Difference between revisions of "1968 AHSME Problems/Problem 33"
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| + | == 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>. | ||
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| + | The correct choice is <math>\fbox{A}</math>. | ||
== See also == | == See also == | ||
Revision as of 15:37, 10 August 2022
Contents
Problem
A number 
has three digits when expressed in base 
. When is expressed in base 
the digits are reversed. Then the middle digit is:
Solution
Call the number 
in base 7.
Then, 
. (Breaking down the number in base-form).
After combining like terms and moving the variables around,
,
. This shows that 
is a multiple of 8 (we only have to find the middle digit under one of the bases). Thus, 
(since 8>6, the largest digit in base 7).
Select 
as our answer.
~hastapasta
Solution 2
If the number in base 7 has reverse digits in base 9, then . When 3-digit numbers' digits reverse, the first and third digits swap places and the middle one doesn't change. So, 
and therefore, .
The correct choice is .
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. 
