2010 AMC 10A Problems/Problem 10

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 $365$ is congruent to $1 \mod{ 7}$ ), 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 count 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

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?

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

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AMC 10 Problems and Solutions

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2010 AMC 10A Problems/Problem 10

Solution

Problem 10

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