Difference between revisions of "2022 AMC 10A Problems/Problem 17"
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==Solution== | ==Solution== | ||
| − | + | We rewrite the given equation, then rearrange: | |
| − | + | <cmath>\begin{align*} | |
| − | + | \frac{100a+10b+c}{999} &= \frac13\left(\frac a9 + \frac b9 + \frac c9\right) \\ | |
| − | + | 100a+10b+c &= 37a + 37b + 37c \\ | |
| − | + | 63a &= 27b+36c \\ | |
| − | ~ | + | 7a &= 3b+4c. |
| + | \end{align*}</cmath> | ||
| + | ~MRENTHUSIASM | ||
== See Also == | == See Also == | ||
Revision as of 02:34, 12 November 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:
~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. 
