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

Revision as of 12:23, 22 February 2014

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 $D = 3,4,5,6,7,8,9 \implies \boxed{\textbf{(C)}\ 7}$

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.

@@ Line 12: / Line 12: @@
 <cmath>D = 3,4,5,6,7,8,9 \implies \boxed{\textbf{(C)}\ 7}</cmath>
+== See also ==
 {{AMC12 box|year=2014|ab=B|num-b=7|num-a=9}}
 {{MAA Notice}}

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

Revision as of 12:23, 22 February 2014

Problem

Solution

See also