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 "1968 AHSME Problems/Problem 33"

(→‎Solution)
(→‎Solution)
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
−
Call the number ''abc'' in base 7.
+
Call the number <math>abc</math> in base 7.
−
Then, 49''a''+7''b''+''c''=81''c''+9''b''+''a''. (Breaking down the number in base-form);
+
Then, <math>49a+7b+c=81c+9b+a</math>. (Breaking down the number in base-form);
 
After combining like terms and moving the variables around,
 
After combining like terms and moving the variables around,
−
48''a''=2''b''+80''c'', ''b''=40''c''-24''a''=8(5''c''-2''a''). This shows that ''b'' is a multiple of 8 (we only have to find the middle digit under ''one'' of the bases). Thus, ''b''=0 (since 8>6, the largest digit in base 7).
+
<math>48a=2b+80c</math>,<math>b=40c-24a=8(5c-2a)</math>. This shows that <math>b</math> is a multiple of 8 (we only have to find the middle digit under ''one'' of the bases). Thus, <math>b=0</math> (since 8>6, the largest digit in base 7).
−
So ''b''=0.
 
 
Select <math>\fbox{A}</math> as our answer.
 
Select <math>\fbox{A}</math> as our answer.
 
I have no idea how to format besides the basics, so please help. Feel free to message me tips on AOPS. Thanks.
 
I have no idea how to format besides the basics, so please help. Feel free to message me tips on AOPS. Thanks.

Revision as of 16:35, 6 January 2022

Problem

A number $N$ has three digits when expressed in base $7$. When $N$ is expressed in base $9$ the digits are reversed. Then the middle digit is:

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } 5$

Solution

Call the number $abc$ in base 7. Then, $49a+7b+c=81c+9b+a$. (Breaking down the number in base-form); After combining like terms and moving the variables around, $48a=2b+80c$,$b=40c-24a=8(5c-2a)$. This shows that $b$ is a multiple of 8 (we only have to find the middle digit under one of the bases). Thus, $b=0$ (since 8>6, the largest digit in base 7). Select $\fbox{A}$ as our answer. I have no idea how to format besides the basics, so please help. Feel free to message me tips on AOPS. Thanks. ~hastapasta

See also

1968 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 32 		Followed by
Problem 34
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=1968_AHSME_Problems/Problem_33&oldid=169446"