Difference between revisions of "2002 AMC 8 Problems/Problem 5"

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==Solution 2 (similar to solution 1)==
 
==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>.
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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>.
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==Video Solution by WhyMath==
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https://youtu.be/ze0yDOAGL4Y
  
 
==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}}

Latest revision as of 13:27, 29 October 2024

Problem

Carlos Montado was born on Saturday, November 9, 2002. On what day of the week will Carlos be 706 days old?

$\text{(A)}\ \text{Monday}\qquad\text{(B)}\ \text{Wednesday}\qquad\text{(C)}\ \text{Friday}\qquad\text{(D)}\ \text{Saturday}\qquad\text{(E)}\ \text{Sunday}$

Solution 1

Days of the week have a cycle that repeats every $7$ days. Thus, after $100$ cycles, or $700$ days, it will be Saturday again. Six more days will make it $\text{Friday} \rightarrow \boxed{C}$

Solution 2 (similar to solution 1)

Building off of solution 1, we can make things simpler by fast-forwarding $101$ cycles ($707$ days) instead of $100$ cycles. Day $707$ would be a Saturday again, and one day before then (Day $706$) would be a Friday. Therefore the answer is $\boxed{(C)}$.

Video Solution by WhyMath

https://youtu.be/ze0yDOAGL4Y

See Also

2002 AMC 8 (ProblemsAnswer KeyResources)
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

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