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

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~MRENTHUSIASM
 
~MRENTHUSIASM
  
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==Solution 2 (fast)==
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Use the method from Solution <math>1</math> to get <math>63a = 27b+36c</math>. Subtract <math>27b+36c</math> from both sides to get <math>63a -27b-36c=0</math>. From here we look at possible cases. Either one (from <math>63a</math>, <math>-27b</math>, or <math>-36c</math>) is positive, and the other 2 are negative, or 2 are positive, and one is negative. There are 6 of these cases (<math>pnn,npn,nnp,ppn,pnp,npp</math>). For all of these cases, there are 2 distinct pairs of <math>a,b, and c</math>, as you can simply switch the values of the 2 positives or negatives (<math>x+y=y+x). This leaves us with </math>6 \cdot2 =12<math>  cases. Then, we add the case where </math>a, b<math>, and </math>c<math> are all 0, giving us </math>12+1=\boxed{\textbf{(D) } 13}<math> ordered triples </math>(a,b,c).$
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2022|ab=A|num-b=16|num-a=18}}
 
{{AMC10 box|year=2022|ab=A|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:17, 14 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 1

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

Solution 2 (fast)

Use the method from Solution $1$ to get $63a = 27b+36c$. Subtract $27b+36c$ from both sides to get $63a -27b-36c=0$. From here we look at possible cases. Either one (from $63a$, $-27b$, or $-36c$) is positive, and the other 2 are negative, or 2 are positive, and one is negative. There are 6 of these cases ($pnn,npn,nnp,ppn,pnp,npp$). For all of these cases, there are 2 distinct pairs of $a,b, and c$, as you can simply switch the values of the 2 positives or negatives ($x+y=y+x). This leaves us with$6 \cdot2 =12$cases. Then, we add the case where$a, b$, and$c$are all 0, giving us$12+1=\boxed{\textbf{(D) } 13}$ordered triples$(a,b,c).$

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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