Maths Olympiad Prep

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Stage 5 · Mixed

10 problems · AIME late · mathsolympiadprep.com

The answer key prints on its own page at the end.

  1. Veronica observes that 81⋅3=24381 \cdot 3=243 and 81⋅4=32481 \cdot 4=324 and wonders how many numbers mm there are with 10≤m≤9910 \leq m \leq 99 such that 3m=ABC3 m=A B C and 4m=CAB4 m=C A B, with A,BA, B and CC decimal digits (the cases in which one or more of the digits A,B,CA, B, C are equal to zero are also considered valid).

    1. A1
    2. B2
    3. C3
    4. D4
    5. E6

    Progetto Olimpiadi della Matematica - GARA di FEBBRAIO

    Number theory Solution and answer checking →

  2. Teresa the bunny has a fair 8-sided die. Seven of its sides have fixed labels 1,2,…,71, 2, \ldots, 7, and the label on the eighth side can be changed and begins as 11. She rolls it several times, until each of 1,2,…,71, 2, \ldots, 7 appears at least once. After each roll, if kk is the smallest positive integer that she has not rolled so far, she relabels the eighth side with kk. The probability that 77 is the last number she rolls is ab\frac{a}{b}, where aa and bb are relatively prime positive integers. Compute 100a+b100a + b.

    HMMT Spring 2021

    Combinatorics Solution and answer checking →

  3. Paul Erdős was one of the most prolific mathematicians of all time and was renowned for his many collaborations. The Erdős number of a mathematician is defined as follows. Erdős has an Erdős number of 00, a mathematician who has coauthored a paper with Erdős has an Erdős number of 11, a mathematician who has not coauthored a paper with Erdős, but has coauthored a paper with a mathematician with Erdős number 11 has an Erdős number of 22, etc. If no such chain exists between Erdős and another mathematician, that mathematician has an Erdős number of infinity. Of the mathematicians with a finite Erdős number (including those who are no longer alive), what is their average Erdős number according to the Erdős Number Project? If the correct answer is XX and you write down AA, your team will receive max⁡(25−⌊100∣X−A∣⌋,0)\max (25-\lfloor 100|X-A|\rfloor, 0) points where ⌊x⌋\lfloor x\rfloor is the largest integer less than or equal to xx.

    Harvard-MIT November Tournament

    Combinatorics Solution and answer checking →

  4. For positive integers nn, mm, kk, let us write n≡m(modk)n \equiv m \pmod k if n−mn - m is divisible by kk. Let AA be the sum of all positive integers aa less than or equal to 20112011 for which a≡1(mod3)a \equiv 1 \pmod 3, and BB be the sum of all positive integers bb less than or equal to 20112011 for which b≡2(mod3)b \equiv 2 \pmod 3. Find the value of A−BA - B.

    Japan Mathematical Olympiad

    Number theory Solution and answer checking →

  5. Graph theory's Four-Color Theorem says that four colors are enough to color the regions in a plane so that no two adjacent regions receive the same color. The theorem was proved in 1976 by Kenneth Appel and Wolfgang Haken, 124 years after the Four-Color Problem was posed.

    Fermat's Last Theorem in Number Theory was proved by Andrew Wiles in 1995, after 358 years of attempts by generations of mathematicians.

    In 2003, Grigori Perelman completed the proof of a conjecture in topology. Considered as one of the seven millennium prize problems, the conjecture says that the sphere is the only type of bounded three-dimensional surface that contains no holes. Mathematicians worked on this conjecture for almost a century. What is the name of this conjecture that earned Perelman the Fields Medal which he refused to accept in 2006?

    Tenth Philippine Mathematical Olympiad

    Geometry Solution and answer checking →

  6. There are 1024 players, ranked from 1 (most skilled) to 1024 (least skilled), participating in a single elimination tournament. In each of the 10 rounds, the remaining players are paired uniformly at random. In each match, the player with a lower rank always wins, and the loser is eliminated from the tournament.

    For each positive integer n∈[1,1024]n \in [1,1024], let f(n)f(n) be the expected number of rounds that the participant with rank nn participates in. Estimate the minimum positive integer NN such that f(N)<2f(N)<2.

    Submit a positive integer EE. If the correct answer is AA, you will receive max (0,20−⌊∣E−A∣2⌋)\left(0,20-\left\lfloor\frac{|E-A|}{2}\right\rfloor\right) points.

    HMMT November 2024

    Combinatorics Solution and answer checking →

  7. Let r1,r2,…,r7r_{1}, r_{2}, \ldots, r_{7} be the distinct complex roots of the polynomial P(x)=x7−7P(x)=x^{7}-7. Let K=∏1≤i<j≤7(ri+rj)K=\prod_{1 \leq i<j \leq 7}\left(r_{i}+r_{j}\right) that is, the product of all numbers of the form ri+rjr_{i}+r_{j}, where ii and jj are integers for which 1≤i<j≤71 \leq i<j \leq 7. Determine the value of K2K^{2}.

    HMMT November

    Algebra Solution and answer checking →

  8. The dimensions of a television screen are 60 cm×45 cm60~\mathrm{cm} \times 45~\mathrm{cm}. A camera films the entire television, and sends the image back onto the television itself, so that inside this television another one can be seen, and so on. The largest television seen inside the screen has an area equal to half the area of the screen. Assuming that a person watches the television seated at a distance such that they cannot distinguish images with an area smaller than 1 cm21~\mathrm{cm}^2, how many televisions does he see inside the screen?

    Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO

    Algebra Solution and answer checking →

  9. We wish to color the integers 1,2,3,…,101,2,3, \ldots, 10 in red, green, and blue, so that no two numbers aa and bb, with a−ba-b odd, have the same color. (We do not require that all three colors be used.) In how many ways can this be done?

    HMMT February

    Combinatorics Solution and answer checking →

  10. Let ABCDABCD be a quadrilateral such that AB=24AB=24, BC=20BC=20, CD=15CD=15, DA=7DA=7, BD=25BD=25. How long is ACAC?

    1. A18
    2. B14214\sqrt{2}
    3. C20
    4. D21
    5. E24

    Progetto Olimpiadi della Matematica

    Geometry Solution and answer checking →

Answer key — Stage 5 · Mixed

Worked solutions for every problem are on the site, one page per problem.

  1. CC open
  2. 104104 open
  3. 4.654.65 open
  4. 13411341 open
  5. PoincareˊConjecturePoincaré Conjecture open
  6. 350350 open
  7. 117649117649 open
  8. 1111 open
  9. 186186 open
  10. CC open

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