Difference between revisions of "2022 AMC 10A Problems/Problem 17"
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<math>\textbf{(A) } 9 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 11 \qquad \textbf{(D) } 13 \qquad \textbf{(E) } 14</math> | <math>\textbf{(A) } 9 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 11 \qquad \textbf{(D) } 13 \qquad \textbf{(E) } 14</math> | ||
| − | ==Solution | + | ==Solution == |
We rewrite the given equation, then rearrange: | We rewrite the given equation, then rearrange: | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
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== See Also == | == See Also == | ||
Revision as of 21:52, 4 December 2022
Problem
How many three-digit positive integers 
are there whose nonzero digits 
and 
satisfy
![\[0.\overline{\underline{a}~\underline{b}~\underline{c}} = \frac{1}{3} (0.\overline{a} + 0.\overline{b} + 0.\overline{c})?\]](http://latex.artofproblemsolving.com/9/e/4/9e448c87f48492e9f1518a6989dee06242cab99f.png)
(The bar indicates repetition, thus 
in the infinite repeating decimal 
)
Solution
We rewrite the given equation, then rearrange:
Now, this problem is equivalent to counting the ordered triples 
that satisfies the equation.
Clearly, the 
ordered triples 
are solutions to this equation.
The expression 
has the same value when:
increases by 
as 
decreases by 
increases by 
- decreases by as increases by
We find 
more solutions from the solutions above: 
Note that all solutions are symmetric about 
Together, we have 
ordered triples 
~MRENTHUSIASM
See Also
| 2022 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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. 
