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 "1967 AHSME Problems/Problem 16"

(Created page with "== Problem == Let the product <math>(12)(15)(16)</math>, each factor written in base <math>b</math>, equals <math>3146</math> in base <math>b</math>. Let <math>s=12+15+16</math>...")
 
Line 9: Line 9:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{B}</math>
+
 
 +
Converting everything into base <math>10</math>, we have <math>(b + 2)(b+5)(b+6) = 3b^3 + b^2 + 4b + 6</math>.  Looking ahead, the constant term of the polynomial will be <math>2*5*6 -6 = 54</math>. By the Rational Root Theorem, the only possible integer roots are <math>1, 2, 3, 6, 9, 18, 27, 54</math>.  Bases <math>1, 2, 3, 6</math> do not have a <math>6</math> as a digit.  Testing <math>b=9</math> gives a solution that works.
 +
 
 +
Therefore, we are working in base <math>9</math>.  Adding the units place in base <math>9</math>, <math>2 + 5 + 6 = 14_9</math>, so we carry the <math>1</math> to get a total of <math>44_9</math>, which is option <math>\fbox{B}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 01:59, 13 July 2019

Problem

Let the product $(12)(15)(16)$, each factor written in base $b$, equals $3146$ in base $b$. Let $s=12+15+16$, each term expressed in base $b$. Then $s$, in base $b$, is

$\textbf{(A)}\ 43\qquad \textbf{(B)}\ 44\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 46\qquad \textbf{(E)}\ 47$

Solution

Converting everything into base $10$, we have $(b + 2)(b+5)(b+6) = 3b^3 + b^2 + 4b + 6$. Looking ahead, the constant term of the polynomial will be $2*5*6 -6 = 54$. By the Rational Root Theorem, the only possible integer roots are $1, 2, 3, 6, 9, 18, 27, 54$. Bases $1, 2, 3, 6$ do not have a $6$ as a digit. Testing $b=9$ gives a solution that works.

Therefore, we are working in base $9$. Adding the units place in base $9$, $2 + 5 + 6 = 14_9$, so we carry the $1$ to get a total of $44_9$, which is option $\fbox{B}$.

See also

1967 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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=1967_AHSME_Problems/Problem_16&oldid=107594"