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 "1997 AHSME Problems/Problem 1"

(Created page with "== Problem== If <math>\texttt{a}</math> and <math>\texttt{b}</math> are digits for which <math> \begin{tabular}{ccc}& 2 & a\\ \times & b & 3\\ \hline & 6 & 9\\ 9 & 2\\ \hline 9 ...")
 
m (→‎Problem)
 
(2 intermediate revisions by 2 users not shown)
Line 2: Line 2:
 
If <math>\texttt{a}</math> and <math>\texttt{b}</math> are digits for which
 
If <math>\texttt{a}</math> and <math>\texttt{b}</math> are digits for which
  
−
<math> \begin{tabular}{ccc}& 2 & a\\ \times & b & 3\\ \hline & 6 & 9\\ 9 & 2\\ \hline 9 & 8 & 9\end{tabular} </math>
+
<math> \begin{array}{ccc}& 2 & a\\ \times & b & 3\\ \hline & 6 & 9\\ 9 & 2\\ \hline 9 & 8 & 9\end{array} </math>
  
 
then <math>\texttt{a+b =}</math>
 
then <math>\texttt{a+b =}</math>
Line 17: Line 17:
  
 
==See Also==
 
==See Also==
−
{{AHSME box|year=1997|before=First Problem|num-a=10}}
+
{{AHSME box|year=1997|before=First Problem|num-a=2}}
 +
{{MAA Notice}}

Latest revision as of 18:48, 10 March 2015

Problem

If $\texttt{a}$ and $\texttt{b}$ are digits for which

$\begin{array}{ccc}& 2 & a\\ \times & b & 3\\ \hline & 6 & 9\\ 9 & 2\\ \hline 9 & 8 & 9\end{array}$

then $\texttt{a+b =}$

$\mathrm{(A)\ } 3 \qquad \mathrm{(B) \ }4 \qquad \mathrm{(C) \ } 7 \qquad \mathrm{(D) \ } 9 \qquad \mathrm{(E) \ }12$

Solution

From the units digit calculation, we see that the units digit of $a\times 3$ is $9$. Since $0 \le a \le 9$ and $a$ is an integer, the only value of $a$ that works is is $a=3$. As a double-check, that does work, since $23 \times 3 = 69$, which is the first line of the multiplication.

The second line of the multiplication can be found by doing the multiplication $23\times b = 92$. Dividing both sides by $23$ gives $b=4$.

Thus, $a + b = 3 + 4 = 7$, and the answer is $\boxed{C}$.

See Also

1997 AHSME (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
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 • 26 • 27 • 28 • 29 • 30
All AHSME 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=1997_AHSME_Problems/Problem_1&oldid=68704"