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**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**+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