1959 AHSME Problems/Problem 50
Revision as of 10:13, 20 July 2020 by Duck master (talk | contribs) (created page for real! (I confused this page initially with the "Astrophysics" page.))
A club with members is organized into four committees in accordance with these two rules:
Then :
Solution
We can label each of the members with the pair of committees that they belong to, which is clearly valid due to rule (1). Then, by rule (2), for each pair of committees, there is exactly one member labeled with that pair. But since we have four committees, there must be members in total. Thusly our choice is , and we are done.
See also
1959 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 49 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.