S2T2

Jim Randell 8:16 am on 7 February 2025 Permalink Reply
Tags: by: Andrew Skidmore ( 85 )   

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  Jim Randell 8:16 am on 7 February 2025 Permalink | Reply

  The number of teams goes:

  n → n + (k + 1) → n + (2k + 1)

  where n is a 2-digit number.

  So the current season has an odd number of teams (≥ 3) more than the number of teams two seasons ago.

  Each pair of teams plays twice, so for n teams the number of matches in a season is:

  M(n) = 2 C(n, 2) = n (n − 1)

  and if the numbers of matches are A → B → C we have:

  C > 3A
  C − B = A; or: A + B = C

  This Python program looks for the first possible solution.

  It runs in 66ms. (Internal runtime is just 38µs).

  from enigma import (irange, inf, printf)
  
  # number of matches for n teams = 2 * C(n, 2)
  M = lambda n: n * (n - 1)
  
  def solve():
    # consider a league with c teams (current season)
    for c in irange(13, inf):
      C = M(c)
      # consider an odd number of teams less than this (two seasons ago)
      for a in irange(c - 3, 10, step=-2):
        A = M(a)
        if 3 * A >= C or a > 99: continue
        b = (a + c + 1) // 2
        B = M(b)
        if A + B == C:
          yield ((a, b, c), (A, B, C))
  
  # find the first solution
  for ((a, b, c), (A, B, C)) in solve():
    printf("{a} teams, {A} matches -> {b} teams, {B} matches -> {c} teams, {C} matches")
    break
  

  Solution: There are now 17 teams in the league.

  And so there are 272 matches in the current season.

  Two years ago there were 10 teams in the league, and the season consisted of 90 matches. (The current season has more than 3× this number of matches).

  One year ago there were 14 teams in the league, and the season consisted of 182 matches, and 182 + 90 = 272.

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  • 

    Frits 4:07 pm on 7 February 2025 Permalink | Reply

    Further analysis (solving a quadratic equation) leads to :

    c = (5a + 1) / 3

    and C greater than 3 * A means a^2 -11a + 1 less than 0.
    a =10 is still valid but already at a = 11 the condition is not met.

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    • 

      John Crabtree 4:58 am on 14 February 2025 Permalink | Reply

      Frits, nicely done. It is useful to know the intermediate result, ie n = 3k – 2

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