+0  
 
0
122
2
avatar+195 

A certain integer has 4 digits when written in base 8. The same integer has d digits when written in base 2. What is the sum of all possible values of d?

Rollingblade  May 30, 2018
 #1
avatar
0

[1 x 8^3] + [0 x 8^2] + [0 x 8^1] + [0 x 8^0] =512 in base 10

1                0               0                0            =1 000 in base 8

512 in base 10 =

[1 x 2^9] + [0 x 2^8] + [0 x 2^7] + [0 x 2^6]......and so on to + [0 x 2^0]=

10 0000 0000 in base 2. The sum of all digits in base 2 = 1

Guest May 30, 2018
 #2
avatar+945 
+2

The smallest integer that has 4 digits in base 8 is \(1000_8\), which stands for \(8^3 = 2^9\). The largest integer that has 4 digits in base 8 is \(7777_8\), which is 1 less than \(10000_8\) and therefore stands for \(8^4-1 = 2^{12}-1.\)

 

Thus, when a 4-digit base-8 integer is written in base 2, its highest place value is either \( 2^9, 2^{10},\) or \(2^{11}\). It follows that the base-2 expression has 10, 11, or 12 digits, so the sum of all possible values for d is \(10+11+12 = \boxed{33}.\)

 

I hope this helped,

 

Gavin

GYanggg  May 31, 2018

10 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.