Determine all three-digit numbers 
having the property that is divisible by 11, and 
is equal to the sum of the squares of the digits of .
Let 
for some digits 
and 
. Then![\[100a + 10b+c = 11m\]](https://latex.artofproblemsolving.com/f/1/2/f1290e6e100fc16188bc4b670ad244f670a79ba5.png)
for some 
. We also have 
. Substituting this into the first equation and simplification, we get
is irreducible for every natural number. Cambodian Olympiad Math is written by Lim Sovanvichet . This book is shared on social media by many students which is very important document for all Khmer students. In this book, there are many International Math Problems which he translated from other foreign books and wrote them in to Khmer language.
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This is a document that I research from the internet and rewrite it to upload at my website, I think It can help you do Math with different language which makes you to be better at Math. Why Vietnames students can challenge with other students around the world.
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When is \(n^2+2021n\) a perfect square?
Let \(m,n\in\mathbb{N}\) such that \(n^2+2021n=m^2\). Since \(2021=43\times47\), we have
\[ 4n^2+4\times2021n=4m^2 \]
Adding \(2021^2\) to both sides,
\[ (2n+2021)^2-4m^2=2021^2 \]
Therefore,
\[ (2n-2m+2021)(2n+2m+2021)=2021^2. \]
Case 1: \(2021^2=1\times2021^2\).
Then \(2n-2m+2021=1\) and \(2n+2m+2021=2021^2\). Adding both equations,
\[ 4n+2\times2021=2021^2+1\Rightarrow4n=(2021-1)^2, \]
so \(n=1020100\).
Case 2: \(2021^2=43\times(43\times47)^2\).
Then \(2n-2m+2021=43\) and \(2n+2m+2021=(43\times47)^2\). Solving similarly, we get \(n=22747\).
Hence, \(n^2+2021n\) is a perfect square when
\[ \boxed{n=1020100\ \text{or}\ n=22747}. \]
Here you can download the PDF file (Solution by Thin Sokkean)Find all positive integers \(n\) such that \(n^8+n+1\) is a prime.
 |
Irish Math Olympiad 2009
SolutionLet \(f(x)=x^8+x+1\). If \(w\) is the third root of unity \((w=e^{2i\pi/3})\), then \(w^2+w+1=0\) and \(w^3=1\), so \(w^8=w^2\). Therefore, \(f(w)=w^8+w+1=w^2+w+1=0\). Hence, \(w\) is a root of both \(f(x)=x^8+x+1\) and \(x^2+x+1\), so \[ f(x)=(x^2+x+1)g(x). \] We have \[ x^8+x+1=(x^2+x+1)(x^6-x^5+x^3-x^2+1). \] Since \(f(1)=1^8+1+1=3\), \(n=1\) gives a prime number. For \(n>1\), \[ n^8+n+1=(n^2+n+1)(n^6-n^5+n^3-n^2+1). \] Both factors are greater than \(1\), so \(n^8+n+1\) is composite for all \(n>1\). Therefore, the only positive integer satisfying the condition is \[ \boxed{n=1}. \] Solution by: Thin Sokkean |
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Prove that \(\frac12+\frac15+\frac18+\frac1{11}+\frac1{20}+\frac1{40}+\frac1{110}+\frac1{1640}=1\).
1. We observe that
\[ \frac12+\frac15+\frac18+\frac1{11}+\frac1{20}+\frac1{40}+\frac1{110}+\frac1{1640} =\frac12+\frac1{20}+\frac15+\frac18+\frac1{11}+\frac1{110}+\frac1{40}+\frac1{1640} \] \[ =\frac{22}{40}+\frac{13}{40}+\frac{11}{110}+\frac{41}{1640} =\frac{22}{40}+\frac{13}{40}+\frac4{40}+\frac1{40} =\frac{40}{40}=1. \]
Hence, \(\frac12+\frac15+\frac18+\frac1{11}+\frac1{20}+\frac1{40}+\frac1{110}+\frac1{1640}=1\).
2. Find the minimum value of \(A=|x-1|+|x-2|+\cdots+|x-100|\).
Using \(|a|+|b|\geq|a-b|\), we have
\[ |x-k|+|x-(101-k)|\geq|101-2k|,\quad k=1,2,\ldots,50. \]
Therefore,
\[ |x-1|+|x-100|\geq99,\ |x-2|+|x-99|\geq97,\ldots,\ |x-50|+|x-51|\geq1. \]
Adding all inequalities,
\[ A\geq99+97+\cdots+1=\frac{50}{2}(1+99)=2500. \]
The equality holds when \(x\in\bigcap_{k=1}^{50}[k,101-k]=[50,51]\).
Hence,
\[ \boxed{A_{\min}=2500\quad\text{when }x\in[50,51].} \]
This book is related to Sequence Problem, the tips to solve Sequence problem, which is written in Khmer language. I hope this book is used for all of you.
Let Z be the set of integers. Determine all function `f: Z \rightarrow Z` such that, for all integers `a` and `b`
`f(2a)+2f(b)=f(f(a+b))` `(1)`
Answer: The solution are `f(n)=0` and `f(n)=2n+k` for any constant `k\in Z`
Substituting `a =0, b= n+1 ` gives `f(f(n+1))=f(0)+2f(n+1)`.
Substituting `a =1, b= n` gives `f(f(n+1))=f(2)+2f(n)` .
In particular, `f(0)+2f(n+1)=f(2)+2f(n)`, and so `f(n+1)-f(n)=1/2(f(2)-f(0))` .
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
