Difference between revisions of "2010 AMC 10A Problems/Problem 10"
(Not-very-elegant solution, but it's a start) |
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− | + | == Problem 9 == | |
+ | 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 | + | <math> |
+ | \mathrm{(A)}\ 20 | ||
+ | \qquad | ||
+ | \mathrm{(B)}\ 21 | ||
+ | \qquad | ||
+ | \mathrm{(C)}\ 22 | ||
+ | \qquad | ||
+ | \mathrm{(D)}\ 23 | ||
+ | \qquad | ||
+ | \mathrm{(E)}\ 24 | ||
+ | </math> | ||
+ | |||
+ | |||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | <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. | ||
For example: | For example: | ||
− | |||
− | |||
− | However, a leap year has 366 days, and 366 = 52 | + | <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 count 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 | ||
− | |||
− | + | == See also == | |
− | + | {{AMC10 box|year=2010|ab=A|num-b=19|num-a=11}} | |
− | |||
− | |||
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Revision as of 20:07, 1 January 2012
Problem 9
A palindrome, such as , is a number that remains the same when its digits are reversed. The numbers and are three-digit and four-digit palindromes, respectively. What is the sum of the digits of ?
Solution
There are days in a non-leap year. There are days in a week. Since (or is congruent to ), the same date (after February) moves "forward" one day in the subsequent year, if that year is not a leap year.
For example:
Tue
Wed
However, a leap year has days, and . So the same date (after February) moves "forward" two days in the subsequent year, if that year is a leap year.
For example: Fri
Sun
You can keep count forward to find that the first time this date falls on a Saturday is in :
Mon
Tue
Wed
Fri
Sat
See also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
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 |