Prove that `(1/2).(3/4)......((2n-1)/(2n))\leq1/sqrt(3n+1)`

•  This is the problem that I picked up from book : 101 Solved Problem in Algebra from The USA IMO with the 43th problem.

What is the coefficient of `x^2` when `(1+x)(1+2x)...(1+2^nx)` is expanded?

•  This is the problem that I picked up from book : 101 Solved Problem in Algebra from The USA IMO with the 32th problem. 

Vietnamese Mathematical Olympiad 2017

•  This is the 2007 Vietnamese Mathematical Olympiad Problem number 6. This is kind of problem that you need to use Permutation Formular . Moreover, you have to know about Sum of Sigma as well. 

Prove that `1/a_1+1/a_2+1/a_3+...+1/a_20` is a natural number

Find function `f(x)` satisfied `f(x)f(y)-f(xy)-90=10(x+y)/xy`

Prove that `a^2+b^2+c^2\geqa^(4/3)+b^(4/3)+c^(4/3)`

Prove that `x, y, z >0` Then `M=x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)\leq3/4`

 

Cambodia National Math 2019, 22/04/2019 Day 02

 

Cambodia National Math 2019, 22/04/2019 Day 02

Math Cambodia 2019 Day 02

• This was the problem that released for Out Standing Student in Cambodia in 2019 for 2nd day of testing.
• There were two days of the testing. This is the first day of exam.
• You all can Click here to download

Cambodia Grade 12, 22/04/2019 Day 01

Cambodia National Math 2019, 22/04/2019

• This was the problem that released for Out Standing Student in Cambodia in 2019.
• There were two days of the testing. This is the first day of exam.

• You can watch solution here:

Prove that the polynomial `x^9999+x^8888+x^7777+...+x^1111+1` is divisible by `x^9+x^8+x^7+....+x+1`

 Problem: 01

Prove that the polynomial `x^9999+x^8888+x^7777+...+x^1111+1` is divisible by `x^9+x^8+x^7+....+x+1`

Solution

Find all function `f(x)` if `(x-y)f(x+y)-(x+y)f(x-y)=4xy(x^2-y^2)`

 Find all function `f(x)` if `(x-y)f(x+y)-(x+y)f(x-y)=4xy(x^2-y^2)`

Solution

Vietnamese Olympiad 2022: Find the value of `1/(a^2023)+1/(b^2023)+1/(c^2023)`

     If `a+b+c=2022` and `1/a+1/b+1/c=1/2022` Find the value of `1/(a^2023)+1/(b^2023)+1/(c^2023)`

Solution

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)`

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`