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

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== Solution 1 ==
 
== Solution 1 ==
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There are either <cmath>65 + 200 = 265</cmath> or <cmath>66 + 200 = 266</cmath> 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> but not <math>265</math>, for both days to be a Tuesday, year <math>N</math> must be a leap year.
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There are either <cmath>65 + 200 = 265</cmath> or <cmath>66 + 200 = 266</cmath> days between the first two dates depending upon whether or not year <math>N+1</math> is a leap year (since the February 29th of the leap year would come between the 300th day of year <math>N</math> and 200th day of year <math>N + 1</math>). Since <math>7</math> divides into <math>266</math> but not <math>265</math>, for both days to be a Tuesday, year <math>N</math> must be a leap year.
  
 
Hence, year <math>N-1</math> is not a leap year, and so since there are <cmath>265 + 300 = 565</cmath> days between the date in years <math>N,\text{ }N-1</math>, this 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>\boxed{\mathbf{(A)} \ \text{Thursday}.}</math>
 
Hence, year <math>N-1</math> is not a leap year, and so since there are <cmath>265 + 300 = 565</cmath> days between the date in years <math>N,\text{ }N-1</math>, this 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>\boxed{\mathbf{(A)} \ \text{Thursday}.}</math>
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~dolphin7
 
~dolphin7
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==Video Solution ==
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https://youtu.be/aFJtbvTAmm4
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https://www.youtube.com/watch?v=YfiUn8hLlpA  ~David
  
 
== See also ==
 
== See also ==

Latest revision as of 06:46, 24 August 2024

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 1

There are either \[65 + 200 = 265\] or \[66 + 200 = 266\] days between the first two dates depending upon whether or not year $N+1$ is a leap year (since the February 29th of the leap year would come between the 300th day of year $N$ and 200th day of year $N + 1$). Since $7$ divides into $266$ but not $265$, for both days to be a Tuesday, year $N$ must be a leap year.

Hence, year $N-1$ is not a leap year, and so since there are \[265 + 300 = 565\] days between the date in years $N,\text{ }N-1$, this leaves a remainder of $5$ upon division by $7$. Since we are subtracting days, we count 5 days before Tuesday, which gives us $\boxed{\mathbf{(A)} \ \text{Thursday}.}$

Solution 2

The $300-29\cdot 7=97^{\text{th}}$ day of year $N$ and the $200-15\cdot 7=95^{\text{th}}$ day of year $N+1$ are Tuesdays. If there were the same number of days in year $N$ and year $N-1,$ the $99^{\text{th}}$ day of year $N-1$ will be a Tuesday. But year $N$ is a leap year because $97-95 \cong 366 \pmod{7},$ so the $98^{\text{th}}$ day of year $N-1$ is a Tuesday. It follows that the $100^{\text{th}}$ day of year $N-1$ is $\boxed{\mathbf{(A)} \ \text{Thursday}.}$

~dolphin7

Video Solution

https://youtu.be/aFJtbvTAmm4

https://www.youtube.com/watch?v=YfiUn8hLlpA ~David

See also

2000 AMC 12 (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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