Difference between revisions of "1965 AHSME Problems/Problem 15"
(Created page with "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...") |
Tecilis459 (talk | contribs) (Add problem statement) |
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| + | == Problem == | ||
| + | |||
| + | The symbol <math>25_b</math> represents a two-digit number in the base <math>b</math>. If the number <math>52_b</math> is double the number <math>25_b</math>, then <math>b</math> is: | ||
| + | |||
| + | <math>\textbf{(A)}\ 7 \qquad | ||
| + | \textbf{(B) }\ 8 \qquad | ||
| + | \textbf{(C) }\ 9 \qquad | ||
| + | \textbf{(D) }\ 11 \qquad | ||
| + | \textbf{(E) }\ 12 </math> | ||
| + | |||
| + | == 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{8}</math> as our final answer. | ||
Latest revision as of 13:50, 16 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.
