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Stage 7 · Algebra

10 problems · National Olympiad second round; IMO P1/P4 · mathsolympiadprep.com

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

  1. Let a1,a2,⋯ ,ana_1,a_2,\cdots,a_n be a permutation of 1,2,⋯ ,n1,2,\cdots,n. Among all possible permutations, find the minimum of ∑i=1nmin⁡{ai,2i−1}.\sum_{i=1}^n \min \{ a_i,2i-1 \}.

    China Team Selection Test

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  2. Find all triples of positive integers (x,y,z)(x,y,z) that satisfy the equation
    2(x+y+z+2xyz)2=(2xy+2yz+2zx+1)2+2023.2(x+y+z+2xyz)^2=(2xy+2yz+2zx+1)^2+2023.

    USAJMO

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  3. Let a1,a2,⋯ ,a41∈R,a_1,a_2,\cdots,a_{41}\in\mathbb{R}, such that a41=a1,∑i=140ai=0,a_{41}=a_1, \sum_{i=1}^{40}a_i=0, and for any i=1,2,⋯ ,40,∣ai−ai+1∣≤1.i=1,2,\cdots,40, |a_i-a_{i+1}|\leq 1. Determine the greatest possible value of
    (1)a10+a20+a30+a40;(1)a_{10}+a_{20}+a_{30}+a_{40};
    (2)a10⋅a20+a30⋅a40.(2)a_{10}\cdot a_{20}+a_{30}\cdot a_{40}.

    China National Olympiad

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  4. For which positive integers mm does there exist an infinite arithmetic sequence of integers a1,a2,...a_1, a_2, . . . and an infinite geometric sequence of integers g1,g2,...g_1, g_2, . . . satisfying the following properties?

    - an−gna_n - g_n is divisible by mm for all integers n≥1n \ge 1;

    - a2−a1a_2 - a_1 is not divisible by mm.

    Holden Mui

    USAJMO

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  5. Find all positive integers n,k1,…,knn, k_1, \dots, k_n such that k1+⋯+kn=5n−4k_1 + \cdots + k_n = 5n-4 and 1k1+⋯+1kn=1. \frac{1}{k_1} + \cdots + \frac{1}{k_n} = 1.

    Putnam

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  6. The function f(n)f(n) is defined on the positive integers and takes non-negative integer values. f(2)=0,f(3)>0,f(9999)=3333f(2)=0,f(3)>0,f(9999)=3333 and for all m,n:m,n: f(m+n)−f(m)−f(n)=0 or 1. f(m+n)-f(m)-f(n)=0 \text{ or } 1. Determine f(1982)f(1982).

    International Mathematical Olympiad

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  7. Determine all functions f:R→Rf:\mathbb{R}\to\mathbb{R} such that for every pair of real numbers xx and yy,
    f(x+y2)=f(x)+∣yf(y)∣.f(x+y^2)=f(x)+|yf(y)|.

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  8. Express
    ∑k=0n(−1)k(n−k)!(n+k)! \sum_{k=0}^{n} (-1)^k (n-k)!(n+k)!
    in closed form.

    USA IMO

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  9. Determine (with proof) whether there is a subset XX of the integers with the following property: for any integer nn there is exactly one solution of a+2b=na + 2b = n with a,b∈Xa,b \in X .

    USAMO

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  10. Let n>1n > 1 be an integer. Find, with proof, all sequences x1,x2,…,xn−1x_1, x_2, \ldots, x_{n-1} of positive integers with the following
    three properties:
    (a). ; (b). for all ; (c). given any two indices and (not necessarily distinct)
    for which , there is an index such
    that .

    USAJMO

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Answer key — Stage 7 · Algebra

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

  1. ∑i=1nmin⁡{n+1−i,2i−1}\sum_{i=1}^n \min \{ n + 1 - i, 2i-1 \} open
  2. (2,3,3)(2, 3, 3) open
  3. 1010 open
  4. m is not squarefree\text{m is not squarefree} open
  5. n=1,k1=1;n=3,(k1,k2,k3)=(2,3,6);n=4,(k1,k2,k3,k4)=(4,4,4,4)n = 1, k_1 = 1; n = 3, (k_1,k_2,k_3) = (2,3,6); n = 4, (k_1,k_2,k_3,k_4) = (4,4,4,4) open
  6. 660660 open
  7. f(x)=ax for a≥0f(x) = ax \text{ for } a \geq 0 open
  8. ∑k=0n(−1)k(n−k)!(n+k)!=(n!)22+(−1)n(2n+1)!2(n+1)\sum_{k=0}^{n} (-1)^k (n-k)!(n+k)! = \frac{(n!)^2}{2} + \frac{(-1)^n (2n+1)!}{2(n+1)} open
  9. Yes, there does exist a subset X X of the integers such that for any integer n n there is exactly one solution to a+2b=n a + 2b = n with a,b∈X a, b \in X . open
  10. The sequence is (2, 4, 6, , 2n-2 ).\text{The sequence is (2, 4, 6, , 2n-2 ).} open

Problems belong to the competitions that set them. Each question on this sheet names the competition it came from, and the licences the material is used under are listed below. Free to copy for classroom use.

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