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

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~hastapasta
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== Solution 2 ==
 
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If the number in base 7 has reverse digits in base 9, then <math>49a+7b+c=81c+9b+a</math>. When 3-digit numbers' digits reverse, the first and third digits swap places and the middle one doesn't change. So, <math>7b=9b</math> and therefore, <math>b=0</math>.
 
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The correct choice is <math>\fbox{A}</math>.
 
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1968|num-b=32|num-a=34}}   
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{{AHSME 35p box|year=1968|num-b=32|num-a=34}}   
  
 
[[Category: Intermediate Algebra Problems]]
 
[[Category: Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:53, 16 August 2023

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 AHSC (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 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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