2022 AMC 10A Problems/Problem 17
Problem
How many three-digit positive integers are there whose nonzero digits and satisfy (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
- 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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