Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
m (→‎Solution)
Line 7: Line 7:
  
 
== Solution ==
 
== Solution ==
−
There are either <math>265 + 300 = 565</math> or <math>266 + 300 = 566</math> days between year <math>N</math> and year <math>N-1</math>, depending on whether year <math>N-1</math> is a leap year. In addition, 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> 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 <math>265 + 300 = 565</math> days between the date in years <math>N,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>\mathrm {Thursday} \text{ (A)}</math>.
+
There are either <cmath>265 + 300 = 565</cmath> or <cmath>266 + 300 = 566</cmath> days between year <math>N</math> and year <math>N-1</math>, depending on whether year <math>N-1</math> is a leap year.  
  
−
Edited by IronicNinja
+
In addition, 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.
 +
 
 +
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>\mathrm {Thursday} \text{ (A)}</math>.
 +
 
 +
Edited by Max0815
  
 
== See also ==
 
== See also ==

Revision as of 12:42, 22 November 2018

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 \[265 + 300 = 565\] or \[266 + 300 = 566\] days between year $N$ and year $N-1$, depending on whether year $N-1$ is a leap year.

In addition, 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 $266$ 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 $\mathrm {Thursday} \text{ (A)}$.

Edited by Max0815

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_18&oldid=98929"