Solve in positive integers the equation \[ n = \varphi(n) + 402 , \] where $ \varphi(n)$ is the number of positive integers less than $ n$ having no common prime factors with $ n$.

Click for solution Substitute $ n$ by $ P_{1}^{\alpha_{1}} \cdot P_{2}^{\alpha_{2}} \cdots P_{x}^{\alpha_{x}}$. As we know, $ \varphi(n) = P_{1}^{\alpha_{1} - 1}\cdot P_{2}^{\alpha_{2} - 1}\cdots P_{x}^{\alpha_{x} - 1}$. So the givven equation rewrites as the following: ${ P_{1}^{\alpha_{1} - 1}\cdot P_{2}^{\alpha_{2} - 1}\cdots P_{x}^{\alpha_{x} - 1}(P_{1}P_{2}\cdots P_{x} - (P_{1} - 1)(P_{2} - 1)\cdots(P_{x} - 1}) = 402$. $ (1)$. If none of the numbers $ P_{1},P_{2},...,P_{x}$ is $ 2$, than in $ (1)$ equality $ LHS$ is odd, while $ 402$ is even-contradiction. So, wlog $ P_{1} = 2$. If $ \alpha_{1} > 1$, than in $ (1)$ equality $ LHS$ is divisible by $ 4$, while $ 402$ isn't divisible by $ 4$. So, $ \alpha_{1} = 1$. Assume there exists the number $ \alpha_{i}$ such that $ \alpha_{i} > 1$. As $ P_{1}^{\alpha_{1} - 1}\cdot P_{2}^{\alpha_{2} - 1}\cdots P_{x}^{\alpha_{x} - 1}|402 = 2\cdot 3\cdot 67$, $ P_{i}$ is either $ P_{i} = 3$ or $ P_{i} = 67$. $ n$ is even. So, $ \varphi(n) < = \frac {n}{2}$, giving $ 402 < = n < = 804$. If wlog $ P_{2} = 67$, than the last inequality gives: $ 3 < = P_{3}\cdots P_{x} < = 6$ which follows $ P_{3 } = P_{x} = 5$. So,$ n = P_{1}P_{2}P_{3} = 2\cdot 67\cdot 5 = 670$. $ \varphi(n) = 132$, but $ 670$ isn't equal to $ 132 + 402$-contradiction. If $ P_{2} = 3$,than the same argument leads us to that: $ 67 < = P_{3} \cdot P_{x} < = 134$. By checking all possible numbers instead of $ P_{3},P_{4},...,P_{x}$, using the fact that if $ i$ doesn't equal to $ j$,where $ 2 < = i,j < = x$, than $ P_{i}$ doesn't equal to $ P_{j}$, we make sure there aren't the solutions to the given equation. Now it remains to take a look to the case $ \alpha_{1} = \alpha_{2} = ... = \alpha{x} = 1$. Our equation rewrites so: $ 2P_{2}P_{3}\cdots P_{x} = (P - {1} - 1)(P_{2} - 1)\cdots(P_{x} - 1) + 402$. $ (2)$. Now we see the two possible cases: $ (a)$ $ x$ is odd; $ (b)$ $ x$ is even; $ (a)$ $ x = 2m + 1$. ($ m$ is positive integer, $ m > = 2$(If $ m = 1$,we have no solutions. If $ m = 0$, we have no solutions). Wlog assume $ P_{1} < P_{2} < ... < P_{x}$. If $ P_{m + 1} > = 19$, than $ 804 > = n > = 2P_{m + 1}P_{m + 2} > = 2\cdot 19\cdot 23 > 804$-contradiction. So, $ P_{m + 1} < = 17$. If $ P_{m + 1} = 17$, than $ 804 > = n > = 2\cdot 3 \cdot 17 \cdot 19 > 804$-contradiction. If $ P_{m + 1} = 13$, than $ 804 > = 2 \cdot 13 \cdot 17 > 804$-contradiction. If $ P_{m + 1} = 11$, than $ 804 > = n > = 2 \cdot 3 \cdot 11 \cdot 13 > 804$-contradiction. If $ P_{m + 1} = 7$, $ 804 > = n > = P_{m - 1}P_{m}P_{m + 1}P_{2m + 1} > = 2 \cdot 3 \cdot 7 \cdot 23 > 804$-contradiction, if $ P_{2m + 1} > = 23$. If not, checking all the cases we get the solution $ n = 546$. If $ p_{m + 1} = 5$, than $ P_{m} = 3$ and $ P_{m - 1} = 2 = P_{1}$.So, $ m = 2$. $ 804 > = P_{1}P_{2}P_{3}P_{4}P_{5} > = 2 \cdot 3 \cdot 5 \cdot7 \cdot 11 > 804$-contradiction. If $ P_{m + 1} = 3$, $ P_{m} = P_{1}$ giving $ m = 1$, which we already checked. (b) $ x = 2m$. If $ m = 1$, $ n = 802$ is the solution. Assume $ m > = 2$. By the same argument as in case $ (a)$ we get there is no more solution. Finally, we have two possible solutions:$ 802$ and $ 546$.

Let $ ABC$ be a triangle and $ K$ and $ L$ be two points on $ (AB)$, $ (AC)$ such that $ BK = CL$ and let $ P = CK\cap BL$. Let the parallel through $ P$ to the interior angle bisector of $ \angle BAC$ intersect $ AC$ in $ M$. Prove that $ CM = AB$.

Click for solution I have a strange solution. WLOG $AB\geq AC$ Let $M$ be a point on ray $CA$ such that $AB=CM$, $l$ is a line parallel through $M$ to angle bisectore of $\angle BAC$ Denote $N, L$ be the intersections of line $l$ and $AB, BC$ respectively. Let's think a point $X$ on edge $ML$. We'll call $P, Q$ as the intersection of $BX, AC$ and $CX, AB$ respectively. And Let $S(X)= CP$, $R(X) = BQ$. It's obvious that $S(X), R(X)$ is the 2-dimension functions of $X$. cuz $S(X)=R(X)$ for $X=L, N$, we can know that two functions are same. Therefore, edge $ML$ is the trace of $P$. $Q.E.D.$

Let $ m\geq n\geq 4$ be two integers. We call a $ m\times n$ board filled with 0's or 1's good if 1) not all the numbers on the board are 0 or 1; 2) the sum of all the numbers in $ 3\times 3$ sub-boards is the same; 3) the sum of all the numbers in $ 4\times 4$ sub-boards is the same. Find all $ m,n$ such that there exists a good $ m\times n$ board.

In a pile you have 100 stones. A partition of the pile in $ k$ piles is good if: 1) the small piles have different numbers of stones; 2) for any partition of one of the small piles in 2 smaller piles, among the $ k + 1$ piles you get 2 with the same number of stones (any pile has at least 1 stone). Find the maximum and minimal values of $ k$ for which this is possible.

Click for solution If $k \geq 14$, then there are at least $1+2+\ldots+14=105>100$ stones, contradiction. So $k \leq 13$. We can take $\{1, 2, \ldots, 11, 12, 22 \}$ for $k=13$; it's easy to check that this works. So the maximum $k$ is $13$. Note that $k=10$ works, taking $\{1, 3, 5, \ldots, 19 \}$: there are $100$ stones, and if we partition a pile, we get an odd number and an even number; the odd number is among the smaller piles. Now we want to show that $k<10$ doesn't work. Suppose that there's no 1-stone pile. Then, since $n=(n-1)+1$, if there's a pile with $n$ stones there must be one with $n-1$. So, if $n$ is number of stones in the biggest pile, there are $n+(n-1)+\ldots+2=(n+1)n/2-1$ stones, which can't be $100$. Then there's a pile with just $1$ stone. Suppose that the biggest pile has $2m+1$ stones. Then at least one of $1, 2m$, one of $2, 2m-1$, ..., one of $m, m+1$ must be present (*), so at least $m+1$ piles in total. If the biggest pile has $2m$ stones, there are at least $m$ piles. Note that (*) can be applied to any pile, in fact. Now, suppose $k \leq 8$. Then the biggest pile has at most $2k=16$ stones. So there are at most $1+16+15+\ldots+10=92<100$ stones, contradiction. Then $k \geq 9$. If $k=9$, the biggest pile has at most $18$ stones. Take the smallest pile $>10$; by (*), there are at least 5 piles $\leq 10$. If the biggest pile has $17$ or fewer stones, there are at most $1+7+8+9+10+14+15+16+17=97<100$ stones, so it has $18$. However, the smallest pile $>1$ can be at most $4$ by (*), so in any case there are at most $1+4+8+9+10+15+16+17+18=98$ stones. Then $k>9$.

Let $ a,b,c,d$ be real numbers with sum 0. Prove the inequality: \[ (ab + ac + ad + bc + bd + cd)^2 + 12\geq 6(abc + abd + acd + bcd). \]

Let $ ABCDEF$ be a convex hexagon such that $ AD = BC + EF$, $ BE = AF + CD$, $ CF = DE + AB$. Prove that: \[ \frac {AB}{DE} = \frac {CD}{AF} = \frac {EF}{BC}. \]