Difference between revisions of "1968 AHSME Problems/Problem 33"
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== Solution == | == Solution == | ||
| − | + | Call the number abc. | |
| + | Then, 49a+7b+c=81c+9b+a. (Breaking down the number in base-form) | ||
| + | Moving everything to one side, | ||
| + | 48a=2b+80c, b=40c-24a=8(5c-2a). This shows that b is a multiple of 8. Thus, b=0 (since 8>7, the base of base 7). | ||
| + | So b=0. Select "A." | ||
| + | I have no idea how to format, so please help. Thanks. | ||
<math>\fbox{A}</math> | <math>\fbox{A}</math> | ||
Revision as of 17:49, 4 January 2022
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 abc. Then, 49a+7b+c=81c+9b+a. (Breaking down the number in base-form) Moving everything to one side, 48a=2b+80c, b=40c-24a=8(5c-2a). This shows that b is a multiple of 8. Thus, b=0 (since 8>7, the base of base 7). So b=0. Select "A." I have no idea how to format, so please help. Thanks.
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. 
