Difference between revisions of "1968 AHSME Problems/Problem 33"

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(Solution)
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== Solution ==
 
== Solution ==
Call the number <math>abc</math> in base 7.
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Call the number <math>\overline{abc}</math> in base 7.
Then, <math>49a+7b+c=81c+9b+a</math>. (Breaking down the number in base-form);
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Then, <math>49a+7b+c=81c+9b+a</math>. (Breaking down the number in base-form).
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After combining like terms and moving the variables around,
 
After combining like terms and moving the variables around,
 
<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).
 
<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).
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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.
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~hastapasta
 
~hastapasta
  

Revision as of 18:30, 2 May 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 $\overline{abc}$ 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.

~hastapasta

See also

1968 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 32
Followed by
Problem 34
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