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

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<math>\boxed{(E)}</math> <math> 2017 </math>
 
<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</math> is congruent to <math>1 \mod{ 7}</math>), the same date (after February) moves "forward" one day in the subsequent year, if that year is not a leap year.
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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:
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<math>5/27/12</math> Sun
 
<math>5/27/12</math> Sun
  
You can keep count forward to find that the first time this date falls on a Saturday is in <math> 2017</math>:
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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>5/27/13</math> Mon
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<math>5/27/17</math> Sat
 
<math>5/27/17</math> Sat
  
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==Video Solution==
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https://youtu.be/P7rGLXp_6es?t=628
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 +
~IceMatrix
  
 
== See also ==
 
== 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}}
 +
{{MAA Notice}}

Latest revision as of 11:59, 29 March 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

See also

2010 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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