Difference between revisions of "2018 AMC 10A Problems/Problem 25"
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<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 \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) | <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) | ||
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==See Also== | ==See Also== | ||
Revision as of 23:44, 8 February 2018
Problem
For a positive integer 
and nonzero digits 
, 
, and 
, let 
be the -digit integer each of whose digits is equal to ; let 
be the -digit integer each of whose digits is equal to , and let 
be the 
-digit (not -digit) integer each of whose digits is equal to . What is the greatest possible value of 
for which there are at least two values of such that 
?
Solution 1
Observe 
; similarly 
and 
. The relation rewrites as
![\[c \cdot \frac{10^{2n} - 1}{9} - b \cdot \frac{10^n - 1}{9} = a^2 \cdot \left(\frac{10^n - 1}{9}\right)^2.\]](http://latex.artofproblemsolving.com/9/8/9/989e308c9a852ec4413d5c1e0c7c68c0cc67394e.png)
Since 
, 
and we may cancel out a factor of 
to obtain
![\[c \cdot (10^n + 1) - b = a^2 \cdot \frac{10^n - 1}{9}.\]](http://latex.artofproblemsolving.com/4/a/5/4a5fe5d8481c23ddbe2ace7cfbf0ec30b17d982f.png)
This is a linear equation in 
. Thus, if two distinct values of satisfy it, then all values of will. Matching coefficients, we need
![\[c = \frac{a^2}{9} \quad \text{and} \quad c - b = -\frac{a^2}{9} \implies b = \frac{2a^2}{9}.\]](http://latex.artofproblemsolving.com/8/6/b/86b1e8d3707a2d603ff4bed6b7c7241e336ccab8.png)
To maximize 
, we need to maximize . Since and must be integers, must be a multiple of 3. If 
then exceeds 9. However, if 
then 
and 
for an answer of 
. (CantonMathGuy)
See Also
| 2018 AMC 10A (Problems • Answer Key • Resources) | ||
| 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 (Problems • Answer Key • Resources) | |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
