Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2014 AMC 12B Problems/Problem 8"

(→‎Solution (Equation Algorithm))
(→‎Solution (Equation Algorithm))
Line 30: Line 30:
 
<math>10^3C + 10C = 0</math>.
 
<math>10^3C + 10C = 0</math>.
  
−
Therefore <math>c = 0</math>.
+
Therefore <math>C = 0</math>.
  
 
This means that <math>A + B = D</math> and <math>D < 10</math> or else there would be parts carried over in the equation. The positive integers that satisfy this equation are a minimum <math>(2, 1)</math> and a maximum of <math>(4, 5)</math>. This means that <math>D = 3</math> <math>, 4</math> <math>, 5</math> <math>, 6</math> <math>, 7</math> <math>, 8</math> <math>, 9</math>. Giving  
 
This means that <math>A + B = D</math> and <math>D < 10</math> or else there would be parts carried over in the equation. The positive integers that satisfy this equation are a minimum <math>(2, 1)</math> and a maximum of <math>(4, 5)</math>. This means that <math>D = 3</math> <math>, 4</math> <math>, 5</math> <math>, 6</math> <math>, 7</math> <math>, 8</math> <math>, 9</math>. Giving  

Revision as of 23:45, 10 August 2024

Problem

In the addition shown below $A$, $B$, $C$, and $D$ are distinct digits. How many different values are possible for $D$?

\[\begin{tabular}{cccccc}&A&B&B&C&B\\ +&B&C&A&D&A\\ \hline &D&B&D&D&D\end{tabular}\]

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

Solution

From the first column, we see $A+B < 10$ because it yields a single digit answer. From the fourth column, we see that $C+D$ equals $D$ and therefore $C = 0$. We know that $A+B = D$. Therefore, the number of values $D$ can take is equal to the number of possible sums less than $10$ that can be formed by adding two distinct natural numbers. Letting $A=1$, and letting $B=2,3,4,5,6,7,8$, we have \[D = 3,4,5,6,7,8,9 \implies \boxed{\textbf{(C)}\ 7}\]

Solution (Equation Algorithm)

It is intuitively obvious, even to the most casual observer that the problem statement can be rewritten as:

$10^4A + 10^4B + 10^3B + 10^3C + 10^2B + 10^2A + 10C + 10D + B + A = 10^4D + 10^3B + 10^2D + 10D + D$. This equation can be simplified into:

$10^4A + 10^4B + 10^3C + 10^2B + 10^2A + 10C + B + A = 10^4D + 10^2D + D$.

Now from here, it should hopefully make sense that $A + B = D$ by looking at the one's digit of both equations. Factoring out $A + B$ gives:

$10^4(A+B) + 10^3C + 10^2(A+B) + 10C + (B + A) = 10^4D + 10^2D + D$.

Which equals: $10^4(D) + 10^3C + 10^2(D) + 10C + D = 10^4D + 10^2D + D$.

This simplifies into: $10^3C + 10C = 0$.

Therefore $C = 0$.

This means that $A + B = D$ and $D < 10$ or else there would be parts carried over in the equation. The positive integers that satisfy this equation are a minimum $(2, 1)$ and a maximum of $(4, 5)$. This means that $D = 3$ $, 4$ $, 5$ $, 6$ $, 7$ $, 8$ $, 9$. Giving $\boxed{\textbf{(C)}\ 7}$ ~PeterDoesPhysics

See also

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_8&oldid=224903"