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6

Rename $m\to x$ and $n\to y$ We see $x\geq 3$, $y\geq 1$. Modulu 3 implies $x$ is odd. For $x\leq 5$ we get only $(3,1)$, $(5,3)$. Say $x\geq 6$, then $$3^y\equiv -5\;({\rm mod}\; 64)$$ It is not difficult to see $$3^{11}\equiv -5\;({\rm mod}\; 64)$$ so $3^{y-11}\equiv 1\;({\rm mod}\; 64)$. Let $r=ord_{64}(3)$, then since $\phi(64)=32$, we have (Euler) $$...


5

Since $$(a_1+a_2+\cdots +a_n)^2=(a_1^2+a_2^2+\cdots +a_n^2)+2\sum_{1 \le i < j \le n} a_{i} a_{j}$$ we can write $$\sum_{1 \le i < j \le n} a_{i} a_{j}=\frac{S_n^2-(a_1^2+a_2^2+\cdots +a_n^2)}{2}$$ So, the hint $$ \sum_{1 \le i < j \le n} a_{i} a_{j} \ge \frac{n(n-1)}{2} $$ is equivalent to $$\frac{S_n^2-(a_1^2+a_2^2+\cdots +a_n^2)}{2}\ge \frac{n(n-...


4

Sum up $a_i^2+a_j^2 \geq 2a_i a_j$ for all pairs ($i,j$). You'll get, $$(n-1)\sum_ia_i^2 \geq 2\sum_{i<j}a_ia_j$$ Implying that, $$\left(\sum_i a_i\right)^2 = \sum_i a_i^2 + 2\sum_{i<j}a_ia_j \geq \left(2+\frac{2}{n-1}\right)\sum_{i<j}a_ia_j=\frac{2n}{n-1}\sum_{i<j}a_ia_j\geq n^2$$ $$\implies \sum_ia_i \geq n$$


4

In modulo arithmetic, "division" means multiplying by the multiplicative inverse, e.g., $b = \frac{1}{a}$ means the value which when multiplied by $a$ gives $1$ modulo the value, e.g., $ba \equiv 1 \pmod n$. Note you may sometimes see $b = a^{-1}$ instead to avoid using explicit "division". This works, and gives a unique value, in any cases where the value ...


3

By Holder $$x^2+y^2=(x^2+y^2)\left(\frac{1}{x}+\frac{8}{y}\right)^2\geq\left(\sqrt[3]{x^2\cdot\left(\frac{1}{x}\right)^2}+\sqrt[3]{y^2\cdot\left(\frac{8}{y}\right)^2}\right)^3=125.$$ The equality occurs for $(x^2,y^2)||\left(\frac{1}{x},\frac{8}{y}\right),$ which says that we got a minimal value.


3

The Lemma below shows how modular division is compatible with integer division. In particular $$\,(b,n)=1,\ c = \frac{a}b\in \Bbb Z\ \Rightarrow\,\bmod n\!:\,\ c \equiv \frac{a\bmod n}{b\bmod n} := (a\bmod n)(b\bmod n)^{-1}\qquad\quad $$ Lemma $\ $ If $\,a,b,c,n\in \Bbb Z\,$ and $\,(b,n)=1\,$ then $\ c = a/b\,\Rightarrow\,c\equiv a/b := ab^{-1}\pmod{\!n}\...


2

There must be at least $6$ participants, since the common language spoken by the members of a triple is spoken by no more than half the participants. If there are exactly $6$ participants, then there are ${6\choose3}=20$ triples, and we can assign a different language to each triple. That is, the members of $\{1,2,3\}$ speak language $1$, and no other ...


1

Division does not always work in modular arithmetic. You may be familiar with the fact that $\mathbb{Z}\setminus n\mathbb{Z}$ is an abelian (commutative) group under addition, so we can add and subtract as usual. But this isn't the case with multiplication. Instead, we must take the set $(\mathbb{Z}\setminus n\mathbb{Z})^\times$, the set of congruence ...


1

Look at the minimum of sizes of each datum. Let $m_i$ be the minimum of sizes of bacteria at $i$th observation. If $m_i+m_{i+2}=2m_{i+1}$ for some $i$, it means that at least one bacterium had size $m_i$ at time $i$ and has growth rate $m_{i+1}-m_i$ since if the $(i+1)$th minimum $m_{i+1}$ was not made by the bacterium it means there was another bacterium ...


1

For $(a,b,c)=(40,5,3)$ we obtain $k<80.$ We'll prove that $k=79$ is valid, for which we need to prove that $$a^2+b^3+c^4+2019\geq79(a+b+c)$$ or $$(a^2-79a+1561)+(b^3-79b+271)+(c^4-79c+187)\geq0,$$ which is true by AM-GM. Can you end it now? The play with these numbers you can make by the following way. You got that for $a=\frac{k}{2}$, $b=\sqrt{\frac{...


1

Let $x=\frac{2}{3}a$, $y=\frac{1}{3}b$ and $z=\frac{2}{9}c$. Thus, $a+b+c=3$ and by AM-GM we obtain: $$\sqrt{\frac{3yz}{3yz + x}} + \sqrt{\frac{3zx}{3zx + 4y}} + \sqrt{\frac{xy}{xy + 3z}}=\sum_{cyc}\sqrt{\frac{bc}{bc+3a}}=$$ $$=\sum_{cyc}\sqrt{\frac{bc}{(a+b)(a+c)}}\leq\frac{1}{2}\sum_{cyc}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)=\frac{1}{2}\sum_{cyc}\left(\...


1

Hint: $$a^3-a=(a-1)a(a+1)$$ is divisible by $6$ being the product of three consecutive integers $$\implies a^3+b^3+c^3+\cdots\equiv a+b+c+\cdots\pmod6$$


1

Since $x^3\equiv_3 x$ for each integer $x$ we have $$3|a+b+c \iff a+b\equiv_3 -c$$ $$\iff (a+b)^3\equiv_3 (-c)^3$$ $$\iff a^3+3ab(a+b)+b^3\equiv_3 -c^3$$ $$\iff a^3+b^3+c^3\equiv_3 0 $$ $$\iff 3\mid a^3+b^3+c^3$$


1

Let's rewrite the question a bit, remembering that $5 = 2^5-3^3$ and putting this in the basic equation such that we start with: $$ 2^m -2^5 = 3^n - 3^3 \tag 1$$ $$ { 2^M-1 \over 3^3} = { 3^N-1 \over 2^5} \tag 2$$ $\qquad \qquad \qquad $ where $m=5+M$ and $n=3+N$. For $M=N=0$ this is our largest known solution. We'll prove now, that assuming $...


1

Here's an alternative argument. Suppose that this is not the case, and the number of people is $n$. If everyone spoke at least three languages then by double counting some language would be spoken by at least $3n/4$. So there is someone who speaks exactly two languages, A and B, say. There is some set $X$ of more than $2n/5$ who don't speak $A$, and a set $Y$...


1

Let $a = \frac{x}{y}, \ b = \frac{y}{z}, \ c = \frac{z}{x}; \ x, y, z > 0$. It suffices to prove that $f(x, y, z)\ge 0$ where \begin{align*} f(x,y, z) &= 4\, x^7\, z^5 - 3\, x^6\, y^3\, z^3 + 4\, x^6\, y^2\, z^4 + 4\, x^5\, y^7 - 3\, x^5\, y^5\, z^2 - 3\, x^5\, y^2\, z^5 + 4\, x^4\, y^6\, z^2 - 6\, x^4\, y^4\, z^4\\ &\quad - 3\, x^3\, y^6\, z^3 - ...


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