Difference between revisions of "1965 AHSME Problems/Problem 15"
Tecilis459 (talk | contribs) (Add problem statement) |
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== Solution == | == Solution == | ||
− | We begin by converting both <math>25_b</math> and <math>52_b</math> to base <math>10</math>. <math>25_b = 2b+5</math> in base <math>10</math> and <math>52_b = 5b+2</math> base <math>10</math>. The problem tells us that <math>5b+2 = 4b+10</math>, yielding <math>\boxed{8}</math> as our final answer. | + | We begin by converting both <math>25_b</math> and <math>52_b</math> to base <math>10</math>. <math>25_b = 2b+5</math> in base <math>10</math> and <math>52_b = 5b+2</math> base <math>10</math>. The problem tells us that <math>5b+2 = 4b+10</math>, yielding <math>\boxed{\textbf{(B) }8}</math> as our final answer. |
Revision as of 13:11, 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 converting both and
to base
.
in base
and
base
. The problem tells us that
, yielding
as our final answer.