Difference between revisions of "2010 AMC 10B Problems/Problem 5"
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==Solution== | ==Solution== | ||
| − | In this month there are four weeks and three remaining days. | + | In this month there are four weeks and three remaining days. Any 7 days must have exactly one Monday and one Wednesday, so it works if the last <math>31 - 4\cdot 7 = 3</math> days have the same number of Mondays and Wednesdays. We have three choices: Monday, Tuesday, Wednesday; Thursday, Friday, Saturday; Friday, Saturday, Sunday. The number of days the month can start on are Monday, Thursday, and Friday, for a final answer of <math>\boxed{\textbf{(B)}\ 3}.</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2010|ab=B|num-b=4|num-a=6}} | {{AMC10 box|year=2010|ab=B|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 01:14, 24 November 2019
Problem
A month with 
days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?
Solution
In this month there are four weeks and three remaining days. Any 7 days must have exactly one Monday and one Wednesday, so it works if the last 
days have the same number of Mondays and Wednesdays. We have three choices: Monday, Tuesday, Wednesday; Thursday, Friday, Saturday; Friday, Saturday, Sunday. The number of days the month can start on are Monday, Thursday, and Friday, for a final answer of 
See Also
| 2010 AMC 10B (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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
