Difference between revisions of "2003 AMC 12B Problems/Problem 8"
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− | Let <math> | + | Let <math>y=\clubsuit (x)</math>. Since <math>x \leq 99</math>, we have <math>y \leq 18</math>. Thus if <math>\clubsuit (y)=3</math>, then <math>y=3</math> or <math>y=12</math>. The 3 values of <math>x</math> for which <math>\clubsuit (x)=3</math> are 12, 21, and 30, and the 7 values of <math>x</math> for which <math>\clubsuit (x)=12</math> are 39, 48, 57, 66, 75, 84, and 93. There are <math>\boxed{E=10}</math> values in all. |
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==See Also== | ==See Also== |
Latest revision as of 15:38, 30 March 2020
- The following problem is from both the 2003 AMC 12B #8 and 2003 AMC 10B #13, so both problems redirect to this page.
Problem
Let denote the sum of the digits of the positive integer . For example, and . For how many two-digit values of is ?
Solution
Let . Since , we have . Thus if , then or . The 3 values of for which are 12, 21, and 30, and the 7 values of for which are 39, 48, 57, 66, 75, 84, and 93. There are values in all.
See Also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 |
2003 AMC 10B (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.