \]

Using \(f(x-1)=\frac{x-1}{x+1}f(x)\), we have:

Case 1: \(f(1)=0\), then \(f(x)=0\) for all \(x\in\mathbb{R}\).

Case 2: \(f(1)=2\), then \(f(x)=x(x+1)\).

Both functions satisfy the original equation.

Therefore, the solutions are:

\[ \boxed{f(x)\equiv0\quad\text{or}\quad f(x)=x(x+1),\ \forall x\in\mathbb{R}.} \]

Solution by: Thin Sokkean

1970 IMO Problems And Solutions

 

Problem 01

Let  be a point on the side  of . Let , and  be the inscribed circles of triangles , and . Let , and  be the radii of the exscribed circles of the same triangles that lie in the angle . Prove that

.

Solution

1979 IMO Problems And Solutions

 

Problem 01

If  and  are natural numbers so thatProve that  is divisible with .

Solution

1963 IMO Problems And Solutions

 

Problem 01

Find all real roots of the equation

,

where  is a real parameter.

Solution

2007 IMO Problems And Solutions

 

Problem 01

Real numbers  are given. For each  () define

and let

.

(a) Prove that, for any real numbers ,

(b) Show that there are real numbers  such that equality holds in (*)

Solution

2011 IMO Problems And Solutions

 

Problem 01

Given any set  of four distinct positive integers, we denote the sum  by . Let  denote the number of pairs  with  for which  divides . Find all sets  of four distinct positive integers which achieve the largest possible value of 

Solution

2019 IMO Problems And Solutions

 

Problem 01

Let  be the set of integers. Determine all functions  such that, for all integers  and ,

Solutions

Solution 1

Let us substitute  in for  to get

Now, since the domain and range of  are the same, we can let  and  equal some constant  to get