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

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== Problem 9 ==
+
== Problem 10 ==
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>?
+
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>
\mathrm{(A)}\ 20
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\mathrm{(A)}\ 2011
 
\qquad
 
\qquad
\mathrm{(B)}\ 21
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\mathrm{(B)}\ 2012
 
\qquad
 
\qquad
\mathrm{(C)}\ 22
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\mathrm{(C)}\ 2013
 
\qquad
 
\qquad
\mathrm{(D)}\ 23
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\mathrm{(D)}\ 2015
 
\qquad
 
\qquad
\mathrm{(E)}\ 24
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\mathrm{(E)}\ 2017
 
</math>
 
</math>
  
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== See also ==
 
== See also ==
{{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}}

Revision as of 20:09, 1 January 2012

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 $365$ 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


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
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