This transfer is possible in two ways: direct transfer and using the decimal system.
first, let\'s make a direct transfer.
let\'s do a direct translation from hexadecimal to binary like this:
AF32C16 = A F 3 2 C = A(=1010) F(=1111) 3(=0011) 2(=0010) C(=1100) = 101011110011001011002
answer: AF32C16 = 101011110011001011002
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
10∙164+15∙163+3∙162+2∙161+12∙160 = 10∙65536+15∙4096+3∙256+2∙16+12∙1 = 655360+61440+768+32+12 = 71761210
got It: AF32C16 =71761210
Translate the number 71761210 в binary like this:
the Integer part of the number is divided by the base of the new number system:
717612 | 2 | | | | | | | | | | | | | | | | | | | |
-717612 | 358806 | 2 | | | | | | | | | | | | | | | | | | |
0 | -358806 | 179403 | 2 | | | | | | | | | | | | | | | | | |
| 0 | -179402 | 89701 | 2 | | | | | | | | | | | | | | | | |
| | 1 | -89700 | 44850 | 2 | | | | | | | | | | | | | | | |
| | | 1 | -44850 | 22425 | 2 | | | | | | | | | | | | | | |
| | | | 0 | -22424 | 11212 | 2 | | | | | | | | | | | | | |
| | | | | 1 | -11212 | 5606 | 2 | | | | | | | | | | | | |
| | | | | | 0 | -5606 | 2803 | 2 | | | | | | | | | | | |
| | | | | | | 0 | -2802 | 1401 | 2 | | | | | | | | | | |
| | | | | | | | 1 | -1400 | 700 | 2 | | | | | | | | | |
| | | | | | | | | 1 | -700 | 350 | 2 | | | | | | | | |
| | | | | | | | | | 0 | -350 | 175 | 2 | | | | | | | |
| | | | | | | | | | | 0 | -174 | 87 | 2 | | | | | | |
| | | | | | | | | | | | 1 | -86 | 43 | 2 | | | | | |
| | | | | | | | | | | | | 1 | -42 | 21 | 2 | | | | |
| | | | | | | | | | | | | | 1 | -20 | 10 | 2 | | | |
| | | | | | | | | | | | | | | 1 | -10 | 5 | 2 | | |
| | | | | | | | | | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
71761210 = 101011110011001011002
answer: AF32C16 = 101011110011001011002