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:
5A2316 = 5 A 2 3 = 5(=0101) A(=1010) 2(=0010) 3(=0011) = 1011010001000112
answer: 5A2316 = 1011010001000112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
5∙163+10∙162+2∙161+3∙160 = 5∙4096+10∙256+2∙16+3∙1 = 20480+2560+32+3 = 2307510
got It: 5A2316 =2307510
Translate the number 2307510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
23075 | 2 | | | | | | | | | | | | | | |
-23074 | 11537 | 2 | | | | | | | | | | | | | |
1 | -11536 | 5768 | 2 | | | | | | | | | | | | |
| 1 | -5768 | 2884 | 2 | | | | | | | | | | | |
| | 0 | -2884 | 1442 | 2 | | | | | | | | | | |
| | | 0 | -1442 | 721 | 2 | | | | | | | | | |
| | | | 0 | -720 | 360 | 2 | | | | | | | | |
| | | | | 1 | -360 | 180 | 2 | | | | | | | |
| | | | | | 0 | -180 | 90 | 2 | | | | | | |
| | | | | | | 0 | -90 | 45 | 2 | | | | | |
| | | | | | | | 0 | -44 | 22 | 2 | | | | |
| | | | | | | | | 1 | -22 | 11 | 2 | | | |
| | | | | | | | | | 0 | -10 | 5 | 2 | | |
| | | | | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
2307510 = 1011010001000112
answer: 5A2316 = 1011010001000112