Difference between revisions of "2003 AMC 10B Problems/Problem 13"
(Created page with "==Problem== Let <math>\clubsuit(x)</math> denote the sum of the digits of the positive integer <math>x</math>. For example, <math>\clubsuit(8)=8</math> and <math>\clubsuit(123)=...") |
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==Solution== | ==Solution== | ||
− | We can divide <math>\clubsuit(x)</math> into two cases so that <math>\clubsuit(\clubsuit(x))=3.</math> The first is where <math>\clubsuit(x)</math> is a one-digit number, and the second is where it is a two | + | We can divide <math>\clubsuit(x)</math> into two cases so that <math>\clubsuit(\clubsuit(x))=3.</math> The first is where <math>\clubsuit(x)</math> is a one-digit number, and the second is where it is a two-digit number. |
− | For <math>\clubsuit(x)</math> to be a one-digit number, <math>x</math>'s digits must add up to be <math>3.</math> This can be done in three ways <math>\ | + | For <math>\clubsuit(x)</math> to be a one-digit number, <math>x</math>'s digits must add up to be <math>3.</math> This can be done in three ways <math>\Rightarrow 30, 21,</math> and <math>12.</math> |
− | For <math>\clubsuit(x)</math> to be a two-digit number, <math>x</math>'s digits must add up to be <math>12,</math> since the sum cannot exceed <math>9+9=18.</math> This can be done in seven ways <math>\ | + | For <math>\clubsuit(x)</math> to be a two-digit number, <math>x</math>'s digits must add up to be <math>12,</math> since the sum cannot exceed <math>9+9=18.</math> This can be done in seven ways <math>\Rightarrow 93, 84, 75, 66, 57, 48,</math> and <math>39.</math> |
− | Add the number of ways together <math>\ | + | Add the number of ways together <math>\Rightarrow 3+7=\boxed{\textbf{(E)}\ 10}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2003|ab=B|num-b=12|num-a=14}} | {{AMC10 box|year=2003|ab=B|num-b=12|num-a=14}} |
Revision as of 19:18, 26 November 2011
Problem
Let denote the sum of the digits of the positive integer . For example, and . For how many two-digit values of is ?
Solution
We can divide into two cases so that The first is where is a one-digit number, and the second is where it is a two-digit number.
For to be a one-digit number, 's digits must add up to be This can be done in three ways and
For to be a two-digit number, 's digits must add up to be since the sum cannot exceed This can be done in seven ways and
Add the number of ways together
See Also
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 |