Difference between revisions of "2013 AMC 10A Problems/Problem 13"
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− | These three digit numbers are of the form <math>xyx</math>. We see that <math>x\ | + | We use a casework approach to solve the problem. These three digit numbers are of the form <math>\overline{xyx}</math>.(<math>\overline{abc}</math> denotes the number <math>100a+10b+c</math>). We see that <math>x\neq 0</math> and <math>x\neq 5</math>, as <math>x=0</math> does not yield a three-digit integer and <math>x=5</math> yields a number divisible by 5. |
The second condition is that the sum <math>2x+y<20</math>. When <math>x</math> is <math>1</math>, <math>2</math>, <math>3</math>, or <math>4</math>, <math>y</math> can be any digit from <math>0</math> to <math>9</math>, as <math>2x<10</math>. This yields <math>10(4) = 40</math> numbers. | The second condition is that the sum <math>2x+y<20</math>. When <math>x</math> is <math>1</math>, <math>2</math>, <math>3</math>, or <math>4</math>, <math>y</math> can be any digit from <math>0</math> to <math>9</math>, as <math>2x<10</math>. This yields <math>10(4) = 40</math> numbers. |
Revision as of 15:44, 5 November 2019
Problem
How many three-digit numbers are not divisible by , have digits that sum to less than , and have the first digit equal to the third digit?
Solution
We use a casework approach to solve the problem. These three digit numbers are of the form .( denotes the number ). We see that and , as does not yield a three-digit integer and yields a number divisible by 5.
The second condition is that the sum . When is , , , or , can be any digit from to , as . This yields numbers.
When , we see that so . This yields more numbers.
When , so . This yields more numbers.
When , so . This yields more numbers.
When , so . This yields more numbers.
Summing, we get
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.