Online math open (OMO) 2012 2017 fall, EN with solutions

The Online Math Open Fall Contest September 24-October 1, 2012 Contest Information Format The test will start Monday September 24 and end Monday October You will have until 7pm EST on October to submit your answers The test consists of 30 short answer questions, each of which has a nonnegative integer answer The problem difficulties range from those of AMC problems to those of Olympiad problems Problems are ordered in roughly increasing order of difficulty Team Guidelines Students may compete in teams of up to four people Participating students must not have graduated from high school International students may participate No student can be a part of more than one team The members of each team not get individual accounts; they will all share the team account Each team will submit its final answers through its team account Though teams can save drafts for their answers, the current interface does not allow for much flexibility in communication between team members We recommend using Google Docs and Spreadsheets to discuss problems and compare answers, especially if teammates cannot communicate in person Teams may spend as much time as they like on the test before the deadline Aids Drawing aids such as graph paper, ruler, and compass are permitted However, electronic drawing aids are not allowed This is includes (but is not limited to) Geogebra and graphing calculators Published print and electronic resources are not permitted (This is a change from last year’s rules.) Four-function calculators are permitted on the Online Math Open That is, calculators which perform only the four basic arithmetic operations (+-*/) may be used Any other computational aids such as scientific and graphing calculators, computer programs and applications such as Mathematica, and online databases is prohibited All problems on the Online Math Open are solvable without a calculator Four-function calculators are permitted only to help participants reduce computation errors Clarifications Clarifications will be posted as they are answered For the Fall 2012-2013 Contest, they will be posted at here If you have a question about a problem, please email OnlineMathOpenTeam@gmail.com with “Clarification” in the subject We have the right to deny clarification requests that we feel we cannot answer Scoring Each problem will be worth one point Ties will be broken based on the highest problem number that a team answered correctly If there are still ties, those will be broken by the second highest problem solved, and so on Results After the contest is over, we will release the answers to the problems within the next day If you have a protest about an answer, you may send an email to OnlineMathOpenTeam@gmail.com (Include “Protest” in the subject) Solutions and results will be released in the following weeks September/October 2012 Fall OMO 2012-2013 Page Calvin was asked to evaluate 37 + 31 × a for some number a Unfortunately, his paper was tilted 45 degrees, so he mistook multiplication for addition (and vice versa) and evaluated 37 × 31 + a instead Fortunately, Calvin still arrived at the correct answer while still following the order of operations For what value of a could this have happened? Petya gave Vasya a number puzzle Petya chose a digit X and said, “I am thinking of a number that is divisible by 11 The hundreds digit is X and the tens digit is Find the units digit.” Vasya was excited because he knew how to solve this problem, but then realized that the problem Petya gave did not have an answer What digit X did Petya chose? Darwin takes an 11 × 11 grid of lattice points and connects every pair of points that are unit apart, creating a 10 × 10 grid of unit squares If he never retraced any segment, what is the total length of all segments that he drew? Let lcm(a, b) denote the least common multiple of a and b Find the sum of all positive integers x such that x ≤ 100 and lcm(16, x) = 16x Two circles have radius and 26 The smaller circle passes through center of the larger one What is the difference between the lengths of the longest and shortest chords of the larger circle that are tangent to the smaller circle? An elephant writes a sequence of numbers on a board starting with Each minute, it doubles the sum of all the numbers on the board so far, and without erasing anything, writes the result on the board It stops after writing a number greater than one billion How many distinct prime factors does the largest number on the board have? Two distinct points A and B are chosen at random from 15 points equally spaced around a circle centered at O such that each pair of points A and B has the same probability of being chosen The probability that the perpendicular bisectors of OA and OB intersect strictly inside the circle can be expressed in the form m n , where m, n are relatively prime positive integers Find m + n In triangle ABC let D be the foot of the altitude from A Suppose that AD = 4, BD = 3, CD = 2, and AB is extended past B to a point E such that BE = Determine the value of CE Define a sequence of integers by T1 = and for n ≥ 2, Tn = 2Tn−1 Find the remainder when T1 + T2 + · · · + T256 is divided by 255 10 There are 29 unit squares in the diagram below A frog starts in one of the five (unit) squares on the top row Each second, it hops either to the square directly below its current square (if that square exists), or to the square down one unit and left one unit of its current square (if that square exists), until it reaches the bottom Before it reaches the bottom, it must make a hop every second How many distinct paths (from the top row to the bottom row) can the frog take? September/October 2012 Fall OMO 2012-2013 Page 11 Let ABCD be a rectangle Circles with diameters AB and CD meet at points P and Q inside the rectangle such that P is closer to segment BC than Q Let M and N be the midpoints of segments AB and CD If ∠M P N = 40◦ , find the degree measure of ∠BP C 12 Let a1 , a2 , be a sequence defined by a1 = and for n ≥ 1, an+1 = a2n − 2an + + Find a513 13 A number is called 6-composite if it has exactly composite factors What is the 6th smallest 6composite number? (A number is composite if it has a factor not equal to or itself In particular, is not composite.) 14 When Applejack begins to buck trees, she starts off with 100 energy Every minute, she may either choose to buck n trees and lose energy, where n is her current energy, or rest (i.e buck trees) and gain energy What is the maximum number of trees she can buck after 60 minutes have passed? 15 How many sequences of nonnegative integers a1 , a2 , , an (n ≥ 1) are there such that a1 · an > 0, n−1 a1 + a2 + · · · + an = 10, and (ai + ai+1 ) > 0? i=1 16 Let ABC be a triangle with AB = 4024, AC = 4024, and BC = 2012 The reflection of line AC over line AB meets the circumcircle of ABC at a point D = A Find the length of segment CD 17 Find the number of integers a with ≤ a ≤ 2012 for which there exist nonnegative integers x, y, z satisfying the equation x2 (x2 + 2z) − y (y + 2z) = a 18 There are 32 people at a conference Initially nobody at the conference knows the name of anyone else The conference holds several 16-person meetings in succession, in which each person at the meeting learns (or relearns) the name of the other fifteen people What is the minimum number of meetings needed until every person knows everyone elses name? 19 In trapezoid ABCD, AB < CD, AB ⊥ BC, AB CD, and the diagonals AC, BD are perpendicular at point P There is a point Q on ray CA past A such that QD ⊥ DC If AP QP + = AP QP AP BP − can be expressed in the form then AP BP m + n m n 51 14 − 2, for relatively prime positive integers m, n Compute 20 The numbers 1, 2, , 2012 are written on a blackboard Each minute, a student goes up to the board, chooses two numbers x and y, erases them, and writes the number 2x + 2y on the board This continues until only one number N remains Find the remainder when the maximum possible value of N is divided by 1000 21 A game is played with 16 cards laid out in a row Each card has a black side and a red side, and initially the face-up sides of the cards alternate black and red with the leftmost card black-side-up A move consists of taking a consecutive sequence of cards (possibly only containing card) with leftmost September/October 2012 Fall OMO 2012-2013 Page card black-side-up and the rest of the cards red-side-up, and flipping all of these cards over The game ends when a move can no longer be made What is the maximum possible number of moves that can be made before the game ends? 22 Let c1 , c2 , , c6030 be 6030 real numbers Suppose that for any 6030 real numbers a1 , a2 , , a6030 , there exist 6030 real numbers {b1 , b2 , , b6030 } such that n an = bgcd(k,n) k=1 and bn = cd an/d d|n for n = 1, 2, , 6030 Find c6030 23 For reals x ≥ 3, let f (x) denote the function f (x) = √ −x + x 4x − Let a1 , a2 , , be the sequence satisfying a1 > 3, a2013 = 2013, and for n = 1, 2, , 2012, an+1 = f (an ) Determine the value of 2012 a3i+1 a1 + a2 + ai+1 + a2i+1 i=1 i 24 In scalene ABC, I is the incenter, Ia is the A-excenter, D is the midpoint of arc BC of the circumcircle of ABC, and M is the midpoint of side BC Extend ray IM past M to point P such that IM = M P Let Q be the intersection of DP and M Ia , and R be the point on the line M Ia such that AR DP QM m a Given that AI AI = 9, the ratio RIa can be expressed in the form n for two relatively prime positive integers m, n Compute m + n 25 Suppose 2012 reals are selected independently and at random from the unit interval [0, 1], and then written in nondecreasing order as x1 ≤ x2 ≤ · · · ≤ x2012 If the probability that xi+1 − xi ≤ 2011 for m i = 1, 2, , 2011 can be expressed in the form n for relatively prime positive integers m, n, find the remainder when m + n is divided by 1000 26 Find the smallest positive integer k such that x + kb 12 ≡ x 12 (mod b) for all positive integers b and x (Note: For integers a, b, c we say a ≡ b (mod c) if and only if a − b is divisible by c.) 27 Let ABC be a triangle with circumcircle ω Let the bisector of ∠ABC meet segment AC at D and circle ω at M = B The circumcircle of BDC meets line AB at E = B, and CE meets ω at P = C The bisector of ∠P M C meets segment AC at Q = C Given that P Q = M C, determine the degree measure of ∠ABC September/October 2012 28 Find the remainder when Fall OMO 2012-2013 216 16 2k (3 · 214 + 1)k (k − 1)2 −1 k k=1 is divided by 16 Page + (Note: It is well-known that 216 + = 65537 is prime.) 29 In the Cartesian plane, let Si,j = {(x, y) | i ≤ x ≤ j} For i = 0, 1, , 2012, color Si,i+1 pink if i is even and gray if i is odd For a convex polygon P in the plane, let d(P ) denote its pink density, i.e the fraction of its total area that is pink Call a polygon P pinxtreme if it lies completely in the region S0,2013 and has at least one vertex on each of the lines x = and x = 2013 Given that the minimum value of d(P ) over all non-degenerate convex pinxtreme polygons P in the plane can be expressed in the form √ (1+ p)2 q2 for positive integers p, q, find p + q 30 Let P (x) denote the polynomial 1209 xk + k=0 146409 xk + k=10 xk k=1210 Find the smallest positive integer n for which there exist polynomials f, g with integer coefficients satisfying xn − = (x16 + 1)P (x)f (x) + 11 · g(x) September/October 2012 Fall OMO 2012-2013 Acknowledgments Contest Directors Ray Li, James Tao, Victor Wang Head Problem Writers Ray Li, Victor Wang Problem Contributors Ray Li, James Tao, Anderson Wang, Victor Wang, David Yang, Alex Zhu Proofreaders and Test Solvers Mitchell Lee, James Tao, Anderson Wang, David Yang, Alex Zhu Website Manager Ray Li Page The Online Math Open January 16-23, 2012 Contest Information Format The test will start Monday January 16 and end Monday January 23 The test consists of 50 short answer questions, each of which has a nonnegative integer answer The problem difficulties range from those of AMC problems to those of Olympiad problems Problems are ordered in roughly increasing order of difficulty Team Guidelines Students may compete in teams of up to four people Participating students must not have graduated from high school International students may participate No student can be a part of more than one team The members of each team not get individual accounts; they will all share the team account The team will submit their final answers through their account Though teams can save drafts for their answers, the current interface does not allow for much flexibility in communication between team members We recommend using Google Docs and Spreadsheets to discuss problems and compare answers, especially if teammates cannot communicate in person Teams may spend as much time as they like on the test before the deadline Aids Drawing aids such as graph paper, ruler, and compass are permitted However, electronic drawing aids are not allowed This is includes (but is not limited to) Geogebra and graphing calculators Published print and electronic resources are permitted Four-function calculators are permitted on the Online Math Open That is, calculators which perform only the four basic arithmetic operations (+-*/) may be used No other computational aids such as scientific and graphing calculators, computer programs and applications such as Mathematica, and online databases are permitted All problems on the Online Math Open are solvable without a calculator Four-function calculators are permitted only to help participants reduce computation errors Scoring Each problem will be worth one point Ties will be broken based on the highest problem number that a team answered correctly If there are still ties, those will be broken by the second highest problem solved, and so on Results After the contest is over, we will release the answers to the problems within the next day If you have a protest about an answer, you may send an email to onlinemathopen2011@gmail.com (Include ”Protest” in the subject) Solutions and results will be released in the following weeks January 2012 OMO 2012 Page The average of two positive real numbers is equal to their difference What is the ratio of the larger number to the smaller one? How many ways are there to arrange the letters A, A, A, H, H in a row so that the sequence HA appears at least once? A lucky number is a number whose digits are only or What is the 17th smallest lucky number? How many positive even numbers have an even number of digits and are less than 10000? Congruent circles Γ1 and Γ2 have radius 2012, and the center of Γ1 lies on Γ2 Suppose that Γ1 and Γ2 intersect at A and B The line through A perpendicular to AB meets Γ1 and Γ2 again at C and D, respectively Find the length of CD Alice’s favorite number has the following properties: • It has distinct digits • The digits are decreasing when read from left to right • It is divisible by 180 What is Alice’s favorite number? A board 64 inches long and inches high is inclined so that the long side of the board makes a 30 degree angle with the ground √ The distance from the highest point on the board to the ground can be expressed in the form a + b c where a, b, c are positive integers and c is not divisible by the square of any prime What is a + b + c? An × × cube is painted red on faces and blue on faces such that no corner is surrounded by three faces of the same color The cube is then cut into 512 unit cubes How many of these cubes contain both red and blue paint on at least one of their faces? At a certain grocery store, cookies may be bought in boxes of 10 or 21 What is the minimum positive number of cookies that must be bought so that the cookies may be split evenly among 13 people? 10 A drawer has pairs of socks Three socks are chosen at random If the probability that there is a pair among the three is m n , where m and n are relatively prime positive integers, what is m + n? 11 If 1 1 1 + 2+ 3+ 4+ + ··· = , x 2x 4x 8x 16x 64 and x can be expressed in the form m n , where m, n are relatively prime positive integers, find m + n 12 A cross-pentomino is a shape that consists of a unit square and four other unit squares each sharing a different edge with the first square If a cross-pentomino is inscribed in a circle of radius R, what is 100R2 ? OMO Spring 2017 Official Solutions Solution Consider two cases: Case 1: the two lines intersect on the positive y-axis Without loss of generality, assume that a1 < a3 < < a2 < a4 , and let p = −a1 , q = a2 , r = −a3 , s = a4 It is not difficult to see that the line v −u2 through (−u, u2 ) and (v, v ) has slope v−(−u) = v − u, and intersect the y-axis at the point (0, uv), which implies that pq = rs If both (or neither) of p < q and r < s are true, then both lines have positive (or negative) slopes, which means that θ < 45◦ Otherwise, we assume p > q and r < s, and therefore θ = 180◦ − tan−1 (p − q) − tan−1 (s − r) Since tan−1 (1) = 45◦ and thus θ < 90◦ , we should make both p − q and s − r as small as possible Since p, q, r, s are all distinct, it is not difficult to see that |(p − q) − (s − r)| ≥ Moreover, we can make one of them and the other by setting (p, q, r, s) = (6, 2, 3, 4) (and θ > 45◦ ), but (p − q, s − r) = (1, 3) or (3, 1) is impossible since it would imply that p, q, r, s are consecutive integers in some order, which 1+4 = 53 cannot satisfy pq = rs Therefore, the maximum possible value of tan θ in this case is − 1−1·4 Case 2: the two lines intersect on the non-positive y-axis In this case, we can see that all four points lie on the same side of y-axis, and so the slopes of both lines are both integers and have the same sign, and thus θ < 45◦ , which we need not consider since we have already found a bigger θ Therefore, the maximum possible value of tan θ is 35 , and the answer is 503 12 Alice has an isosceles triangle M0 N0 P , where M0 P = N0 P and ∠M0 P N0 = α◦ (The angle is measured in degrees.) Given a triangle Mi Nj P for nonnegative integers i and j, Alice may perform one of two elongations: −−→ • an M -elongation, where she extends ray P Mi to a point Mi+1 where Mi Mi+1 = Mi Nj and then removes the point Mi −−→ • an N -elongation, where she extends ray P Nj to a point Nj+1 where Nj Nj+1 = Mi Nj and then removes the point Nj After a series of elongations, k of which were M -elongations, Alice finds that triangle Mk N5−k P is an isosceles triangle Given that 10α is an integer, compute 10α Proposed by Yannick Yao Answer 264 ∠Mi+1 = ∠Mi ∠Nj m n After one M -elongation, we see that m/2 m m i+1 ∠Mi /2 and ∠Nj = ∠Nj + ∠Mi /2, which means that ∠M ∠Nj = n+m/2 = 2n+m = n+(n+m) one N -elongation will change the ratio to m+(m+n) Notice that since initially the ratio is 11 , n Solution For each triangle P Mi Nj , consider the ratio = Similarly, and each elongation will double the sum of numerator and denominator, the sum of the numerator and denominator after t elongations is 2t+1 (and since both numerator and denominator are odd, this also implies that they will always be relatively prime) Moreover, if we express both m and n in binary, the t-th elongation will add a at the (t + 1) − th digit from the right on either m or n It is not difficult to see that any pair of (m, n) that are both odd with sum 2t+1 is achievable with some combinations of M - and N - elongations After elongations, the sum of the numerator and denominator is 26 = 64, and since the final triangle is isosceles, ∠P must be equal to one of the other two angles (since m = n starting with the first elogation, ∠Mi = ∠Nj ) Therefore, the sum of 64 and one of the m and n needs to be an odd factor of 10 · 180 = 1800, and since this sum is between 64 + = 65 and 64 + 63 = 127, the only possibility is 64 + 11 = 75, achieved when (m, n) = (11, 53) or (53, 11) Thus we can see that the desired ratio is 11 α = 180 · 75 = 26.4 and therefore 10α = 264 OMO Spring 2017 Official Solutions 13 On a real number line, the points 1, 2, 3, , 11 are marked A grasshopper starts at point 1, then jumps to each of the other 10 marked points in some order so that no point is visited twice, before returning to point The maximal length that he could have jumped in total is L, and there are N possible ways to achieve this maximum Compute L + N Proposed by Yannick Yao Answer 28860 Solution It’s not difficult to see that the longest length is L = 2+4+6+8+10+10+8+6+4+2 = 60, by counting the maximal number of times each unit segment gets covered Moreover, one can show that each jump must either jump from one of the first points to the last points, or vice versa, or jumping to/from point (The point before and after must also belong to different sides) Using this, we may ultimately compute that the number of ways is N = 2(5!)2 = 28800 Therefore the answer is N + L = 28860 14 Let ABC be a triangle, not right-angled, with positive integer angle measures (in degrees) and circumcenter O Say that a triangle ABC is good if the following three conditions hold: • There exists a point P = A on side AB such that the circumcircle of P OA is tangent to BO • There exists a point Q = A on side AC such that the circumcircle of QOA is tangent to CO • The perimeter of AP Q is at least AB + AC Determine the number of ordered triples (∠A, ∠B, ∠C) for which ABC is good Proposed by Vincent Huang Answer 59 Solution The first two conditions imply that ∠OBP = ∠BAO = ∠P OB, which means that BP O is similar to triangle BOA, and analogously triangle CQO is similar to COA This requires that AB, AC > R or equivalently, ∠B, ∠C > 30◦ We can also see that BP = P O and similarly CQ = QO Therefore AB +AC = AP +P O+QO+AQ ≥ AP + P Q + AQ, so ABC is good if and only if equality holds, i.e P, O, Q are collinear in that order In particular, O is on or inside triangle ABC so ABC is acute or right But collinearity in this order occurs if and only if ∠P OB + ∠BOC + ∠COQ = 180◦ , i.e (90◦ − ∠C) + 2∠A + (90◦ − ∠B) = 180◦ , which is equivalent to ∠A = 60◦ Since ∠B, ∠C > 30◦ , ∠B can range from 31◦ to 89◦ , resulting in 59 good triangles 15 Let φ(n) denote the number of positive integers less than or equal to n which are relatively prime to n Over all integers ≤ n ≤ 100, find the maximum value of φ(n2 + 2n) − φ(n2 ) Proposed by Vincent Huang Answer 72 Solution It’s well known that φ(n2 ) = nφ(n) When n is odd, φ(n2 + 2n) = φ(n)φ(n + 2), and when n is even, φ(n2 + 2n) = 2φ(n)φ(n + 2) Let f (n) = φ(n2 + 2n) − φ(n2 ) Then when n is even, f (n) = φ(n)[2φ(n + 2) − n] Since n + is even, φ(n + 2) ≤ n+2 with equality if and only if n + is a power of When n + is not a power of 2, then f (n) ≤ When n + is a power of 2, f (n) = 2φ(n) ≤ n Since n + ≤ 102, the largest power of that n + can obtain is 64, giving f (n) ≤ 62 When n is odd, f (n) = φ(n)[φ(n + 2) − n] Note that φ(n + 2) ≤ n + with equality if and only if n + is a prime When n + is not a prime, then f (n) ≤ When n + is a prime, f (n) = φ(n) ≤ n OMO Spring 2017 Official Solutions Note that when n + = 101, 97, 89, 83, 79, we get that f (n) = 60, 72, 56, 54, 60, respectively Otherwise, n + ≤ 73 =⇒ f (n) ≤ 71 Hence the maximum value of f (n) is achieved when n = 95, giving an answer of 72 16 Let S denote the set of subsets of {1, 2, , 2017} For two sets A and B of integers, define A ◦ B as the symmetric difference of A and B (In other words, A ◦ B is the set of integers that are an element of exactly one of A and B.) Let N be the number of functions f : S → S such that f (A◦B) = f (A)◦f (B) for all A, B ∈ S Find the remainder when N is divided by 1000 Proposed by Michael Ren Answer 112 Solution Consider each element A of S as a 2017-dimensional vector vA with entries in F2 , such that the ith entry of vA equal to if i ∈ A and otherwise Define wA similarly with respect to f (A) Note that we have the condition wA+B = wA◦B = wA ◦ wB = wA + wB , so the problem now becomes determining the number of linear maps on F2017 By setting A = B = ∅, we have w∅ = w∅+∅ = w∅ + w∅ = 0, so the empty set must map to itself Moreover, the vectors v{1} , v{2} , , v{2017} are the basis of F2017 , so one can assign each of w{1} , w{2} , , w{2017} to any one of the 22017 vectors which also 2 determines all other mappings consistently (because of linearity) Thus there are (22017 )2017 = 22017 possible functions, and we can reduce 22017 ≡ 289 ≡ (27 )12 · 25 ≡ 312 · 32 ≡ 112 (mod 125) and thus 22017 ≡ 112 (mod 1000) A more combinatorial way to phrase this solution would be to note that empty set must go to empty set by setting A and B to be the empty set and that setting the outputs of {1}, {2}, , {2017} uniquely determines the entire function 17 Let ABC be a triangle with BC = 7, AB = 5, and AC = Let M, N be the midpoints of sides AC, AB respectively, and let O be the circumcenter of ABC Let BO, CO meet AC, AB at P and Q, respectively If M N meets P Q at R and OR meets BC at S, then the value of OS can be written in the form m n where m, n are relatively prime positive integers Find 100m + n Proposed by Vincent Huang Answer 240607 Solution By the Law of Cosines, ∠A = 60◦ Since ∠BOC = 120◦ = 180◦ − A we know A, P, O, Q are concyclic Then the Simson line of O with respect to triangle AP Q must be line M N , which meets P Q at R, implying OR ⊥ P Q Now define S as the point on BC with B, S , O, Q concyclic By Miquel’s Theorem on triangle ABC and points S , P, Q, we know that C, P, O, S are concyclic as well It’s easy to see ∠QS O = ∠QBO = ∠OAQ = ∠OP Q = 90◦ − ∠C and similarly we deduce 90◦ − ∠B = ∠P QO = ∠P S O, implying O is the orthocenter of triangle P QS , hence OS ⊥ P Q Therefore S = S Furthermore, from the angle-chasing we know that SP Q ∼ ABC Let H be the orthocenter of ABC Since ∠A = 60◦ it’s well-known and easy to prove that AH = AO Therefore by the similarity, OS is equal to the circumradius of P QS Let R, R be the circumradii of triangles ABC, P OQ Since O is the orthocenter of P QS we know that (P OQ), (P SQ) are congruent, so it suffices to find the circumradius of (AP OQ) By the Law of Sines in AP O, we know that R = AO = 2R sin ∠AP O = 2R sin(C + 30◦ ) √ √ 11 By standard methods we can compute [ABC] = 10 3, R = √73 , cos C = 14 , sin C = 5143 , so it’s not hard to find OS = 2401 507 , yielding an answer of 240607 OMO Spring 2017 Official Solutions 18 Let p be an odd prime number less than 105 Granite and Pomegranate play a game First, Granite picks a integer c ∈ {2, 3, , p − 1} Pomegranate then picks two integers d and x, defines f (t) = ct + d, and writes x on a sheet of paper Next, Granite writes f (x) on the paper, Pomegranate writes f (f (x)), Granite writes f (f (f (x))), and so on, with the players taking turns writing The game ends when two numbers appear on the paper whose difference is a multiple of p, and the player who wrote the most recent number wins Find the sum of all p for which Pomegranate has a winning strategy Proposed by Yang Liu Answer 65819 Solution Let’s say that c, d are already chosen Let f0 be the sequence defined by f0 = x and d d fi+1 = cfi + d Then fi = − c−1 · ci To prevent losing, Pomegranate would of course first + x + c−1 d choose and x = − c−1 (Or else fi is a constant sequence) Otherwise, the sequence fi repeats with period equal to ordp (c) So for Granite to win, he needs ordp (c) to be odd Since p > c > (a condition), we need for p − to have an odd factor > This happens unless p is a Fermat prime So the sum of all possible primes is + + 17 + 257 + 65537 = 65819 19 For each integer ≤ j ≤ 2017, let Sj denote the set of integers ≤ i ≤ 22017 − such that an odd integer Let P be a polynomial such that  1≤j≤2017 is  1 − P (x0 , x1 , , x22017 −1 ) = i 2j−1 xi  i∈Sj Compute the remainder when P (x0 , , x22017 −1 ) ∈{0,1}22017 (x0 , ,x22017 −1 ) is divided by 2017 Proposed by Ashwin Sah Answer 1840 Solution First of all, the set Sj is exactly the set of all integers in [0, 22017 − 1] whose j-th rightmost digit in binary is odd The value of P is equal to if and only if for each j, there is at least one integer i ∈ Sj such that xi = If we consider a bijection between all the integers in [0, 22017 − 1] with the set of all subsets T0 , T1 , , T22017 −1 of T = {1, 2, , 2017} such that j ∈ Ti if and only if i ∈ Sj , and consider each tuple (x0 , x1 , , x22017 −1 ) as a way of choosing a subset of {T0 , T1 , , T22017 −1 } (where correspond to chosen and correspond to not chosen), then for P to be equal to 1, the union of these chosen subsets must be equal to T itself Therefore it suffices to count the number of ways to pick such a collection of subsets The number of ways to pick a collection of subsets whose union is a subset of a fixed (2017 − s)-element 2017−s 2017−s 2017 22 ≡ subset is equal to 22 So by PIE, we find that the number of ways is s=0 (−1)s 2017 s 2017 2017 2 −2 (mod 2017) We notice ≡ (mod 2016) so that we find −2 ≡ 1840 (mod 2017) 20 Let n be a fixed positive integer For integer m satisfying |m| ≤ n, define Sm = i−j=m 0≤i,j≤n 2 lim S−n + S−n+1 + + Sn2 n→∞ 2i+j Then OMO Spring 2017 Official Solutions can be expressed in the form p for relatively prime positive integers p, q Compute 100p + q q Proposed by Vincent Huang Answer 8027 Solution Let = an a1 )2 + (an a0 )2 We wish to consider the expression (a0 an )2 +(a0 an−1 +a1 an )2 + +(an−1 a0 + 2i Each parenthesis consists of terms of the form aj with i − j fixed So if we expand, we get something of the form aj ak al The key observation is that we can also write this sum in the form i−j=k−l aj ak al , and grouping these terms by the value of i + l = j + k, the expression becomes i+l=j+k (a0 a0 )2 + (a0 a1 + a1 a0 )2 + + (an−1 an + an an−1 )2 + (an an )2 = (n − 1)2 + + 2n n+2 4 As n approaches infinity, this sum approaches i≥0 (n + 1)2 n2 + + + + n+1 + n 4 4 80 (i + 1)2 , which evaluates to by standard methods, i 27 so the answer is 8027 21 Let Z≥0 be the set of nonnegative integers Let f : Z≥0 → Z≥0 be a function such that, for all a, b ∈ Z≥0 : f (a)2 + f (b)2 + f (a + b)2 = + 2f (a)f (b)f (a + b) Furthermore, suppose there exists n ∈ Z≥0 such that f (n) = 577 Let S be the sum of all possible values of f (2017) Find the remainder when S is divided by 2017 Proposed by Zack Chroman Answer 1191 Solution Note that P (0, 0) =⇒ f (0) = Then, letting f (1) = k, P (1, m) =⇒ f (m + 1)2 − 2kf (m)f (m + 1) = − k − f (m)2 By P (1, m − 1), this quadratic is satisfied by f (m − 1), so either f (m + 1) = f (m − 1) or f (m + 1) = 2kf (m) − f (m − 1) If f (2) = 1, f (3) is k in both cases, and iterating this we see that the function goes 1, k, 1, k, 1, k, It turns out this function works, so we can have f (2017) = 577 by taking k = 577 Otherwise, we have f (2) = 2k − Then, f (3) ∈ {k, 4k − 3k} If f (3) = k, f (4) is one of 2k − and However, P (2, 2) =⇒ f (4) = or f (4) = 2f (2)2 − = 8k − 8k + If this is also 2k − 1, solving the resulting quadratic gives k = ±1, but then f (2) = which puts us back in the first case Then we take f (4) = Then we immediately get f (5) = f (3) = k, and P (4, 2) tells us that f (6) = 2k − 1, not In general, we get f repeats the sequence {1, k, 2k − 1, k} Noting that 577 = · 172 − 1, and 2017 ≡ (mod 4), this gives the possible value of 17 for f (2017) In the final case, we take f (3) = 4k − 3k I claim that from here on out, f (n) is defined as f (n) = 2kf (n − 1) − f (n − 2) This is clear upon computing P (1, n − 1) − P (2, n − 2), which is linear in f (n) It remains to determine which of these sequences 577 belongs to We already know that k = 17 works, and clearly k = 577 works It turns√out f (4) = 577 when k = 3, and these are all the cases Noting √ that f (n) = 21 (3 + 2)n + (3 − 2)n gives f (2017) ≡ (mod 2017), and similarly when k = 17 and k = 577, f (2017) ≡ 17 (mod 2017) and f (2017) ≡ 577 (mod 2017) respectively But in this case f (2017) = 17 or 577, so they need to be included separately The sum of all answers modulo 2017 is 577 + 17 + (3 + 17 + 577) = 1191 OMO Spring 2017 Official Solutions 22 Let S = {(x, y) | −1 ≤ xy ≤ 1} be a subset of the real coordinate plane If the smallest real number that is greater than or equal to the area of any triangle whose interior lies entirely in S is A, compute the greatest integer not exceeding 1000A Proposed by Yannick Yao Answer 5828 Solution It is clear that the boundary of S is composed of two hyperbolas, namely xy = and xy = −1 For ease of reference, we call the four branches of the two hyperbolas by the quadrant number they lie in (i.e branch 1, branch 2, etc.) Note that the affine transformation (x, y) → (kx, y/k) preserves both hyperbola and any area for any positive number k Also note that the triangle with maximal area necessarily has at least one side tangent to the boundary of S (otherwise we would have three vertices on three different branches, and we could always slide one of them away from the opposite side, increasing the area until one side is tangent), so we may assume that the side is tangent at T (1, 1) (after rotation and the aforementioned transformation) √ √ √ √ The line x + y = intersects branch and at points P (1 − 2, + 2) and Q(1 + 2, − 2) respectively, and if we let R = (−1, −1), we may see that P R and QR are also tangent to branch and branch at P and Q respectively From this, we may deduce that if we choose P on the interior of P T and Q on the interior of QT , then the tangent line to branch through P and the tangent line to branch through Q always intersect outside S (in Quadrant 3) As a result, we can always choose a point R on branch such that either R P is tangent to branch or R Q is tangent to branch WLOG assume the former case is true, then we can make sure that R is to the left of R, and so either (1) we can move Q to Q and get a larger area, or (2) in the process of doing (1), we got stuck at a point where R Q is tangent to branch We show that (2) is impossible In fact, if we the affine transformation to make the tangent point on branch (−1, 1), then similar to the argument in the preceeding paragraph, we can show that Q needs to be outside S since it’s the intersection of two tangent lines, which contradicts the assumption of (2) itself Therefore, a maximal triangle necessarily have two vertices on the adjacent branches With this in mind, we restart and WLOG let these two vertices be A(a, 1/a) (on branch 1) and √ B(b, −1/b) (on branch 4) Notice that the tangent line to branch through A has equation y = 3+2 a2 (x − a) + a , √ √ √ which intersects branch at (( − 1)a, −( + 1)/a), implying that ab ≥ − Similarly we can √ √ +b2 show that ab ≥ − Therefore, we have a ab = ab + ab ∈ [2, 2] It is obvious that the maximal triangle when points A and B are fixed has the third vertex C being the intersection of the line tangent to branch through A and the line tangent to branch through B (If we place C to the right of AB where AC and BC are tangent to branch and branch respectively, then we may extend rays CA and CB to intersect branch and branch at A and B respectively √ and get a larger area.) Since the equations of the two tangent lines are y = 3−2 (x − a) + a1 and a2 √ (x − b) − 1b , we get that the intersection is y = −3+2 b2 √ ab(a + b) √ b−a C = (−(2 + 2) , (2 − 2) ) a +b a + b2 And the area is equal to √ 1 a2 + b2 8ab ((b − xC )( − yC ) − (a − xC )(− − yC )) = ( + 2) + a b ab a + b2 Due to the result we established in the previous paragraph, it is not difficult to see that√the right-hand √ +b2 side is maximized when a ab = 2, or when a = b, and the maximum is 21 (2 + 82 + 2) = + 2 This maximum √ can be achieved√with the triangle whose vertices are at A = (1, 1), B = (1, −1), and C = (−2 − 2, 0) Since + 2 ≈ 5.8284, the final answer is 5828 OMO Spring 2017 Official Solutions 23 Determine the number of ordered quintuples (a, b, c, d, e) of integers with ≤ a < b < c < d < e ≤ 30 for which there exist polynomials Q(x) and R(x) with integer coefficients such that xa + xb + xc + xd + xe = Q(x)(x5 + x4 + x2 + x + 1) + 2R(x) Proposed by Michael Ren Answer 5208 Solution Work in F2 First, we claim that x5 + x4 + x2 + x + is irreducible If it were reducible, it would have to be divisible by an irreducible polynomial of degree or Thus, we only have to consider divisibility by x, x + 1, and x2 + x + But x5 + x4 + x2 + x + = x(x + 1)2 (x2 + x + 1) + 1, so it is not divisible by any of those, as desired Now consider a root z of x5 + x4 + x2 + x + Since x5 + x4 + x2 + x + is irreducible, z is an element of F32 Note that the order of a nonzero element of F32 divides 31, so it is either or 31 Since the only element with order is 1, z must have order 31, which means that it is a primitive root in F32 Hence, 1, z, z , , z 30 are the nonzero elements of F32 Furthermore, x5 +x4 +x2 +x+1 | xa +xb +xc +xd +xe if and only if z a + z b + z c + z d + z e = Hence, we just want to find the number of ways distinct elements of F32 can add to We will first suppose that they are ordered and divide by 5! at the end Note that we can just choose random distinct elements and the fifth will be uniquely determined (it is actually their sum) This results in 31 · 30 · 29 · 28 ways The only catch is that the fifth element might be the same as one of the first four or To resolve this, we first count the number of ways distinct elements of F32 can add to By choosing random distinct elements and taking their sum as the third, we have that there are 31 · 30 ways It is not possible for their sum to be equal to one of them, because that implies that the other is Now, the number of overcounted quintuples is simply 31 · 30 · · 28, since there are ways to choose three of the first four entries to sum to and 28 ways to choose the element for the remaining entry and the last entry Also, note that the number of ways for four elements to sum to is 31 · 30 · 29 − 31 · 30 by a similar argument to what we had before Hence, our answer is 31·30·29·28−31·30·4·28−31·30·29+31·30 = 31·30·24·28 = 5208 5! 120 24 For any positive integer n, let Sn denote the set of positive integers which cannot be written in the form an + 2017b for nonnegative integers a and b Let An denote the average of the elements of Sn if the cardinality of Sn is positive and finite, and otherwise Compute ∞ An 2n n=1 Proposed by Tristan Shin Answer 840 Solution Let m = 2017 It is clear that An > if and only if (m, n) = and n ≥ Now, fix n that works I first claim that an integer k is not in Sn if and only if we can express k = xm + yn with x, y ∈ N0 and x ≤ n − The if direction is trivial Assume that the only if direction is false Take k = x0 m + y0 n with x0 ≥ n and let z = xn0 We then have that k = x0 m + y0 n = (x0 − nz) m + (y0 + mz) n It is obvious that x0 − nz ∈ N0 , so this is a contradiction (take x = x0 − nz and y = y0 + mz) Thus, the claim is true Now, what does this mean? This means that in the expression + xm + x2m + + x(n−1)m + xn + x2n + = 10 − xmn · , − xm − xn OMO Spring 2017 Official Solutions the generating function for the set of k with k = xm + yn for x, y ∈ N0 and x ≤ n − 1, is also the generating function for the integers that are in Sn Thus, 1 − xmn − − x (1 − xm ) (1 − xn ) is the generating function for the integers that are in Sn (Side note: combining these terms into one fraction and comparing the degrees of the numerator and denominator gives us the so-called Chicken McNugget Theorem.) Thus, if − xmn − , − x (1 − xm ) (1 − xn ) F (x) = then |Sn | = lim F (x) and the sum of the elements of Sn is lim F (x) x→1 x→1 We have that n−1 n−1 xmi n−1 xmi + xmi+1 i=0 i=0 − i=0 n = F (x) = 1−x 1−x − x − xn + xn+1 To find the limit of F as x → 1, we apply L’Hopital’s rule twice: − xn − n−1 n−1 −n (n − 1) xn−2 + (n + 1) nxn−1 x→1 = i=1 i=1 lim F (x) = lim x→1 mi (mi + 1) xmi−1 mi (mi − 1) xmi−2 + −n (n − 1) xn−2 − n−1 2n −n (n − 1) + 2m i = i=1 (−n (n − 1) + mn (n − 1)) 2n (m − 1) (n − 1) = Therefore, the size of Sn is |Sn | = (m−1)(n−1) Now, rewrite F as n−1 xmi F (x) = + i=0 1−x xn − Then n−1 n−1 mixmi−1 F (x) = (x − 1) xmi (xn − 1) − n i=0 i=1 + (xn − 1) n−1 2 i x n−1 + i=0 mix i=0 xmi (x − 1) − n xn−1 i=0 n−1 n−1 (mi − n) xmi+n−1 − + n (xn − 1) xi xn−1 (x − 1) n−1 mi−1 i=1 n−1 xmi i=0 (x − 1) (xn − 1) n−1 = (xn − 1) (x − 1) − n i=1 = = n−1 mixmi−1 (xn − 1) + xn−1 i=1 mixmi−1 − nxn−1 i=1 x2n − 2xn + 11 OMO Spring 2017 Official Solutions It follows from applying L’Hopital’s rule twice that n−1 ix i−1 n−1 i=1 n−1 xi +2 n−1 i (i − 1) xi−2 i=0 (mi − n) (mi + n − 1) (mi + n − 2) xmi+n−3 + i=2 i=1 n−1 mi (mi − 1) (mi − 2) xmi−3 − n (n − 1) (n − 2) xn−3 − i=1 lim F (x) = lim x→1 2n (2n − 1) x2n−2 − 2n (n − 1) xn−2 x→1  n2 (n − 1) n2 (n − 1) (2n − 1) + − n (n − 1) − n (n − 1) (n − 2)     = 2  n−1 2n   + (mi − n) (mi + n − 1) (mi + n − 2) − mi (mi − 1) (mi − 2)  i=1 = (n − 1) (n − 1) (2n − 1) n − (n − 1) (n − 2) + − − + 2n 2n n−1 m2 ni2 − mn2 i − n (n − 1) (n − 2) i=1 (n − 1) (2n − 1) n − (n − 1) (n − 2) m (n − 1) (2n − 1) (n − 1) + − − + 2n 12 = mn (n − 1) (n − 1) (n − 2) − − 2n (m − 1) (n − 1) (2mn − m − n − 1) = 12 after mass simplification Thus, the sum of the elements of Sn is is 2mn−m−n−1 (m−1)(n−1)(2mn−m−n−1) , 12 so the average of the elements of Sn Thus, if y = 12 , ∞ ∞ An (2m − 1) n − (m + 1) n m − = y − y− n 6 n=1 n=1 ∞ =− ∞ k=1 (2m − 1) mk − (m + 1) mk y ∞ m − 2m − m+1 m (2m − 1) + ny n − yn − 12 6 n=1 n=1 ∞ k k (y m ) + k=1 m+1 ∞ k (y m ) k=1 m − (2m − 1) y (m + 1) y m (2m − 1) y m (m + 1) y m =− + − − + 2 12 (1 − y) (1 − y m ) (1 − y) (1 − y m ) m m − 2m − m + m (2m − 1) m+1 =− + − − + m 12 6 (2m − 1) (2 − 1) As − m − 2m − m + 5m − + − = = 840 + 12 12 12 and − 2017 · 4033 · 22017 2018 11 Then, 98 891 98 · 891 87318 100 11 by the theory of Farey Sequences, the fraction with the smallest denominator between and is 891 98 111 111 100 1 111 , and − = = > The largest fraction in F4734 that is less than 989 989 891 989 · 891 881199 989 100 + · 111 433 433 100 is = Note that − = < Hence, the desired fraction in S is 891 + · 989 3858 3858 891 1145826 3858 3858 = Hence, our answer is 3862291 433 + 3858 4291 3858 989 Note: Our given expression is about 0.89909092, while ≈ 0.89909112 and ≈ 0.89909091 4291 1100 √ √ 29 Let ABC be a triangle with AB = 6, BC = 5, CA = 26, midpoint M of BC, circumcircle Ω, and orthocenter H Let BH intersect AC at E and CH intersect AB at F Let R be the midpoint of EF and let N be the midpoint of AH Let AR intersect the circumcircle of AHM again at L Let the 16 OMO Spring 2017 Official Solutions circumcircle of AN L intersect Ω and the circumcircle of BN C at J and O, respectively Let circles AHM and JM O intersect again at U , and let AU intersect the circumcircle of AHC again at V = A m The square of the length of CV can be expressed in the form for relatively prime positive integers n m and n Find 100m + n Proposed by Michael Ren Answer 1376029 Solution Consider Ψ, the inversion at A with power AH ∗ AD I claim that Ψ(G) = S, Ψ(M ) = Q This is pretty straightforward to verify, in particular noting that Ψ(BC) = (AH) Then Ψ(L) = AR ∩ Ψ(H)Ψ(M ) = AR ∩ DQ Call this point La Then since Ψ(N ) is the reflection of A over BC, call it Na , Ψ(J) = Na La ∩ EF lemma : Ψ(J) lies on AM This would imply that J = AM ∩ Ω proof : First, note that B, H, Q, C, Na are cyclic on the reflection γ of Ω over BC Then, −1 = H (D, S; B, C) = (Na , Q; B, C)γ = (A, Q∗; B, C)Ω Here Q∗ is the reflection of Q over BC, which by the above lies on the A−symmedian In particular, A, R, Q collinear Let I = AR ∩ BC ∩ Na Q Then, working over RP2 , N L Q M (A, Q; Ψ(J), M ) =a (D, I; Na La ∩BC, M ) =a (D, A; Na , M La ∩AD) = (I, A; Q , La ) = (Na , A; ∞AD , D) = −1 Therefore, Ψ(J) lies on the polar of M with respect to circle (AEHF ), which is just line EF , as desired This completes that lemma As a corollary, note that J is the reflection of Q over M Lemma : O is the reflection of R over M Proof : By repeated power of a point, M N ∗ M O = AM ∗ M J = M B ∗ M C = M E = M R ∗ M N Lemma : J, M, O, S are cyclic on the circle of diameter M S proof : By the previous two lemmas, it suffices to show that S, R, Q, M, G are cyclic on the circle of diameter (SM ) This follows from inversion with respect to the circle of diameter (BC), which sends S → D, R → N, Q → A, G → H Lemma : AG, AU are isogonal in ∠BAC Proof : Let AU intersect (JM O) again at U1 , and let M1 be the antipode of M on (AHM ) Note that M , U, S collinear Then ∠S M U1 = ∠S U U1 = ∠AU M = ∠AM M = ∠DM H = ∠SM G Therefore, since G lies on (SM ), we have U1 = G , the reflection of G over the perpendicular bisector of BC Since GG BC, and G, G ∈ Ω, it follows that lines AG, AG = AU are isogonal in ∠BAC, and in particular BG = CG Lemma : BG = CV Proof : Consider the circle at C with radius BG, call it ω By the previous lemma U1 = G lies on this circle Clearly so does the reflection G of G over AC G lies on (AHC) since G lies on (ABC) Let P be the second intersection of ω, (AHC) By construction ∠P AC = ∠CAG P ∈ AG Therefore, P = V , and V ∈ ω, as desired Since SBG and SAC are similar by antiparallels, BG = will find BS, AC, and AS in terms of x, y, and z Note that DM · DS = DB · DC, so DS = x2 + 4y z (y−z)2 = 2 2 x (y−z) +4y z (y−z)2 2 2yz z−y BS·AC AS = ∠CAG , so Let AD = x, BD = y, CD = z We Then BS = DS − y = y +yz z−y , and AS = DS + AD2 = We also clearly have AC = x2 + z +yz) (x +z ) 2 Thus, BG2 = (y x2 (y−z)2 +4y z Now it suffices to compute x, y, z We have y+z = and y −z = −2, so 23 27 529 y−z = − This means that y = 10 and z = 10 The Pythagorean Theorem gives x = 24− 100 = 1871 100 17 OMO Spring 2017 Official Solutions Now, we just have to plug everything in for the answer, which is 232 ·252 ·26 393125 = 232 ·26 629 = 13754 629 232 ·502 ·26 1871·42 +4·232 ·272 = 232 ·252 ·26 1871·4+232 ·272 = Therefore the answer is 1376029 30 Let p = 2017 be a prime Given a positive integer n, let T be the set of all n × n matrices with entries in Z/pZ A function f : T → Z/pZ is called an n-determinant if for every pair ≤ i, j ≤ n with i = j, f (A) = f (A ), where A is the matrix obtained by adding the jth row to the ith row Let an be the number of n-determinants Over all n ≥ 1, how many distinct remainders of an are (pp − 1)(pp−1 − 1) possible when divided by ? p−1 Proposed by Ashwin Sah Answer 12106 Solution Notice that prime (pp −1)(pp−1 −1) p−1 = (pp−1 − 1) pp −1 p−1 , and that those two terms are relatively Let B, C be two rows of the matrix Then an n-determinant remains constant when we transform them to (B + C, C) and (B + 2C, C), and so on Thus (B + (p − 1)C, C) is equivalent, so (B − C, C) is as well and then (B − C, B) is equivalent so (−C, B) is as well after adding p − times Thus we can ”swap rows,” to some extent Now if xy ≡ (mod p) then we can send (B, C) to (B + xC, C) to (B + xC, C − y(B + xC)) = (B + xC, −yB) to (B + xC + x(−yB), −yB) = (xC, −yB) to (yB, xC) Thus we can ”scale rows,” to some extent Thus, we can easily perform steps equivalent to row reduction and reduce any matrix to a near-reduced row echelon form with the exact same n-determinant value If the rows are linearly dependent, we will obtain an all zero row and using it to scale other rows, we can reduce fully to reduced row echelon form Otherwise, we can reduce the matrix to the identity matrix, except the upper left element is the nonzero determinant of the original matrix, since the operations we perform preserve the determinant Furthermore, since the reduced row echelon form is unique, we can easily show that any matrix reduces to precisely one of these end matrices Thus the number of functions as desired is simply pSn , where Sn is the number of such final matrices n−1 −(k Simple counting shows that S = (p − 1) + p 2) s , counting the nonzero determinant matrices n k=0 k with p−1 and counting the other rref forms with the latter sum, where sk = a1 , ,ak ∈{0,1, ,n−1} pa1 +···+ak where we sum over distinct (s0 = since we must count the all zero matrix.) n Then pSn (mod pp−1 − 1) reduces to p2 (mod −1 (mod pp−1 − 1) while pSn (mod p pp −1 p−1 ) reduces to pn−1 p −1 p−1 ) Now the value of n (mod p) determines the latter value - and all p possibilities are distinct -, while the value of 2n (mod p − 1) determines the former There are p possibilities for the first For the second, let v = v2 (p − 1) and d = ord p−1 (2) Then n = 1, , v − give v − distinct values while 2v n = v, v + 1, , v + d − give d values which repeat forever This gives a total of pd + (v − 1) residues for the answer by the Chinese Remainder Theorem a bunch of times, since gcd(d, p) = Now for p = 2017 we have v = 5, d = Thus we find 2017 · + = 12106 18 ... the Online Math Open Any other computational aids, including scientific calculators, graphing calculators, or computer programs is prohibited All problems on the Online Math Open are solvable without... the Online Math Open Any other computational aids, including scientific calculators, graphing calculators, or computer programs is prohibited All problems on the Online Math Open are solvable without... the Online Math Open Any other computational aids, including scientific calculators, graphing calculators, or computer programs is prohibited All problems on the Online Math Open are solvable without