How to find the binary equivalent of 111(10)?

Question

If possible please tell how to add binary numbers and to find the equivalent of a binary digit [ like 1101(2) ]. Thanks.....

Answer 1

1101(2)

The rightmost digit is the number of ones you add in. This spot represents (2^0).
The second digit from the right is the number of twos you add in. This spot represents (2^1)
The third digit from the right is the number of fours you add in. This spot represents (2^2)
The fourth digit from the right is the number of eights you add in. This spot represents (2^3).

For 1101(2), you thus have one eight, one four, no twos, and one one = 8 + 4 + 1 = 13


Any number written in base two can be analyzed in this fashion. Each digit represents a power of two. As you have longer numbers in base two, you just have more powers of two to consider adding to your solution.

For additional digits:
(2^4) = 16
(2^5) = 32
(2^6) = 64
(2^7) = 128 etc.

To go the other way, from base 10 to binary, first determine how many digits you will have. 111(10) is less than (2^7) = 128, so you will have only seven digits in your binary answer.

Start with:
1000000(2) = 64(10)

This is still less than 111(10).

The next power of two you could possibly add in is (2^5) = 32. Will this put you over 111(10)? No - 64 + 32 = 96. So we have:

1100000(2) = 96(10)

The next power of two you could possibly add in is (2^4) = 16. Will this put you over 111(10)? Yes - 96 + 16 = 112. So that spot has a zero in binary.

The next power of two you could possibly add in is (2^3) = 8. Will this put you over 111(10)? No- 96 + 8 = 104. So we have:

1101000(2) = 104(10)

The next power of two you could possibly add in is (2^2) = 4. Will this put you over 111(10)? No- 104 + 4 = 108. So we have:

1101100(2) = 108(10)

The next power of two you could possibly add in is (2^1) = 2. Will this put you over 111(10)? No- 108 + 2 = 110. So we have:

1101110(2) = 110(10)

For the last spot, do you need to add a (2^0), or a 1, to that to get your match? Yes, and you're done!

1101111(2) = 111(10)

Good luck - hope this helps!

Answer 2

If you have a number in base 10 and you want to transform it into another base, divide the number by the base and note the remainder. The first remainder is the right-most digit in the new base.

e.g. 111 (base 10) to be transformed to base 2:

111 / 2 = 55 + remainder 1
55/2 = 27 + remainder 1
27 / 2 = 13 + remainder 1
13 / 2 = 6 + remainder 1
6 / 2 = 3 + remainder 0
3/2 = 1 + remainder 1
1 / 2 = 0 + remainder 1

stop (having arrived at 0)

working backwards, the number is 1101111

This is not a "short method" but it has the advantage of working with any integer, from any base to any other base.

For example, 111(10) to base 7:
111 / 7 = 15 + remainder 6
15 / 7 = 2 + remainder 1
2 / 7 = 0 + remainder 2

answer (remember: backwards) = 216

111(10) to base 16:

111 / 16 = 6 + remainder 15 (F)
6 / 16 = 0 + remainder 6

answer = 6F

(digits in base 16 are
0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F)

Answer 3

Binary numbers can be added. Follow one simple rule 1+1 =0, carry the 1.

01101
+1011
---------
11000

like in the base 10 system, the "place values" follow an exponential sequence.
in base 10, we have 1000's place, 100's place, 10's, 1's
same as 10^3 place, 10^2, 10^1, 10^0.

in base two the places are still exponential, but are base 2:
2^5, 2^4, 2^3, 2^2, 2^1, 2^0
or 32's, 16's, 8's, 4's, 2's 1's place.

so 110101 = 1(32) + 1(16) + 0(8) + 1(4) + 0(2) + 1(1) = 53

Answer 4

Start from the far right and begin with 2^0 =1 then 0=2^1=2 0x2=0
1x2^2=4 1x2^3=8==1+0+4+8=13--answer