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

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~hastapasta
 
~hastapasta
 
== Solution 2 ==
 
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>.
 
 
The correct choice is <math>\fbox{A}</math>.
 
 
~ForestNYoung
 
  
 
== See also ==
 
== See also ==
{{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 (ProblemsAnswer KeyResources)
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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