Difference between revisions of "1969 AHSME Problems/Problem 3"

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== Solution ==
 
== Solution ==
<math>\fbox{E}</math>
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<math>11000_2-1_2=10111_2</math>. <math>\fbox{E}</math>
  
 
== See also ==
 
== See also ==
{{AHSME box|year=1969|num-b=2|num-a=3}}   
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{{AHSME 35p box|year=1969|num-b=2|num-a=3}}   
  
 
[[Category: Introductory Algebra Problems]]
 
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:31, 10 July 2015

Problem

If $N$, written in base $2$, is $11000$, the integer immediately preceding $N$, written in base $2$, is:

$\text{(A) } 10001\quad \text{(B) } 10010\quad \text{(C) } 10011\quad \text{(D) } 10110\quad \text{(E) } 10111$

Solution

$11000_2-1_2=10111_2$. $\fbox{E}$

See also

1969 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 3
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
All AHSME Problems and Solutions

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