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:
1AC2F16 = 1 A C 2 F = 1(=0001) A(=1010) C(=1100) 2(=0010) F(=1111) = 110101100001011112
answer: 1AC2F16 = 110101100001011112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
1∙164+10∙163+12∙162+2∙161+15∙160 = 1∙65536+10∙4096+12∙256+2∙16+15∙1 = 65536+40960+3072+32+15 = 10961510
got It: 1AC2F16 =10961510
Translate the number 10961510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
109615 | 2 | | | | | | | | | | | | | | | | |
-109614 | 54807 | 2 | | | | | | | | | | | | | | | |
1 | -54806 | 27403 | 2 | | | | | | | | | | | | | | |
| 1 | -27402 | 13701 | 2 | | | | | | | | | | | | | |
| | 1 | -13700 | 6850 | 2 | | | | | | | | | | | | |
| | | 1 | -6850 | 3425 | 2 | | | | | | | | | | | |
| | | | 0 | -3424 | 1712 | 2 | | | | | | | | | | |
| | | | | 1 | -1712 | 856 | 2 | | | | | | | | | |
| | | | | | 0 | -856 | 428 | 2 | | | | | | | | |
| | | | | | | 0 | -428 | 214 | 2 | | | | | | | |
| | | | | | | | 0 | -214 | 107 | 2 | | | | | | |
| | | | | | | | | 0 | -106 | 53 | 2 | | | | | |
| | | | | | | | | | 1 | -52 | 26 | 2 | | | | |
| | | | | | | | | | | 1 | -26 | 13 | 2 | | | |
| | | | | | | | | | | | 0 | -12 | 6 | 2 | | |
| | | | | | | | | | | | | 1 | -6 | 3 | 2 | |
| | | | | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | | | | 1 | | |
|
the result of the conversion was:
10961510 = 110101100001011112
answer: 1AC2F16 = 110101100001011112