2022 AMC 10A Problems/Problem 17
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 1
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
Solution 2 (fast)
Use the method from Solution 
to get 
. Subtract 
from both sides to get 
. From here we look at possible cases. Either one (from 
, 
, or 
) is positive, and the other 2 are negative, or 2 are positive, and one is negative. There are 6 of these cases (
). For all of these cases, there are 2 distinct pairs of 
, and , as you can simply switch the values of the 2 positives or negatives (
). This leaves us with 
cases. Then, we add the case where 
, and are all 0, giving us 
ordered triples
~iluvme Note: This solution does not work because a, b, and c are positive integers...
See Also
| 2022 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
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| All AMC 10 Problems and Solutions | ||
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