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 "2010 AMC 10A Problems/Problem 10"

m (→‎See also)
m
 
(3 intermediate revisions by 3 users not shown)
Line 13: Line 13:
 
\mathrm{(E)}\ 2017
 
\mathrm{(E)}\ 2017
 
</math>
 
</math>
−
 
−
 
  
 
==Solution==
 
==Solution==
−
 
−
<math>\boxed{(E)}</math> <math> 2017 </math>
 
  
 
There are <math>365</math> days in a non-leap year. There are <math>7</math> days in a week. Since <math>365 = 52 \cdot 7 + 1</math> (or <math>365\equiv 1 \pmod{ 7}</math>), the same date (after February) moves "forward" one day in the subsequent year, if that year is not a leap year.
 
There are <math>365</math> days in a non-leap year. There are <math>7</math> days in a week. Since <math>365 = 52 \cdot 7 + 1</math> (or <math>365\equiv 1 \pmod{ 7}</math>), the same date (after February) moves "forward" one day in the subsequent year, if that year is not a leap year.
Line 46: Line 42:
  
 
<math>5/27/17</math> Sat
 
<math>5/27/17</math> Sat
 +
 +
<math>\boxed{(E) 2017}</math>
  
 
==Video Solution==
 
==Video Solution==
Line 52: Line 50:
 
~IceMatrix
 
~IceMatrix
  
−
== = Good luck all! ==
+
== See also ==
 
{{AMC10 box|year=2010|ab=A|num-b=9|num-a=11}}
 
{{AMC10 box|year=2010|ab=A|num-b=9|num-a=11}}
 
{{AMC12 box|year=2010|ab=A|num-b=1|num-a=3}}
 
{{AMC12 box|year=2010|ab=A|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:45, 30 October 2024

Problem 10

Marvin had a birthday on Tuesday, May 27 in the leap year $2008$. In what year will his birthday next fall on a Saturday?

$\mathrm{(A)}\ 2011 \qquad \mathrm{(B)}\ 2012 \qquad \mathrm{(C)}\ 2013 \qquad \mathrm{(D)}\ 2015 \qquad \mathrm{(E)}\ 2017$

Solution

There are $365$ days in a non-leap year. There are $7$ days in a week. Since $365 = 52 \cdot 7 + 1$ (or $365\equiv 1 \pmod{ 7}$), the same date (after February) moves "forward" one day in the subsequent year, if that year is not a leap year.

For example:

$5/27/08$ Tue

$5/27/09$ Wed

However, a leap year has $366$ days, and $366 = 52 \cdot 7 + 2$ . So the same date (after February) moves "forward" two days in the subsequent year, if that year is a leap year.

For example: $5/27/11$ Fri

$5/27/12$ Sun

You can keep counting forward to find that the first time this date falls on a Saturday is in $2017$:

$5/27/13$ Mon

$5/27/14$ Tue

$5/27/15$ Wed

$5/27/16$ Fri

$5/27/17$ Sat

$\boxed{(E) 2017}$

Video Solution

https://youtu.be/P7rGLXp_6es?t=628

~IceMatrix

See also

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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
2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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

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=2010_AMC_10A_Problems/Problem_10&oldid=230977"