Evaluate the value of: \(A=\frac{3}{1!+2!+3!}+\frac{4!}{2!+3!+4!}+......+\frac{n}{(n-2)!+(n-1)!+n!}\).
Suppose that \(\frac{k}{(k-2)!+(k-1)!+k!}=\frac{k}{(k-2)![1+(k-1)+k(k-1)]}=\frac{k}{(k-2)!k^2}=\frac{1}{(k-2)!k}=\frac{k-1}{(k-2)!(k-1)k}=\frac{k-1}{k!}=\frac{1}{(k-1)!}-\frac{1}{k!}\). Therefore, \(A=\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+...+\left(\frac{1}{(n-1)!}-\frac{1}{n!}\right)=\frac{1}{2!}-\frac{1}{n!}=\frac{n!-2}{2n!}\).
Evaluate the value of: \(A=\frac{3}{1!+2!+3!}+\frac{4!}{2!+3!+4!}+......+\frac{n}{(n-2)!+(n-1)!+n!}\).
Solution:
If \(n=0\), then:
\(A_0=2^{6(0)+1}+9^{0+1}=2+9=11\equiv0\pmod{11}\)
Therefore, \(A_0\) is divisible by \(11\).
We prove that \(A_n=2^{6n+1}+9^{n+1}\) is divisible by \(11\) for all natural numbers \(n\).
Step 1: Base case
If \(n=0\), then:
\(A_0=2^{6(0)+1}+9^{0+1}=2+9=11\equiv0\pmod{11}\)
Therefore, \(A_0\) is divisible by \(11\).
Step 2: Induction hypothesis
Assume that for \(n=k\):
\(A_k=2^{6k+1}+9^{k+1}\equiv0\pmod{11}\)
Step 3: Prove for \(n=k+1\)
We have:
\[ A_{k+1}=2^{6k+7}+9^{k+2} \]
\[ A_{k+1}=2^6(2^{6k+1})+9^{k+1}\cdot9 \]
\[ A_{k+1}=64(2^{6k+1}+9^{k+1})+9^{k+1}(9-64) \]
\[ A_{k+1}=64A_k-55\cdot9^{k+1} \]
Since \(A_k\equiv0\pmod{11}\) and \(55\equiv0\pmod{11}\), we get:
\[ A_{k+1}\equiv64(0)-0\cdot9^{k+1}\equiv0\pmod{11} \]
Therefore, \(A_{k+1}\) is divisible by \(11\).
Hence, by mathematical induction, \(A_n=2^{6n+1}+9^{n+1}\) is divisible by \(11\) for all natural numbers \(n\).
It is given that:
\(E_n=831^n+709^n-743^n-610^n\)
for all natural numbers \(n\). Prove that \(E_n\) is divisible by \(189\) for all natural numbers \(n\).
Hint: Using modulo formula and \(\gcd(9,21)=189\).
Prove that for all natural numbers \(n\), we have:
\[ 1+\frac1{\sqrt2}+\frac1{\sqrt3}+...+\frac1{\sqrt{n+1}}<2\sqrt{n+1} \]
Hint: Using Mathematical Induction.
It is given a natural real sequence satisfied that:
\[ U_0=\sqrt2 \]
\[ U_{n+1}=\sqrt{2+U_n} \]
a. Find \(U_n\) as a function of \(n\).
b. Find the product:
\[ P_n=U_0U_1U_2...U_n \]
There is a 4-digit number with every single digit arranged as:
\(aabb\)
Find those numbers if they are perfect squares.
It is given that:
\[ 33^2=1089 \]
\[ 333^2=110889 \]
\[ 3333^2=11108889 \]
\[ 33333^2=1111088889 \]
From the given examples, find the general term and prove it.
a. Prove that:
\[ 1+\frac{1}{\cos x}=\frac{\cot(x/2)}{\cot x} \]
b. Calculate the product:
\[ P_n=(1+\frac1{\cos a})(1+\frac1{\cos(a/2)}) (1+\frac1{\cos(a/2^2)})\cdots (1+\frac1{\cos(a/2^n)}) \]
Calculate the value of:
\[ S=\cos^3(\frac{\pi}{9})-\cos^3(\frac{4\pi}{9}) +\cos^3(\frac{7\pi}{9}) \]
Calculate the sum:
\[ S_n=9+99+999+\cdots+\underbrace{99\cdots9}_{n\text{ digits}} \]
Find all pairs of integers \((m,n)>2\) satisfied that for all positive integers \(a\):
\[ \frac{a^m+a-1}{a^n+a^2-1} \]
is an integer.
Solution: \((m,n)=(5,3)\)
It is given three positive integers \(a,b,c\) satisfied that:
\[ a+b+c=10 \]
Find the minimum value of:
\[ P=a\times b\times c \]
Solution: \(P=36\)
Find the exact value of:
\[ \sin(\frac{\pi}{10}) \quad \text{and} \quad \cos(\frac{\pi}{10}) \]
It is given two positive real numbers \(a\) and \(b\). Prove that:
\[ (1+a)(1+b)\geq(1+\sqrt{ab})^2 \]
From the result above, find the minimum value of the function:
\[ f(x)=(1+4^{\sin^2x})(1+4^{\cos^2x}) \]
for all real numbers \(x\).
It is given three real numbers \(a,b,c\). Prove that:
\[ a^2+b^2+c^2\geq ab+bc+ac \]
It is given \(n\) positive real numbers:
\[ a_1,a_2,a_3,\ldots,a_n \]
satisfied that:
\[ a_1a_2a_3\cdots a_n=1 \]
Prove that:
\[ (1+a_1)(1+a_2)(1+a_3)\cdots(1+a_n)\geq2^n \]
It is given \(m,n\) are positive integers. Prove that for all positive real numbers \(x\):
\[ \frac{x^{mn}-1}{m}\geq\frac{x^n-1}{x} \]
It is given that: \[ E_n=831^n+709^n-743^n-610^n \] for all natural number \(n\).
Prove that: \[ E_n \] is divided by \(189\) for all natural number \(n\).
Hint: Using modulo formula and \(\gcd(9,21)=189\).
Prove that for all natural number \(n\): \[ 1+\frac1{\sqrt2}+\frac1{\sqrt3}+...+\frac1{\sqrt{n+1}} <2\sqrt{n+1} \]
Hint: Using Mathematical Induction.
It is given the real sequence: \[ U_0=\sqrt2 \] and \[ U_{n+1}=\sqrt{2+U_n} \]
a. Find \(U_n\) as a function of \(n\).
b. Find the product: \[ P_n=U_0U_1U_2...U_n \]
There is a 4 digit number with every single digit in the order: \[ aabb \] Find those numbers if they are perfect squares.
It is given:
\[ 33^2=1089 \]
\[ 333^2=110889 \]
\[ 3333^2=11108889 \]
\[ 33333^2=1111088889 \]
Find the general term and prove it.
a. Prove that: \[ 1+\frac1{\cos x}=\frac{\cot(x/2)}{\cot x} \]
b. Calculate: \[ P_n= (1+\frac1{\cos a}) (1+\frac1{\cos(a/2)}) (1+\frac1{\cos(a/2^2)}) ... (1+\frac1{\cos(a/2^n)}) \]
Calculate the value: \[ S=\cos^3(\frac{\pi}{9}) -\cos^3(\frac{4\pi}{9}) +\cos^3(\frac{7\pi}{9}) \]
Calculate the sum: \[ S_n=9+99+999+...+999 \] where the last number contains \(n\) digits of 9.
Find all pairs of integers \((m,n)>2\) such that for every positive integer \(a\):
\[ \frac{a^m+a-1}{a^n+a^2-1} \]
is an integer.
It is given three positive integers \(a,b,c\) satisfying: \[ a+b+c=10 \]
Find the minimum value of: \[ P=a\times b\times c \]
Find the exact value of:
\[ \sin(\frac{\pi}{10}) \] and \[ \cos(\frac{\pi}{10}) \]
It is given two positive real numbers \(a\) and \(b\). Prove that:
\[ (1+a)(1+b)\geq(1+\sqrt{ab})^2 \]
From the proven result, find the minimum value of:
\[ f(x)=(1+4^{\sin^2x})(1+4^{\cos^2x}) \]
for all real numbers \(x\).
It is given three real numbers \(a,b,c\).
Prove that:
\[ a^2+b^2+c^2\geq ab+bc+ac \]
It is given \(n\) positive real numbers:
\[ a_1,a_2,a_3,...,a_n \]
satisfying:
\[ a_1a_2a_3...a_n=1 \]
Prove that:
\[ (1+a_1)(1+a_2)(1+a_3)...(1+a_n)\geq2^n \]
It is given positive integers \(m,n\). Prove that for all positive real numbers \(x\):
\[ \frac{x^{mn}-1}{m}\geq\frac{x^n-1}{x} \]
For all real numbers \(x\), prove that:
\[ (1+\sin x)(1+\cos x) \leq \frac32+\sqrt2 \]
It is given:
\[ x_n=2^{2^n}+1 \]
for \(n=1,2,3,...\)
Prove that:
\[ \frac1{x_1} +\frac2{x_2} +\frac{2^3}{x_3} +... +\frac{2^{n-1}}{x_n} <\frac13 \]
It is given the function:
\[ y=\frac{x^2+2mx+3m-8}{2(x^2+1)} \]
where \(x\) is a real number and \(m\) is a parameter.
Is it possible to find a value of \(m\) to make the function \(y\) be the value of cosine of one single angle?
It is given the function:
\[ f(x,y)= \frac{(x^2-y^2)(1-x^2y^2)} {(1+x^2)^2(1+y^2)^2} \]
where \(x,y\) are real numbers.
Prove that:
\[ |f(x,y)|\leq\frac14 \]
It is given:
\[ 0<\theta<\frac{\pi}{2} \]
Prove that:
\[ (\sin\theta)^{\cos\theta} + (\cos\theta)^{\sin\theta} >1 \]
There are three real numbers: \[ a>0,\quad b>0,\quad c>0 \]
Prove that:
\[ ab(a+b)+bc(b+c)+ac(a+c)\geq6abc \]
Solve the equation:
\[ 9^{(x^2-x)}+3^{(1-x^2)} = 3^{(x-1)^2}+1 \]
Find both functions \(f(x)\) and \(g(x)\) satisfying:
\[ f(2x-1)+2g(3x+1)=x^2 \]
\[ f(4x-3)-g(6x-2)=-2x^2+2x+1 \]
Find the sum:
\[ S_n= \tan a+\frac12\tan\frac a2 +\frac1{2^2}\tan\frac a{2^2} +... +\frac1{2^n}\tan\frac a{2^n} \]
It is a third degree polynomial \(P(x)\) satisfying:
\[ P(x)+2 \] is divisible by \[ (x+1)^2 \]
and
\[ P(x)-2 \] is divisible by \[ (x-1)^2 \]
Determine the polynomial \(P(x)\).
Let:
\[ A= (\frac1{\sqrt3}+i)^n - (\frac1{\sqrt3}-i)^n \]
for all natural numbers \(n\).
Prove that:
\[ A= i\frac{2^{n+1}}{(\sqrt3)^n} \sin\frac{n\pi}{3} \]
Solve the equation in integer set:
\[ 47x+29y=1 \]
Find all possible functions:
\[ f(x) \]
satisfying:
\[ f(x+\sqrt{x^2-2x+1}) = \frac{x^2-1}{x^2+1} \]
It is given a sequence of real numbers:
\[ (a_n),\quad n\geq1 \]
satisfying:
\[ a_1=1,\quad a_2=3 \]
and
\[ a_{n+2}=(n+3)a_{n+1}-(n+2)a_n \]
Evaluate the value of \(n\) if:
\[ a_n\equiv0\pmod{11} \]
Prove that for all positive integers \(n\):
\[ 3^n+n^3 \]
is divisible by \(7\) if and only if:
\[ 3^n n^3+1 \]
is divisible by \(7\).
Find the sum:
\[ S_n= \frac{3}{1!+2!+3!} +\frac{4}{2!+3!+4!} +... +\frac{n+2}{n!+(n+1)!+(n+2)!} \]
Prove that:
\[ 16< \sum_{k=1}^{80}\frac1{\sqrt{k}} <17 \]
(China 1992)
Find all real numbers \(x\) satisfying:
\[ 2^x+3^x-4^x+6^x-9^x=1 \]
(Korean 2000)
It is given positive real numbers:
\[ X_1,X_2,X_3,...,X_n \]
satisfying:
\[ \sum_{i=1}^{n}X_i=1 \]
Prove that:
\[ \left(\sum_{i=1}^{n}\sqrt{X_i}\right) \left(\sum_{i=1}^{n}\frac1{\sqrt{1+X_i}}\right) \leq \frac{n^2}{\sqrt{n+1}} \]
(China Team Selection Test 2006)
There are \(a,b,c\) non-negative real numbers satisfying:
\[ ab+bc+ca=\frac13 \]
Prove that:
\[ \frac1{a^2-bc+1} + \frac1{b^2-ac+1} + \frac1{c^2-ab+1} \leq\frac13 \]
(China Team Selection Test 2005)
It is given the sequence of real numbers:
\[ a_1=1,\quad a_2=5 \]
and:
\[ a_{n+1} = \frac{a_na_{n-1}} {\sqrt{a_n^2+a_{n-1}^2+1}}, \quad n\geq2 \]
Determine the general term of \((a_n)\).
(China 2002)
Find all functions:
\[ f(x):\mathbb{R}\rightarrow\mathbb{R} \]
such that:
\[ f([x]y)=f(x)[f(y)] \]
is true for all \(x,y\in\mathbb{R}\).
where \([a]\) is the greatest integer less than or equal to \(a\).
(IMO 2010)
It is given:
\[ f(x)=\frac{x+4}{x+1} \]
where \(x\neq-1\).
Evaluate:
\[ f_n[f[...f[f(x)]...]] \]
It is given the function relation:
\[ 2f(\frac{\pi}{2}-x) + f(\frac{\pi}{2}+x) = \sin x+3\sqrt3\cos x \]
Find the values of \(\theta\) and \(r\) if:
\[ f(x)=r\sin(x+\theta) \]
Find all pairs of integers \((a,b)\) satisfying that:
\[ \frac{x^2y+x+y}{xy^2+y+7} \]
is an integer.
Let \((a_n)\) be a sequence of real numbers satisfying \(a_0=1\) and \(a_{n+1}=a_0a_1\cdots a_n+4\) for all natural numbers \(n\). Prove that \(a_n-\sqrt{a_{n+1}}=2\) for all \(n>0\).
We observe that \(a_k>0\) for all natural numbers \(k\).
From \(a_{n+1}=a_0a_1\cdots a_n+4\), replacing \(n\) by \(n+1\), we get:
\(a_{n+2}=a_0a_1\cdots a_{n+1}+4=(a_0a_1\cdots a_n)(a_0a_1\cdots a_n+4)+4\)
\(a_{n+2}=(a_0a_1\cdots a_n)^2+4(a_0a_1\cdots a_n)+4=(a_0a_1\cdots a_n+2)^2\)
\(\sqrt{a_{n+2}}=a_0a_1\cdots a_n+2=a_{n+1}+2\)
\(\sqrt{a_{n+1}}=a_n+2\)
Therefore, \(a_n-\sqrt{a_{n+1}}=a_n-(a_n+2)=-2\).
It is given function \(f(x)\) for all real number \(x\) such that: \(x(2x+1)f(x)+f(\frac{1}{x})=x+1\).
From \(x(2x+1)f(x)+f(\frac{1}{x})=x+1\), replacing \(x\) by \(\frac{1}{x}\) gives \(f(\frac{1}{x})+\frac{x^2}{x+2}f(x)=\frac{x^2+x}{x+2}\). Subtracting, \(\frac{2x(x^2+1)}{x+2}f(x)=\frac{2x+2}{x+2}\), hence \(f(x)=\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}\). Therefore, \(S=\sum_{k=1}^{2022}f(k)=\sum_{k=1}^{2022}(\frac{1}{k}-\frac{1}{k+1})=1-\frac{1}{2022}=\frac{2021}{2022}\).
It is given polynomial \(P(x)=(x\sin a+\cos a)^n\) for natural number \(n\). Find the remainder when \(P(x)\) is divided by \(x^2+1\).
Let \(R(x)\) be the remainder when \(P(x)\) is divided by \(x^2+1\). If \(n=1\), then \(P(x)=x\sin a+\cos a\), so \(R(x)=x\sin a+\cos a\). For \(n\ge2\), let \(P(x)=(x^2+1)Q(x)+Ax+B\). Taking \(x=i\), we get \((i\sin a+\cos a)^n=(\cos a+i\sin a)^n=\cos(na)+i\sin(na)=B+iA\), hence \(A=\sin(na)\) and \(B=\cos(na)\). Therefore, the remainder is \(R(x)=x\sin(na)+\cos(na)\).
It is given three positive real numbers \(x,y,z\) such that: \(\cos x+\cos y+\cos z=0\) and \(\cos 3x+\cos 3y+\cos 3z=0\).
Given \( \cos x+\cos y+\cos z=0 \) and \( \cos3x+\cos3y+\cos3z=0 \), prove that \( \cos 2x \cdot \cos 2y \cdot \cos 2z \leq 0 \).
Solution: Using \( \cos3a=4\cos^3a-3\cos a \), we get \( \cos^3x+\cos^3y+\cos^3z=0 \). Since \( \cos x+\cos y+\cos z=0 \), we have \( \cos^3x+\cos^3y+\cos^3z=3\cos x\cos y\cos z \), hence \( \cos x\cos y\cos z=0 \). Assume \( \cos x=0 \), then \( \cos y=-\cos z \), so \( \cos2x\cos2y\cos2z=(2\cos^2x-1)(2\cos^2y-1)(2\cos^2z-1)=-(2\cos^2z-1)^2\leq0 \). Therefore, \( \cos2x\cos2y\cos2z\leq0 \). Q.E.D.
Finding the sum of \(S_n=1\cdot1!+2\cdot2!+3\cdot3!+\cdots+n\cdot n!\), where \(n!=1\cdot2\cdot3\cdots n\).
Finding the sum of the following problem:
\(S_n=1\cdot1!+2\cdot2!+3\cdot3!+\cdots+n\cdot n!\)
where \(n!=1\cdot2\cdot3\cdots n\).
Solution:
In order to solve this kind of problem, we need to start with the general term of the sequence.
The general term of our problem is:
\(k\cdot k!\)
We can rewrite it as:
\(k\cdot k!=(k+1-1)k!=(k+1)k!-k!=(k+1)!-k!\)
Then we replace the value of \(k\) following our main problem:
\(1\cdot1!=2!-1!\)
\(2\cdot2!=3!-2!\)
\(3\cdot3!=4!-3!\)
\(\vdots\)
\(n\cdot n!=(n+1)!-n!\)
Adding all equations side by side:
\(S_n=(2!-1!)+(3!-2!)+(4!-3!)+\cdots+((n+1)!-n!)\)
\(S_n=(n+1)!-1!\)
Hence:
\(\boxed{S_n=(n+1)!-1!}\)
Solution by: Thin Sokkean
It is given two positive real numbers \(x,y\) which satisfy \(4x+3y=11\). Find the maximum value of \(f(x,y)=(x+6)(y+7)(3x+2y)\).
Find the maximum value of the following function:
\(f(x,y)=(x+6)(y+7)(3x+2y)\)
To do so, we use the AM-GM inequality:
\(a+b+c\geq3\sqrt[3]{abc}\)
We have:
\(\frac{(x+6)+(y+7)+(3x+2y)}{3}\geq\sqrt[3]{(x+6)(y+7)(3x+2y)}\)
Therefore:
\(\frac{4x+3y+13}{3}\geq\sqrt[3]{f(x,y)}\)
Hence:
\(f(x,y)\leq\left(\frac{4x+3y+13}{3}\right)^3\)
But we know that:
\(4x+3y=11\)
Therefore:
\(f(x,y)\leq\left(\frac{11+13}{3}\right)^3=512\)
Hence, the maximum value of \(f(x,y)=(x+6)(y+7)(3x+2y)\) is:
\(\boxed{512}\)
Solution by: Thin Sokkean
Prove that if \(x,y,z\) are positive numbers satisfying \(x^2+y^2=z^2\), then \(x+y+z\) divides \(xyz\).
Solution
From the given condition:
\(x^2+y^2=z^2\)
We can rewrite it as:
\(x^2+y^2+2xy=z^2+2xy\)
\((x+y)^2=z^2+2xy\)
\(2xy=(x+y+z)(x+y-z)\)
We observe that \(x+y+z\) and \(x+y-z\) have the same parity, so \(x+y-z\) is divisible by \(2\).
Let:
\(x+y-z=2k\)
Then:
\(2xy=(x+y+z)2k\)
\(xy=k(x+y+z)\)
Multiplying both sides by \(z\), we get:
\(xyz=kz(x+y+z)\)
Therefore, \(xyz\) is divisible by \(x+y+z\).
Hence, the problem is proved.
Solution by: Thin Sokkean
In other to compare two of number, we have several methods to prove . In here, I will use the division operation to see which number is bigger.
If A is bigger than B, we can write in the fraction A/B is bigger than 1.
Noticed that: All IMO Problems are not too difficult for you to solve, It is just because you either rarely study or never face with kind of those problems. For my opinion, you should research more about those problems to make sure that you have ability to join to contest
Slovakian Math Olympiad | 2015-2016
Noticed that: All IMO Problems are not too difficult for you to solve, It is just because you either rarely study or never face with kind of those problems. For my opinion, you should research more about those problems to make sure that you have ability to join to contest
This is the problem that I picked up from 1985 International Mathematics Olympiad Long List Problems. I will make the solution for you by my Microsoft Word and convert it to images for you to see my solution.
Noticed that: All IMO Problems are not too difficult for you to solve, It is just because you either rarely study or never face with kind of those problems. For my opinion, you should research more about those problems to make sure that you have ability to jion to contest.
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What Is The Remainder When 2022^2020 Is Divided By 20?
This problem is a bit harder If you do not know about basic Theory Numbers as well as Modulo's Theory.
Why those theories are important?
Remember that: All dividing problem which is finding the remainder, you must to understand clear the division of natural number. It starts from 2 to 11 or more than that. Exponential formula is the important once for you to use as the power. After that, modulo's theory as I have uploaded already either in this site or my YouTube Channel.
Here is the solution for you:
Why 1=2
In Mathematics operation and logical 1 is equal to 1, can not equal to 2. But, the question here is 1=2? How do you think about this statements?
It is the common problem that you guy should think about its mistake. For me, I can not believe it. I will find the proof for you which is the statement above is truth.
Let's 100=100
Click Here To Find More Math Problem
As you see in the solution, and the conclusion 1 is equal to 2.
Let's give any idea about this problem and find the mistake of the solution.
Thank you!
Vietnamese's Mathematics Problem
These problems I have researched in many Vietnam Math Books to share in my blog. But, I hope these problems can help you to fulfill your shortages of Math. If you want to be good at Math, the only thing you must do is Do More Math Problems Every Day. Push yourself, motivate yourself to be crazy with study, especially Math.
Let's try to solve these problems:
I will release the link for you to download the book so that you can find the solutions by yourself.
Click Here To Download This Book
I hope my efforts can be a small factor of your studies.
Thank you!
Enjoy your study
Find The Last Digit Number Of 2023 The Power of 2023
There are many way to find the last digit number of complicated Math Problems. But, In here I want to shower the best solution to find the last digit number of every single Mathematics amount by just using Dividing Operation.
If you want to be out-standing with these type of problem, you should understand clearly about Modulo Theory.
Solution:
Here Is The Video That I have Made For You.
