Difference between revisions of "2000 AMC 12 Problems/Problem 18"

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== Solution ==
 
== Solution ==
There are either <math>65 + 200 = 265</math> or <math>66 + 200 = 266</math> days between the first two dates depending upon whether or not year <math>N</math> is a leap year. Since <math>7</math> divides into <math>266</math>, then it is possible for both dates to be Tuesday; hence year <math>N+1</math> is a leap year and <math>N-1</math> is not a leap year. There are <math>265 + 300 = 565</math> days between the date in years <math>N,N-1</math>, which leaves a remainder of <math>5</math> upon division by <math>7</math>. Since we are subtracting days, we count 5 days before Tuesday, which gives us Thursday <math>\mathrm{(A)}</math>.
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There are either <math>65 + 200 = 265</math> or <math>66 + 200 = 266</math> days between the first two dates depending upon whether or not year <math>N</math> is a leap year. Since <math>7</math> divides into <math>266</math>, then it is possible for both dates to be Tuesday; hence year <math>N+1</math> is a leap year and <math>N-1</math> is not a leap year. There are <math>265 + 300 = 565</math> days between the date in years <math>N,N-1</math>, which leaves a remainder of <math>5</math> upon division by <math>7</math>. Since we are subtracting days, we count 5 days before Tuesday, which gives us <math>\mathrm Thursday{(A)}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 01:00, 3 February 2015

The following problem is from both the 2000 AMC 12 #18 and 2000 AMC 10 #25, so both problems redirect to this page.

Problem

In year $N$, the $300^{\text{th}}$ day of the year is a Tuesday. In year $N+1$, the $200^{\text{th}}$ day is also a Tuesday. On what day of the week did the $100$th day of year $N-1$ occur?

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

Solution

There are either $65 + 200 = 265$ or $66 + 200 = 266$ days between the first two dates depending upon whether or not year $N$ is a leap year. Since $7$ divides into $266$, then it is possible for both dates to be Tuesday; hence year $N+1$ is a leap year and $N-1$ is not a leap year. There are $265 + 300 = 565$ days between the date in years $N,N-1$, which leaves a remainder of $5$ upon division by $7$. Since we are subtracting days, we count 5 days before Tuesday, which gives us $\mathrm Thursday{(A)}$.

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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 12 Problems and Solutions
2000 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Problem
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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