1968 AHSME Problems/Problem 33: Difference between revisions
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== Solution == | == Solution == | ||
Call the number abc | Call the number <math>\overline{abc}</math> in base 7. | ||
<math>\fbox{A}</math> | 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, | |||
<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). | |||
Select <math>\fbox{A}</math> as our answer. | |||
~hastapasta | |||
== See also == | == See also == | ||
{{AHSME box|year=1968|num-b=32|num-a=34}} | {{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 
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
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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