Difference between revisions of "2003 AMC 12B Problems/Problem 8"
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==Solution== | ==Solution== | ||
| − | 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}</math> = <math>\boxed{10}</math> values in all. |
| − | + | ~Andrew_Lu/Mismatchedcubing | |
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==See Also== | ==See Also== | ||
Latest revision as of 10:40, 1 August 2024
- 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.
~Andrew_Lu/Mismatchedcubing
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. 
