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 "1965 AHSME Problems/Problem 15"

m (put letter choice in final answer)
(made proof more clear and removed inaccurate statement about base 10)
Line 11: Line 11:
 
== Solution ==
 
== Solution ==
  
−
We begin by converting both <math>25_b</math> and <math>52_b</math> to base <math>10</math>. <math>25_b = 2b+5</math> in base <math>10</math> and <math>52_b = 5b+2</math> base <math>10</math>. The problem tells us that <math>5b+2 = 4b+10</math>, yielding <math>\boxed{\textbf{(B) }8}</math> as our final answer.
+
We begin by noting that <math>25_b = 2b+5</math> and <math>52_b = 5b+2</math>. The problem tells us that <math>2*25_b=52_b</math>, so <math>2(2b+5)=5b+2</math>. Solving for b yields the answer <math>\boxed{\textbf{(B) }8}</math>.

Revision as of 12:14, 18 July 2024

Problem

The symbol $25_b$ represents a two-digit number in the base $b$. If the number $52_b$ is double the number $25_b$, then $b$ is:

$\textbf{(A)}\ 7 \qquad \textbf{(B) }\ 8 \qquad \textbf{(C) }\ 9 \qquad \textbf{(D) }\ 11 \qquad \textbf{(E) }\ 12$

Solution

We begin by noting that $25_b = 2b+5$ and $52_b = 5b+2$. The problem tells us that $2*25_b=52_b$, so $2(2b+5)=5b+2$. Solving for b yields the answer $\boxed{\textbf{(B) }8}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1965_AHSME_Problems/Problem_15&oldid=222994"