Difference between revisions of "2000 AMC 8 Problems/Problem 5"
Loverslane22 (talk | contribs) (Created page with '==Problem== Each principal of Lincoln High School serves exactly one <math>3</math>-year term. What is the maximum number of principals this school could have during an <math>8<…') |
(→See Also) |
||
(6 intermediate revisions by 6 users not shown) | |||
Line 10: | Line 10: | ||
term, then in the next six years two more principals would serve, and the last year of the | term, then in the next six years two more principals would serve, and the last year of the | ||
period would be the first year of the fourth principal's term. Therefore, the maximum | period would be the first year of the fourth principal's term. Therefore, the maximum | ||
− | number of principals who can serve during an <math>8</math>-year period is <math>4</math> | + | number of principals who can serve during an <math>8</math>-year period is <math>4</math>, so the answer is <math>\boxed{C}</math> |
− | + | if the terms are divided <math>1\ |\ 2\ 3\ 4\ |\ 5\ 6\ 7\ |\ 8</math> | |
− | <math>\boxed{1|2 3 4|5 6 7|8 | ||
==See Also== | ==See Also== | ||
− | |||
{{AMC8 box|year=2000|num-b=4|num-a=6}} | {{AMC8 box|year=2000|num-b=4|num-a=6}} | ||
+ | <math>\text{{{MAA\,\,Notice}}}</math> |
Latest revision as of 06:04, 13 January 2023
Problem
Each principal of Lincoln High School serves exactly one -year term. What is the maximum number of principals this school could have during an
-year period?
Solution
If the first year of the -year period was the final year of a principal's
term, then in the next six years two more principals would serve, and the last year of the
period would be the first year of the fourth principal's term. Therefore, the maximum
number of principals who can serve during an
-year period is
, so the answer is
if the terms are divided
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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 | ||
All AJHSME/AMC 8 Problems and Solutions |