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Difference between revisions of "2010 AMC 10A Problems/Problem 10"
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== Problem 9 ==
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== Problem 10 ==
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A <i>palindrome</i>, such as <math>83438</math>, is a number that remains the same when its digits are reversed. The numbers <math>x</math> and <math>x+32</math> are three-digit and four-digit palindromes, respectively. What is the sum of the digits of <math>x</math>?
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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?
  
 
<math>
 
<math>
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\mathrm{(A)}\ 20
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\mathrm{(A)}\ 2011
 
\qquad
 
\qquad
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\mathrm{(B)}\ 21
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\mathrm{(B)}\ 2012
 
\qquad
 
\qquad
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\mathrm{(C)}\ 22
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\mathrm{(C)}\ 2013
 
\qquad
 
\qquad
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\mathrm{(D)}\ 23
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\mathrm{(D)}\ 2015
 
\qquad
 
\qquad
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\mathrm{(E)}\ 24
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\mathrm{(E)}\ 2017
 
</math>
 
</math>
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==Solution==
 
==Solution==
  
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<math>\boxed{(E)}</math> <math> 2017 </math>
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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.
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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</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.
 
  
 
For example:
 
For example:
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<math>5/27/12</math> Sun
 
<math>5/27/12</math> Sun
  
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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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<math>\boxed{(E) 2017}</math>
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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 ==
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{{AMC10 box|year=2010|ab=A|num-b=19|num-a=11}}
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{{AMC10 box|year=2010|ab=A|num-b=9|num-a=11}}
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{{AMC12 box|year=2010|ab=A|num-b=1|num-a=3}}
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{{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

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