2022 AMC 10A Problems/Problem 17: Difference between revisions

Latest revision as of 21:44, 25 November 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

Video Solution 3 by SpreadTheMathLove

https://www.youtube.com/watch?v=yLkGjQ-atzU&list=PLmpPPbOoDfggaByZonvjv_0Wy7XftFA9L&index=63

@@ Line 3: / Line 3: @@
 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>
-(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>
-==Solution 1==
+==Solution==
 We rewrite the given equation, then rearrange:
@@ Line 32: / Line 32: @@
 ~MRENTHUSIASM
-==Solution 2 (fast)==
+==Remark==
-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</math>, and <math>c</math>, as you can simply switch the values of the 2 positives or negatives (<math>x+y=y+x</math>). 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).</math>
+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>.
-~iluvme
+==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
+==Video Solution 3 by SpreadTheMathLove==
+https://www.youtube.com/watch?v=yLkGjQ-atzU&list=PLmpPPbOoDfggaByZonvjv_0Wy7XftFA9L&index=63
 == 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
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