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what is the binary representation of 2015?

Guest May 15, 2015

Best Answer 

 #5
avatar+19653 
+10

what is the binary representation of 2015 ?

$$\begin{array}{rcrr}
& & \rm{q o u t i e n t} & \rm{r e m a i n d e r} \\
2015 & : 2 = & 1007 & 1 \\
1007 & : 2 = & 503 & 1 \\
503 & : 2 = & 251 & 1 \\
251 & : 2 = & 125 & 1 \\
125 & : 2 = & 62 & 1 \\
62 & : 2 = & 31 & 0 \\
31 & : 2 = & 15 & 1 \\
15 & : 2 = & 7 & 1 \\
7 & : 2 = & 3 & 1 \\
3 & : 2 = & 1 & 1 \\
1 & : 2 = & 0 & \textcolor[rgb]{1,0,0}{1} \\
\end{array}$$

 

$$2015_2~=~\textcolor[rgb]{1,0,0}{1}~1~1~1~1~0~1~1~1~1~1$$

heureka  May 15, 2015
 #1
avatar+1841 
+5

2015 in binary is 11111011111

 

I do not know of an easy way to explain why 2015 in binary is 11111011111; however, I found a video that can explain it way better than I can.  Here is the web address:  https://www.khanacademy.org/math/pre-algebra/applying-math-reasoning-topic/alternate-number-bases/v/number-systems-introduction

gibsonj338  May 15, 2015
 #2
avatar+87309 
+5

201510  = 111110111112

 

CPhill  May 15, 2015
 #3
avatar+26753 
+10

Here's a detailed breakdown of the steps needed to turn 2015 decimal into binary:

 binary 1

 binary 2

.

Alan  May 15, 2015
 #4
avatar+92805 
+10

I am going to say the same as Alan and the video clip  LOL

1024 is the biggest power of 2 that goes into 2015, so I will start there 

 

2015

= 1024 with 991 remaining

=1024+512+479 remaining

=1024+512+256+223 remaining

=1024+512+256+128+95remaining

=1028+512+256+128+64+31remaining

=1028+512+256+128+64+16+15remaining

=1028+512+256+128+64+16+8+7remaining

=1028+512+256+128+64+16+8+4+3remaining

=1028+512+256+128+64+16+8+4+2+1

 

$$\\=2^{10}+2^9+2^8+2^7+2^6\qquad+2^4+2^3+2^2+2^1+2^0\\
=11111011111$$

 

Of course you could just use this converter

http://www.binaryhexconverter.com/decimal-to-binary-converter        

Melody  May 15, 2015
 #5
avatar+19653 
+10
Best Answer

what is the binary representation of 2015 ?

$$\begin{array}{rcrr}
& & \rm{q o u t i e n t} & \rm{r e m a i n d e r} \\
2015 & : 2 = & 1007 & 1 \\
1007 & : 2 = & 503 & 1 \\
503 & : 2 = & 251 & 1 \\
251 & : 2 = & 125 & 1 \\
125 & : 2 = & 62 & 1 \\
62 & : 2 = & 31 & 0 \\
31 & : 2 = & 15 & 1 \\
15 & : 2 = & 7 & 1 \\
7 & : 2 = & 3 & 1 \\
3 & : 2 = & 1 & 1 \\
1 & : 2 = & 0 & \textcolor[rgb]{1,0,0}{1} \\
\end{array}$$

 

$$2015_2~=~\textcolor[rgb]{1,0,0}{1}~1~1~1~1~0~1~1~1~1~1$$

heureka  May 15, 2015
 #6
avatar+92805 
+5

I would not have thought to do it your way Heureka but I really like your approach.

 

I have put this thread in the "Great answers to Learn from" sticky thread.     

Melody  May 26, 2015

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