Question

1. convert 1011101 two to

a base ten number
a base eight number
2. convert each of the following to base ten
2341 six and 30001 four ​

Answer 1

1a) 93ten

1b) 135eight

2a) 565ten

2b) 769ten

Explanation:

1a)

1011101

7 digits long the first digit (the one that is most far right) is the ones digit or 2^0 digit and so the last digit (the most far left) is 2^6.

So to put the number in base 10 we do:

1(2^6)+0(2^5)+1(2^4)+1(2^3)+1(2^2)+0(2^1)+1(2^0)

=64+0+16+8+4+0+1

=64+16+8+4+1

=(64+16)+(8+4+1)

=80+13

=93

1b)

93

What is the hightest power of 8 in 93?

8^x=93

8^1=8

8^2=64

8^3 would be a factor of 8 too much

So how many 8^2 's are in 93?

64 goes into 93 how many times?

It's once with a remainder of 93-64=29.

So now 29...what is the highest power of 8 that goes into 29? 8^1=8.

How many times? 3(8)=24 with remainder 5.

So 93=1(8^2)+3(8^1)+5(8^0).

93=135eight

2a) 2341six

2(6^3)+3(6^2)+4(6^1)+1(6^0)

2(216)+3(36)+4(6)+1

432+108+24+1

540+25

565

2b) 30001four

3(4^4)+0(4^3)+0(4^2)+0(4^1)+1(4^0)

3(256)+0+0+0+1

768+1

769