Math IOQM

10 × 10 Chessboard: (row, column, (column − row) mod 10) 10 × 10 chessboard: (r, c, (c − r) mod 10) Each square shows its row, column on the first line and its remainder on the second. For example, row 1, column 1 has remainder 0, so its label is 1,1,0 . Highlight remainder group: All groups Show all squares Showing all 100 squares. Tip: click any square to highlight its 10-square group. All squares in a highlighted group lie in different rows and columns. Negative differences: the remainder is computed as ((c - r) % 10 + 10) % 10 . So square (9, 1) has c − r = −8 , giving remainder 2 .

 Q1. S1. First think if a regular pentagon. If you choose any 3 of its vertices they will form an isosceles triangle. Why? Let's say it is ABCDE. Let's say you pick any side like AB or AD. Its distance of any vertex is either the length side of the pentagon or its diagonal length. In any triangle made up of 3 vertices of a pentagon  - by PHP 2 of its sides will have equal length. So it's guaranteed to be isosceles whenever we pick 3 vertices from a pentagon to make a triangle. Now from the regular 20-gon let's group the vertices so that they form regular pentagons: {1,5,9,13,17} {2,6,10,14,18} {3,7,11,15,19} {4,8,12,16,20} We have 17 vertices which are blue or green. WLOG at least 9 are blue. At least 3 of them will fall in one pentagon group above. And they will make an isosceles triangle. H.P. Q2. S2. Try 3x3 You have to place 7 rooks and show that there will be 3 rooks which won't attack each other. Explore the 10x10  chessboard . Group the rooks by c-r mod 10 Fo...

 Q1. S1. If there are infinitely many positive integers then all we have to do is to show one particular kind of family which achieves this. For n^2 + 1 to divide n! we should be able to factorize it into smaller factors which are <= n. Because n^2 + 1 itself is larger than n so it won't be able to divide a single number in the factorial by itself. Is it enough to factorize it into 2 factors which are <= n? That is actually not possible since multiplying them will give at max n^2 but we need n^2 + 1. So we need at least 3 factors <= n. Let's say n-k is a number smaller than n. When we do long division and use remainder theorem we see that it will leave the remainder k^2 + 1. So n^2 + 1 = (n-k)*Q + (k^2 + 1) So Q = (n+k) => n^2 + 1 = (n-k)(n+k) + (k^2 + 1) If (n-k) has to completely divide n^2 + 1 without leaving any remainder then it has to divide (k^2 + 1) as well. Simplest way to achieve that is to do: n-k = k^2 + 1 => n = k^2 + k + 1 => n^2 + 1 = (k^2 + k +...

 Q1. S1. Method 1: a1 + a2 + a3 + a4 + a5  = ai1 + ai2 .. ai5 => (ai1 - a1) + (ai2 - a2) ... + (ai5 - a5) = 0 = even__[1] Now if the product of all the terms in [1] is odd => each of the terms is odd. But if each of the terms is odd then sum has to be odd not even. Hence proof by contradiction. Method 2: PHP WLOG at least 3 numbers of these 5 are even(odd will also work). No matter how you permute them, at least one position will have both numbers even. Hence the product will be even. Q2. S2. There are n possible counts of handshakes: 0,1,2...n-1 But 0 and n-1 can't occur together. So n pigeons and n-1 holes. H.P. Q3. S3. at least 3 of them will leave the same mod wrt 3 (which can be only 0,1) So adding those 3 will give a number divisible by 3. Q4. S4. ab(a-b)(a+b) (a-b) and (a+b) are even. Now we need a power of 5. Let's create 3 remainder groups mod 5. 0 1,4 2,3 If any of the integers is 0 mod 5. Pick that. Else at least 2 integers will belong to the same group. If th...

 Q1. S1. p is odd. why? Since q(p-1) = perfect square p is either 2 or an odd prime. p = 2 => q = perfect square but q is prime. So p has to be odd. Now what? p - q is perfect square => p >= q If p = q then p^2 - p is perfect square. => p(p-1) is perfect square which is not possible since p is a prime and (p-1) can't have p as a factor. => p > q With trial and error we find (3,2) = (p,q) as a valid solution. p - q = a^2___[2] pq - q = b^2__[1] a,b are positive integers. b^2 > a^2 why? pq - q > p - q => pq > p which is true since q >= 2 [1] - [2] => pq - p = b^2 - a^2 p(q-1) = (b-a)(b+a)___[3] p is a prime. It will divide at least one of (b-a) and (b+a). a^ = p - q < p < p^2 => 0 < a < p. b^2 = pq - q < pq < p^2 => 0< b < p. => 0 < b - a < p and => a + b < 2p So p can't divide (b-a), it will have to divide (b+a). But a + b < 2p => a + b = p since that's the only number smaller than 2p and...

Q1. Find all f: R -> R s.t. for all real x,y  f(x + y) + f(xy) = x + y + xy + 2 S1: plug in y = 0 => f(x) + f(0) = x + 2. Plugin x = 0 => f(0) = 1 => f(x) = x + 1 Check the answer in original equation. f(x+y) = x + y + 1 f(xy) = xy + 1 They satisfy the original equation. Q2. Find all f: R -> R s.t. for all real x,y f(2x + y) = f(-y) + 2f(x) + 4y S2. We want to find f(x) but other two f() are in our way. Remove them. 2x + y = -y => -y = x so use this substitution y = -x f(x) = f(x) + 2f(x) + 4y => f(x) = -2y = 2x Check answer in the original equation. It matches. Q3. Find all f: R -> R s.t. for all real x,y f(f(x) + y) = f(x-y) + 2y S3. Making inputs inside f() equal => f(x) + y = x - y  => 2y = x - f(x) => f(x) = x And this matches in the original equation. Q4. Find all f: R -> R s.t. for all real x,y 2f(x) + f(1-x) = 4x + 1 S4. Replace x with 1-z 2f(1-z) + f(z) = 4(1-z) + 1 Now we have 2 vars f(z), f(1-z) and 2 equations. Solve to get f(x)....