Difference between revisions of "2002 AMC 8 Problems/Problem 5"
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<math> \text{(A)}\ \text{Monday}\qquad\text{(B)}\ \text{Wednesday}\qquad\text{(C)}\ \text{Friday}\qquad\text{(D)}\ \text{Saturday}\qquad\text{(E)}\ \text{Sunday} </math> | <math> \text{(A)}\ \text{Monday}\qquad\text{(B)}\ \text{Wednesday}\qquad\text{(C)}\ \text{Friday}\qquad\text{(D)}\ \text{Saturday}\qquad\text{(E)}\ \text{Sunday} </math> | ||
| − | ==Solution== | + | ==Solution 1== |
Days of the week have a cycle that repeats every <math>7</math> days. Thus, after <math>100</math> cycles, or <math>700</math> days, it will be Saturday again. Six more days will make it <math>\text{Friday} \rightarrow \boxed{C}</math> | Days of the week have a cycle that repeats every <math>7</math> days. Thus, after <math>100</math> cycles, or <math>700</math> days, it will be Saturday again. Six more days will make it <math>\text{Friday} \rightarrow \boxed{C}</math> | ||
| + | |||
| + | ==Solution 2 (similar to solution 1)== | ||
| + | |||
| + | Building off of solution 1, we can make things simpler by fast-forwarding <math>101</math> cycles (<math>707</math> days) instead of <math>100</math> cycles. Day <math>707</math> would be a Saturday again, and one day before then (Day <math>706</math>) would be a Friday. Therefore the answer is <math>\boxed{(C)}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2002|num-b=4|num-a=6}} | {{AMC8 box|year=2002|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 00:12, 9 July 2024
Problem
Carlos Montado was born on Saturday, November 9, 2002. On what day of the week will Carlos be 706 days old?
Solution 1
Days of the week have a cycle that repeats every 
days. Thus, after 
cycles, or 
days, it will be Saturday again. Six more days will make it 
Solution 2 (similar to solution 1)
Building off of solution 1, we can make things simpler by fast-forwarding 
cycles (
days) instead of cycles. Day would be a Saturday again, and one day before then (Day 
) would be a Friday. Therefore the answer is 
.
See Also
| 2002 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
