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

(Not-very-elegant solution, but it's a start)
 
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(E) 2017
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== Problem 9 ==
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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>?
  
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.
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<math>
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\mathrm{(A)}\ 20
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\qquad
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\mathrm{(B)}\ 21
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\qquad
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\mathrm{(C)}\ 22
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\qquad
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\mathrm{(D)}\ 23
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\qquad
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\mathrm{(E)}\ 24
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</math>
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==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</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:
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.
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<math>5/27/08</math> Tue
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<math>5/27/09</math> Wed
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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
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<math>5/27/11</math> Fri
5/27/12 Sun
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<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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<math>5/27/13</math> Mon
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<math>5/27/14</math> Tue
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<math>5/27/15</math> Wed
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<math>5/27/16</math> Fri
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<math>5/27/17</math> Sat
  
You can keep count forward to find that the first time this date falls on a Saturday is in 2017:
 
  
5/27/13 Mon
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== See also ==
5/27/14 Tue
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{{AMC10 box|year=2010|ab=A|num-b=19|num-a=11}}
5/27/15 Wed
 
5/27/16 Fri
 
5/27/17 Sat
 

Revision as of 20:07, 1 January 2012

Problem 9

A palindrome, such as $83438$, is a number that remains the same when its digits are reversed. The numbers $x$ and $x+32$ are three-digit and four-digit palindromes, respectively. What is the sum of the digits of $x$?

$\mathrm{(A)}\ 20 \qquad \mathrm{(B)}\ 21 \qquad \mathrm{(C)}\ 22 \qquad \mathrm{(D)}\ 23 \qquad \mathrm{(E)}\ 24$


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 19
Followed by
Problem 11
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All AMC 10 Problems and Solutions
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