1957 AHSME Problems/Problem 19

The base of the decimal number system is ten, meaning, for example, that $123 = 1\cdot 10^2 + 2\cdot 10 + 3$ . In the binary system, which has base two, the first five positive integers are $1,\,10,\,11,\,100,\,101$ . The numeral $10011$ in the binary system would then be written in the decimal system as:

$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 40\qquad \textbf{(C)}\ 10011\qquad \textbf{(D)}\ 11\qquad \textbf{(E)}\ 7$

Solution

Numbers in binary work similar to their decimal counterparts, where the multiplier associated with each place is multiplied by two every single place to the left. For example, $1111_2$ ( $1111$ in base $2$ ) would equate to $1 * 2^3 + 1 * 2^2 + 1 * 2^1 + 1 * 2^0 = 8+4+2+1 = 15$ .

Using this same logic, $10011_2$ would be $1*2^4 + 1*2^1 + 1 * 2^0 = \boxed{\textbf{(A) }19}$ .

1957 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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1957 AHSME Problems/Problem 19

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