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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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