Difference between revisions of "2018 AMC 10A Problems/Problem 25"

(Solution 3)
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<math>\textbf{(A)} \text{ 12} \qquad \textbf{(B)} \text{ 14} \qquad \textbf{(C)} \text{ 16} \qquad \textbf{(D)} \text{ 18} \qquad \textbf{(E)} \text{ 20}</math>
 
<math>\textbf{(A)} \text{ 12} \qquad \textbf{(B)} \text{ 14} \qquad \textbf{(C)} \text{ 16} \qquad \textbf{(D)} \text{ 18} \qquad \textbf{(E)} \text{ 20}</math>
  
== Solution 1==
+
== Solution ==
  
 
Observe <math>A_n = a(1 + 10 + \dots + 10^{n - 1}) = a \cdot \tfrac{10^n - 1}{9}</math>; similarly <math>B_n = b \cdot \tfrac{10^n - 1}{9}</math> and <math>C_n = c \cdot \tfrac{10^{2n} - 1}{9}</math>. The relation <math>C_n - B_n = A_n^2</math> rewrites as
 
<cmath>c \cdot \frac{10^{2n} - 1}{9} - b \cdot \frac{10^n - 1}{9} = a^2 \cdot \left(\frac{10^n - 1}{9}\right)^2.</cmath>Since <math>n > 0</math>, <math>10^n > 1</math> and we may cancel out a factor of <math>\tfrac{10^n - 1}{9}</math> to obtain
 
<cmath>c \cdot (10^n + 1) - b = a^2 \cdot \frac{10^n - 1}{9}.</cmath>This is a linear equation in <math>10^n</math>. Thus, if two distinct values of <math>n</math> satisfy it, then all values of <math>n</math> will. Matching coefficients, we need
 
<cmath>c = \frac{a^2}{9} \quad \text{and} \quad c - b = -\frac{a^2}{9} \implies b = \frac{2a^2}{9}.</cmath>To maximize <math>a + b + c = a + \tfrac{a^2}{3}</math>, we need to maximize <math>a</math>. Since <math>b</math> and <math>c</math> must be integers, <math>a</math> must be a multiple of 3. If <math>a = 9</math> then <math>b</math> exceeds 9. However, if <math>a = 6</math> then <math>b = 8</math> and <math>c = 4</math> for an answer of <math>\boxed{\textbf{(D)} \text{ 18}}</math>. (CantonMathGuy)
 
 
== Solution 2 (quicker?) ==
 
 
Immediately start trying <math>n = 1</math> and <math>n = 2</math>. These give the system of equations <math>11c - b = a^2</math> and <math>1111c - 11b = (11a)^2</math> (which simplifies to <math>101c - b = 11a^2</math>). These imply that <math>a^2 = 9c</math>, so the possible <math>(a, c)</math> pairs are <math>(9, 9)</math>, <math>(6, 4)</math>, and <math>(3, 1)</math>. The first puts <math>b</math> out of range but the second makes <math>b = 8</math>. We now know the answer is at least <math>6 + 8 + 4 = 18</math>.
 
 
We now only need to know whether <math>a + b + c = 20</math> might work for any larger <math>n</math>. We will always get equations like <math>100001c - b = 11111a^2</math> where the <math>c</math> coefficient is very close to being nine times the <math>a</math> coefficient. Since the <math>b</math> term will be quite insignificant, we know that once again <math>a^2</math> must equal <math>9c</math>, and thus <math>a = 9, c = 9</math> is our only hope to reach <math>20</math>. Substituting and dividing through by <math>9</math>, we will have something like <math>100001 - b/9 = 99999</math>. No matter what <math>n</math> really was, <math>b</math> is out of range (and certainly isn't <math>2</math> as we would have needed).
 
 
The answer then is <math>\boxed{\textbf{(D)} \text{ 18}}</math>.
 
 
== Solution 3 ==
 
 
Notice that <math>(0.\overline{3})^2 = 0.\overline{1}</math> and <math>(0.\overline{6})^2 = 0.\overline{4}</math>. Setting <math>a = 3</math> and <math>c = 1</math>, we see <math>b = 2</math> works for all possible values of <math>n</math>. Similarly, if <math>a = 6</math> and <math>c = 4</math>, then <math>b = 8</math> works for all possible values of <math>n</math>. The second solution yields a greater sum of <math>\boxed{\textbf{(D)} \text{ 18}}</math>.
 
Notice that <math>(0.\overline{3})^2 = 0.\overline{1}</math> and <math>(0.\overline{6})^2 = 0.\overline{4}</math>. Setting <math>a = 3</math> and <math>c = 1</math>, we see <math>b = 2</math> works for all possible values of <math>n</math>. Similarly, if <math>a = 6</math> and <math>c = 4</math>, then <math>b = 8</math> works for all possible values of <math>n</math>. The second solution yields a greater sum of <math>\boxed{\textbf{(D)} \text{ 18}}</math>.
  

Revision as of 13:01, 19 February 2018

Problem

For a positive integer $n$ and nonzero digits $a$, $b$, and $c$, let $A_n$ be the $n$-digit integer each of whose digits is equal to $a$; let $B_n$ be the $n$-digit integer each of whose digits is equal to $b$, and let $C_n$ be the $2n$-digit (not $n$-digit) integer each of whose digits is equal to $c$. What is the greatest possible value of $a + b + c$ for which there are at least two values of $n$ such that $C_n - B_n = A_n^2$?

$\textbf{(A)} \text{ 12} \qquad \textbf{(B)} \text{ 14} \qquad \textbf{(C)} \text{ 16} \qquad \textbf{(D)} \text{ 18} \qquad \textbf{(E)} \text{ 20}$

Solution

Notice that $(0.\overline{3})^2 = 0.\overline{1}$ and $(0.\overline{6})^2 = 0.\overline{4}$. Setting $a = 3$ and $c = 1$, we see $b = 2$ works for all possible values of $n$. Similarly, if $a = 6$ and $c = 4$, then $b = 8$ works for all possible values of $n$. The second solution yields a greater sum of $\boxed{\textbf{(D)} \text{ 18}}$.

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Problem
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
2018 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Problem
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 12 Problems and Solutions

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