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Difference between revisions of "2022 AMC 10A Problems/Problem 17"

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Now, this problem is equivalent to counting the ordered triples <math>(a,b,c)</math> that satisfies the equation.
 
Now, this problem is equivalent to counting the ordered triples <math>(a,b,c)</math> that satisfies the equation.
  
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Clearly, the nine ordered triples <math>(a,b,c)=(1,1,1),(2,2,2),\ldots,(9,9,9)</math> are solutions to this equation.
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Clearly, the <math>9</math> ordered triples <math>(a,b,c)=(1,1,1),(2,2,2),\ldots,(9,9,9)</math> are solutions to this equation.
  
 
The expression <math>3b+4c</math> has the same value when:
 
The expression <math>3b+4c</math> has the same value when:
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* <math>b</math> decreases by <math>4</math> as <math>c</math> increases by <math>3.</math>
 
* <math>b</math> decreases by <math>4</math> as <math>c</math> increases by <math>3.</math>
  
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We find four more solutions from the nine solutions above: <math>(a,b,c)=(4,8,1),(5,1,8),(5,9,2),(6,2,9).</math> Note that all solutions are symmetric about <math>(a,b,c)=(5,5,5).</math>
+
We find <math>4</math> more solutions from the <math>9</math> solutions above: <math>(a,b,c)=(4,8,1),(5,1,8),(5,9,2),(6,2,9).</math> Note that all solutions are symmetric about <math>(a,b,c)=(5,5,5).</math>
  
 
Together, we have <math>9+4=\boxed{\textbf{(D) } 13}</math> ordered triples <math>(a,b,c).</math>
 
Together, we have <math>9+4=\boxed{\textbf{(D) } 13}</math> ordered triples <math>(a,b,c).</math>

Revision as of 03:06, 12 November 2022

Problem

How many three-digit positive integers $\underline{a} \ \underline{b} \ \underline{c}$ are there whose nonzero digits $a,b,$ and $c$ satisfy \[0.\overline{\underline{a}~\underline{b}~\underline{c}} = \frac{1}{3} (0.\overline{a} + 0.\overline{b} + 0.\overline{c})?\] (The bar indicates repetition, thus $0.\overline{\underline{a}~\underline{b}~\underline{c}}$ in the infinite repeating decimal $0.\underline{a}~\underline{b}~\underline{c}~\underline{a}~\underline{b}~\underline{c}~\cdots$)

$\textbf{(A) } 9 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 11 \qquad \textbf{(D) } 13 \qquad \textbf{(E) } 14$

Solution

We rewrite the given equation, then rearrange: \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*} Now, this problem is equivalent to counting the ordered triples $(a,b,c)$ that satisfies the equation.

Clearly, the $9$ ordered triples $(a,b,c)=(1,1,1),(2,2,2),\ldots,(9,9,9)$ are solutions to this equation.

The expression $3b+4c$ has the same value when:

  • $b$ increases by $4$ as $c$ decreases by $3.$
  • $b$ decreases by $4$ as $c$ increases by $3.$

We find $4$ more solutions from the $9$ solutions above: $(a,b,c)=(4,8,1),(5,1,8),(5,9,2),(6,2,9).$ Note that all solutions are symmetric about $(a,b,c)=(5,5,5).$

Together, we have $9+4=\boxed{\textbf{(D) } 13}$ ordered triples $(a,b,c).$

~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

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