Difference between revisions of "1965 AHSME Problems/Problem 15"
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We begin by noting that <math>25_b = 2b+5</math> and <math>52_b = 5b+2</math>. The problem tells us that <math>2*25_b=52_b</math>, so <math>2(2b+5)=5b+2</math>. Solving for b yields the answer <math>\boxed{\textbf{(B) }8}</math>. | We begin by noting that <math>25_b = 2b+5</math> and <math>52_b = 5b+2</math>. The problem tells us that <math>2*25_b=52_b</math>, so <math>2(2b+5)=5b+2</math>. Solving for b yields the answer <math>\boxed{\textbf{(B) }8}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1965|num-b=14|num-a=16}} | ||
| + | {{MAA Notice}} | ||
| + | |||
| + | [[Category:Introductory Number Theory Problems]] | ||
Latest revision as of 15:58, 18 July 2024
Problem
The symbol 
represents a two-digit number in the base 
. If the number 
is double the number , then is:
Solution
We begin by noting that 
and 
. The problem tells us that 
, so 
. Solving for b yields the answer 
.
See Also
| 1965 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
