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2010 AMC 10A Problems/Problem 10: Difference between revisions

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(E) 2017
== Problem 10 ==
Marvin had a birthday on Tuesday, May 27 in the leap year <math>2008</math>. In what year will his birthday next fall on a Saturday?


There are 365 days in a non-leap year. There are 7 days in a week. Since 365 = 52 * 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.
<math>
\mathrm{(A)}\ 2011
\qquad
\mathrm{(B)}\ 2012
\qquad
\mathrm{(C)}\ 2013
\qquad
\mathrm{(D)}\ 2015
\qquad
\mathrm{(E)}\ 2017
</math>
 
==Solution==
 
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.


For example:
For example:
5/27/08 Tue
5/27/09 Wed


However, a leap year has 366 days, and 366 = 52 * 7 + 2. So the same date (after February) moves "forward" '''two''' days in the subsequent year, if that year is a leap year.
<math>5/27/08</math> Tue
 
<math>5/27/09</math> Wed
 
However, a leap year has <math>366</math> days, and <math>366 = 52 \cdot 7 + 2</math> . So the same date (after February) moves "forward" '''two''' days in the subsequent year, if that year is a leap year.


For example:
For example:
5/27/11 Fri
<math>5/27/11</math> Fri
5/27/12 Sun
 
<math>5/27/12</math> Sun
 
You can keep counting forward to find that the first time this date falls on a Saturday is in <math> 2017</math>:
 
<math>5/27/13</math> Mon
 
<math>5/27/14</math> Tue
 
<math>5/27/15</math> Wed
 
<math>5/27/16</math> Fri
 
<math>5/27/17</math> Sat
 
<math>\boxed{(E) 2017}</math>
 
==Video Solution==
https://youtu.be/P7rGLXp_6es?t=628


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


5/27/13 Mon
== See also ==
5/27/14 Tue
{{AMC10 box|year=2010|ab=A|num-b=9|num-a=11}}
5/27/15 Wed
{{AMC12 box|year=2010|ab=A|num-b=1|num-a=3}}
5/27/16 Fri
{{MAA Notice}}
5/27/17 Sat

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

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