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2014 AMC 12B Problems/Problem 8: Difference between revisions

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From the first column, we see <math>A+B < 10</math> because it yields a single digit answer.  From the fourth column, we see that <math>C+D</math> equals <math>D</math> and therefore <math>C = 0</math>.  We know that <math>A+B = D</math>.  Therefore, the number of values <math>D</math> can take is equal to the number of possible sums less than <math>10</math> that can be formed by adding two distinct natural numbers.  Letting <math>A=1</math>, and letting <math>B=2,3,4,5,6,7,8</math>, we have  
From the first column, we see <math>A+B < 10</math> because it yields a single digit answer.  From the fourth column, we see that <math>C+D</math> equals <math>D</math> and therefore <math>C = 0</math>.  We know that <math>A+B = D</math>.  Therefore, the number of values <math>D</math> can take is equal to the number of possible sums less than <math>10</math> that can be formed by adding two distinct natural numbers.  Letting <math>A=1</math>, and letting <math>B=2,3,4,5,6,7,8</math>, we have  
<cmath>D = 3,4,5,6,7,8,9 \implies \boxed{\textbf{(C)}\ 7}</cmath>
<cmath>D = 3,4,5,6,7,8,9 \implies \boxed{\textbf{(C)}\ 7}</cmath>
==Solution (Equation Algorithm)==
It is intuitively obvious, even to the most casual observer that the problem statement can be rewritten as:
<math>\overset{4}{10}A + \overset{4}{10}B + \overset{3}{10}B + \overset{3}{10}C + \overset{2}{10}B + \overset{2}{10}A + 10C + 10D + B + A = \overset{4}{10}D + \overset{3}{10}B + \overset{2}{10}D + 10D + D</math>. This equation can be simplified into:
<math>\overset{4}{10}A + \overset{4}{10}B + \overset{3}{10}C + \overset{2}{10}B + \overset{2}{10}A + 10C + B + A = \overset{4}{10}D + \overset{2}{10}D + D</math>.
Now from here, it should hopefully make sense that <math>A + B = D</math> by looking at the one's digit of both equations. Factoring out <math>A + B</math> gives:
<math>\overset{4}{10}(A+B) + \overset{3}{10}C \overset{2}{10}(A+B) + 10C + (B + A) = \overset{4}{10}D + \overset{2}{10}D + D</math>.
Which equals:
<math>\overset{4}{10}(D) + \overset{3}{10}C \overset{2}{10}(D) + 10C + D = \overset{4}{10}D + \overset{2}{10}D + D</math>.
This simplifies into:
<math>\overset{3}{10}C + 10C = 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
<math>\boxed{\textbf{(C)}\ 7}</math>
~PeterDoesPhysics


== See also ==
== See also ==
{{AMC12 box|year=2014|ab=B|num-b=7|num-a=9}}
{{AMC12 box|year=2014|ab=B|num-b=7|num-a=9}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 23:38, 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:

$\overset{4}{10}A + \overset{4}{10}B + \overset{3}{10}B + \overset{3}{10}C + \overset{2}{10}B + \overset{2}{10}A + 10C + 10D + B + A = \overset{4}{10}D + \overset{3}{10}B + \overset{2}{10}D + 10D + D$. This equation can be simplified into:

$\overset{4}{10}A + \overset{4}{10}B + \overset{3}{10}C + \overset{2}{10}B + \overset{2}{10}A + 10C + B + A = \overset{4}{10}D + \overset{2}{10}D + D$.

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:

$\overset{4}{10}(A+B) + \overset{3}{10}C \overset{2}{10}(A+B) + 10C + (B + A) = \overset{4}{10}D + \overset{2}{10}D + D$.

Which equals: $\overset{4}{10}(D) + \overset{3}{10}C \overset{2}{10}(D) + 10C + D = \overset{4}{10}D + \overset{2}{10}D + D$.

This simplifies into: $\overset{3}{10}C + 10C = 0$.

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

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