Find all polynomials `P(x)` such that: `P(x-1).P(x+1)=P(x^2-1)`

 Find all polynomials `P(x)` such that: `P(x-1).P(x+1)=P(x^2-1)`

Solution

Suppose that `\alpha` is a root of `P(x)`  then, `P(\alpha)=0`

Therefore, `P((\alpha+1)-1)=0`

Find the last two digits of : `N=(1!+2!+3!+.......+101!)^101`

Find the last two digits of : `N=(1!+2!+3!+.......+101!)^101` 

Note: This is equivalent to finding `N(mod11)`.

ie: The remainder when dividing `N` by `100`.

Observation: `10!\equiv0(mod100)` Because, `10!=10...5...2`

Therefore, `N\equiv(1!+2!+3!+......+9!)^101(mod100)`

                   `N\equiv(1+2+6+24+20+20+40+20+80)^101(mod100)`

                   `N\equiv13^101(mod100)`

Prove that: `A_n=3^(n+3)-4^(4n+2)` is divided by `11`

 Prove that: `A_n=3^(n+3)-4^(4n+2)` is divided by `11`

Solution

We will prove by Mathematics Induction:

If `n=0` then `A_n=3^3-4^2=27-16=11` It is true that `A_n` is divided with `11`

Assuming that: `A_n` is divided with `11` for `n=k`        `(1)`

We will prove it is true for `n=k+1` then `A_(k+1)` is divided with `11`

Prove that: `forall x\inR` : `:|acosx+bsinx|\leqsqrt(a^2+b^2)`

a. Prove that: `forall x\inR` : `:|acosx+bsinx|\leqsqrt(a^2+b^2)`

b. Find the maximum and minimum of `f(x)=20cosx+21sinx+27`

Solution

a. Prove `|acosx+bsinx|\leqsqrt(a^2+b^2)` `forall x\inR`

Assuming we have: 

`acosx+bsinx=sqrt(a^2+b^2)(a/sqrt(a^2+b^2)cosx+b/sqrt(a^2+b^2)sinx)`

`X_1;X_2` are the root of : `X^2-(2cost+3sint)X-11sin^2t=0` Find the minimum of `A=X_1^2+X_1X_2+X_2^2`.

`X_1;X_2` are the root of : `X^2-(2cost+3sint)X-11sin^2t=0` 

Find the minimum of `A=X_1^2+X_1X_2+X_2^2`.

Solution

We can see that: `A=X_1^2+X_1X_2+X_2^2`

                                 `=(X_1+X_2)^2-X_1X_2`

Following Vieta's Formulas `X^2-SX+P=0`    `(1)`

Find all Polynomials `P(x)` which is satisfied that: `(x-2010)P(x+67)=xP(x)` (2010 Baltic Way)

 Find all Polynomials `P(x)` which is satisfied that:

`(x-2010)P(x+67)=xP(x)`

(2010 Baltic Way) 

Prove that: `tan^3x/(1-3tan^2x)=1/8(tan3x-3tanx)`

 a. Prove that: `tan^3x/(1-3tan^2x)=1/8(tan3x-3tanx)`

As we knew: `tan3x=(3tanx-tan^3x)/(1-3tan^2x)`

                                  `=(3tanx-9tan^3x+8tan^3x)/(1-3tan^2x)`

                                  `=(8tan^3x+3tanx(1-3tan^2x))/(1-3tan^2x)`

                                  `=(8tan^3x)/(1-3tan^2x)+3tanx`

Then,             `tan3x-3tanx=(8tan^3x)/(1-3tan^2x)`

Find the limit of `S_n=2/(1.3)+2/(3.5)+2/(5.3)+.......+2/((2n+1)(2n+3))`

 Find the limit of `S_n=2/(1.3)+2/(3.5)+2/(5.3)+.......+2/((2n+1)(2n+3))`

Solution

We can see that general term of this sequence is:

                    `2/((2k+1)(2k+3))=1/(2k+1)-1/(2k+3)`

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

 It is given function `f(x)` determine on `R` , satisfied that:

            `f(1)=1`

            `f(x+5)\geqf(x)+5`

            `f(x+1)\leqf(x)+1`

If `g(x)=f(x)+1-x` , find the value of `g(2020)`   `\forallx,y\inR`

If `(1+2x+3x^2)^10=a_0+a_1x+a_2x^2+.......+a_(20)x^(20)` then find the coefficients of `a_1` ; `a_2` ; `a_3` and `a_20`

1. If `(1+2x+3x^2)^10=a_0+a_1x+a_2x^2+.......+a_(20)x^(20)` then find the coefficients of `a_1` ; `a_2` ; `a_3` and `a_20`

Solution

        From our hypothesis we already had:

                        `(1+2x+3x^2)^10=a_0+a_1x+a_2x^2+.......+a_(20)x^(20)`

        We will find the coefficients of `a_1` ; `a_2` ; `a_3` and `a_20`

        Which meant they are the coefficients of `x` ; `x^2` ; `x^3` and `x^(20)` .

        We can rewrite `(1+2x+3x^2)^10=[1+x(2+3x)]^10`

        We will use the Newton's Formula  

            `(a+b)^n`=`\sum_{i=0}^nC(n,i)``a^(n-i).b^i`

        Then, `[1+x(2+3x)]^10=C(10,0)+C(10,1)x(2+3x)+C(10,2)x^2.(2+3x)^2`

                                            `+C(10,3)x^3.(2+3x)^3+....+C(10,10)x^10.(2+3x)^10`

        We observe that: 

            `Coef(x)=2C(10,1)=20` then, `a_1=20`

            `Coef(x^2)=2^2C(10,2)=4.45=180` then, `a_2=180`

           `Coef(x^3)=12C(10,2)+8C(10,3)=45.12+120.8=1500` then `a_3=1500`

            `Coef(x^20)=C(10,10).3^10=3^10` then, `a_20=3^10`     

Prove `(21n+4)/(14n+3)` Is Irreducible For Every Natural Number `n`

Prove `(21n+4)/(14n+3)` Is Irreducible For Every Natural Number `n`. 

Solution

 Solution 01

Denoting the greatest common divisor (GCD) of `a` and `b` `(a,b)` and we will use Euclidean Algorithm Theory.

                `(21n+4,14n+3)=(7n+1,14n+3)=(7n+1,1)=1`

It follows that `(21n+4)/(14n+3)` is irreducible. Q.E.D

 Solution 02

 

Vietnam Math Out Standing Student 2012-13

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

Problem 02: It is given `(U_n)` which is determined by: 

                    `U_1=3`

                    `U_(n+1)=1/3(2U_n+3/U_n^2)`;     for all `n\in N`.

                    Let's finding the limit of: `\lim_{n\rightarrow\infty}(u_n)`.

Problem 03: It is given three non-negative real numbers `x,y,z` , Prove that:

                    `1/x+1/y+1/z``>36/(9+x^2y^2+y^2^2+z^2x^2)`.

Problem 04: Find roots positive numbers of equation:

                    `\sqrt{x+2\sqrt3}=\sqrt y+\sqrt z`

Solution

Find all functions `f:R\rightarrow R` such that: `f(x)f(y)=f(xy-1)+xf(y)+yf(x)` `\forall` `x,y` `\inR`

Find all functions `f: R \rightarrow R` Such that:
`f(x)f(y)=f(xy-1)+xf(y)+yf(x)`    `\forall` `x ,y ``\inR`

Solution

Replace : `y=0` we will see that: `f(x).f(0)=f(-1)+x.f(0)`

Notice that: `f(0)` not equal to `0` then, `f(x)=x+c`     which `c` is a constant. We replaced to our hypothesis unsatisfactory. Then, `f(0)=0` from that we can see that: `f(-1)=0`.

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