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

(Created page with "==Problem== How many three-digit positive integers <math>\underline{a}</math> <math>\underline{b}</math> <math>\underline{c}</math> are there whose nonzero digits <math>a,b,<...")
 
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==Problem==
 
==Problem==
  
−
How many three-digit positive integers <math>\underline{a}</math> <math>\underline{b}</math> <math>\underline{c}</math> are there whose nonzero digits <math>a,b,</math> and <math>c</math> satisfy
+
How many three-digit positive integers <math>\underline{a} \ \underline{b} \ \underline{c}</math> are there whose nonzero digits <math>a,b,</math> and <math>c</math> satisfy
 
<cmath>0.\overline{\underline{a}~\underline{b}~\underline{c}} = \frac{1}{3} (0.\overline{a} + 0.\overline{b} + 0.\overline{c})?</cmath>
 
<cmath>0.\overline{\underline{a}~\underline{b}~\underline{c}} = \frac{1}{3} (0.\overline{a} + 0.\overline{b} + 0.\overline{c})?</cmath>
−
(The bar indicates repetition, thus <math>0.\overline{\underline{a}~\underline{b}~\underline{c}}</math> in the infinite repeating decimal <math>0.\underline{a}~\underline{b}~\underline{c}~\underline{a}~\underline{b}~\underline{c}~\cdots</math>
+
(The bar indicates repetition, thus <math>0.\overline{\underline{a}~\underline{b}~\underline{c}}</math> is the infinite repeating decimal <math>0.\underline{a}~\underline{b}~\underline{c}~\underline{a}~\underline{b}~\underline{c}~\cdots</math>)
  
 
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 11 \qquad \textbf{(D) } 13 \qquad \textbf{(E) } 14</math>
 
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 11 \qquad \textbf{(D) } 13 \qquad \textbf{(E) } 14</math>
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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>
 +
Now, this problem is equivalent to counting the ordered triples <math>(a,b,c)</math> that satisfies the equation.
 +
 +
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:
 +
 +
* <math>b</math> increases by <math>4</math> as <math>c</math> decreases by <math>3.</math>
 +
 +
* <math>b</math> decreases by <math>4</math> as <math>c</math> increases by <math>3.</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>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
 +
 +
==Remark==
 +
One way to solve the Diophantine Equation, <math>7a=3b+4c</math> is by taking <math>\pmod{7}</math>, from which the equation becomes <math>0\equiv 3b-3c\pmod{7} \Longrightarrow b\equiv c\pmod{7}</math> so either <math>b=c</math> or WLOG <math>b<c, b+7=c</math>.
 +
 +
==Video Solution 1==
 +
https://youtu.be/p6IauwE8cX8
 +
 +
==Video Solution (⚡️Lightning Fast⚡️)==
 +
https://youtu.be/mgcHM0ATUks
 +
 +
~Education, the Study of Everything
 +
 +
==Video Solution 2==
 +
https://www.youtube.com/watch?v=YAazoVATYQA&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=4
  
 
== See Also ==
 
== See Also ==

Latest revision as of 20:24, 1 May 2024

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 $0.\overline{\underline{a}~\underline{b}~\underline{c}}$ is 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

Remark

One way to solve the Diophantine Equation, $7a=3b+4c$ is by taking $\pmod{7}$, from which the equation becomes $0\equiv 3b-3c\pmod{7} \Longrightarrow b\equiv c\pmod{7}$ so either $b=c$ or WLOG $b<c, b+7=c$.

Video Solution 1

https://youtu.be/p6IauwE8cX8

Video Solution (⚡️Lightning Fast⚡️)

https://youtu.be/mgcHM0ATUks

~Education, the Study of Everything

Video Solution 2

https://www.youtube.com/watch?v=YAazoVATYQA&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=4

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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