Difference between revisions of "2003 AMC 10B Problems/Problem 22"
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Every day, there will be <math>24</math> half-hours and <math>2(1+2+3+\cdots+12)</math> chimes according to the arrow, resulting in <math>24+156=180</math> total chimes. | Every day, there will be <math>24</math> half-hours and <math>2(1+2+3+\cdots+12)</math> chimes according to the arrow, resulting in <math>24+156=180</math> total chimes. | ||
− | On <math>\text{February | + | On <math>\text{February 26},</math> the number of chimes that still need to occur is <math>2003-91=1912.</math> <math>1912 \div 180=10 \text{R}112.</math> Rounding up, it is <math>11</math> days past <math>\text{February 26},</math> which is <math>\boxed{\textbf{(B) \ } \text{March 9}}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2003|ab=B|num-b=21|num-a=23}} | {{AMC10 box|year=2003|ab=B|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:50, 4 May 2019
Problem
A clock chimes once at minutes past each hour and chimes on the hour according to the hour. For example, at there is one chime and at noon and midnight there are twelve chimes. Starting at on on what date will the chime occur?
Solution
First, find how many chimes will have already happened before midnight (the beginning of the day) of half-hours have passed, and the number of chimes according to the hour is The total number of chimes is
Every day, there will be half-hours and chimes according to the arrow, resulting in total chimes.
On the number of chimes that still need to occur is Rounding up, it is days past which is
See Also
2003 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 |
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