Difference between revisions of "2010 AMC 10A Problems/Problem 10"

Revision as of 23:54, 9 January 2021

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

$\boxed{(E)}$ $2017$

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

Video Solution

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

~IceMatrix

Good luck all!

2010 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AMC 12 Problems and Solutions

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

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 ~IceMatrix
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 {{AMC10 box|year=2010|ab=A|num-b=9|num-a=11}}
 {{AMC12 box|year=2010|ab=A|num-b=1|num-a=3}}
 {{MAA Notice}}

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