The method you must use to get to your answer is like this:
Perform a modulo operation of 2 on the number, write the digit down and perform an integer division of 2 on your number.
Then, repeat every step by writing the result of the modulo before your other modulo-digits. Your last number after the integer division will be a 1; write that number down as a digit before your modulo-digits. The number that consists of 0's en 1's is your decimal number in binary.
For example with the decimal number 345125:
345125%2=1
(345125)dec = (..1)bin
345125/2=172562
172562%2=0
(345125)dec = (..01)bin
172562/2=86281
86281%2=1
(345125)dec = (..101)bin
86281/2=43140
43140%2=0
(345125)dec = (..0101)bin
43140/2=21570
21570%2=0
(345125)dec = (..00101)bin
21570/2=10785
10785%2=1
(345125)dec = (..100101)bin
10785/2=5392
5392%2=0
(345125)dec = (..0100101)bin
5392/2=2696
2696%2=0
(345125)dec = (..00100101)bin
2696/2=1348
1348%2=0
(345125)dec = (..000100101)bin
1348/2=674
674%2=0
(345125)dec = (..0000100101)bin
674/2=337
337%2=1
(345125)dec = (..10000100101)bin
337/2=168
168%2=0
(345125)dec = (..010000100101)bin
168/2=84
84%2=0
(345125)dec = (..0010000100101)bin
84/2=42
42%2=0
(345125)dec = (..00010000100101)bin
42/2=21
21%2=1
(345125)dec = (..100010000100101)bin
21/2=10
10%2=0
(345125)dec = (..0100010000100101)bin
10/2=5
5%2=1
(345125)dec = (..10100010000100101)bin
5/2=2
2%2=0
(345125)dec = (..010100010000100101)bin
2/2=1
(345125)dec = (10100010000100101)bin
So the decimal number 345125 in binary is 10100010000100101.
Binary what? Binary numbers? Binary stars? Binary fission?
No, binary is a number system.A binary digit is called a bit.
Infinite (and binary).
Binary trees are commonly used to implement binary search tree and binary heaps.
binary fission
The Binary for ten in 8-bit binary is: 00001010
The sum of binary numbers is also a binary number.
It is 10111111 in binary. Try a search for '191 to binary'.
100011 is 35 in binary.
The binary of 15001 is 11101010011001
111010 is 58 in binary.
128 in binary is 10000000.