Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1957 AHSME Problems/Problem 19"

(Created page with "The base of the decimal number system is ten, meaning, for example, that <math>123 = 1\cdot 10^2 + 2\cdot 10 + 3</math>. In the binary system, which has base two, the first fi...")
 
m (typo fix)
 
Line 7: Line 7:
 
Using this same logic, <math>10011_2</math> would be <math>1*2^4 + 1*2^1 + 1 * 2^0 = \boxed{\textbf{(A) }19}</math>.
 
Using this same logic, <math>10011_2</math> would be <math>1*2^4 + 1*2^1 + 1 * 2^0 = \boxed{\textbf{(A) }19}</math>.
 
==See Also==
 
==See Also==
−
{{AHSME box|year=1957|num-b=18|num-a=20}}
+
{{AHSME 50p box|year=1957|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 
[[Category:AHSME]][[Category:AHSME Problems]]
 
[[Category:AHSME]][[Category:AHSME Problems]]

Latest revision as of 08:54, 25 July 2024

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 $1*2^4 + 1*2^1 + 1 * 2^0 = \boxed{\textbf{(A) }19}$.

See Also

1957 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1957_AHSME_Problems/Problem_19&oldid=223606"