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:
1231516 = 1 2 3 1 5 = 1(=0001) 2(=0010) 3(=0011) 1(=0001) 5(=0101) = 100100011000101012
answer: 1231516 = 100100011000101012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
1∙164+2∙163+3∙162+1∙161+5∙160 = 1∙65536+2∙4096+3∙256+1∙16+5∙1 = 65536+8192+768+16+5 = 7451710
got It: 1231516 =7451710
Translate the number 7451710 в binary like this:
the Integer part of the number is divided by the base of the new number system:
74517 | 2 | | | | | | | | | | | | | | | | |
-74516 | 37258 | 2 | | | | | | | | | | | | | | | |
1 | -37258 | 18629 | 2 | | | | | | | | | | | | | | |
| 0 | -18628 | 9314 | 2 | | | | | | | | | | | | | |
| | 1 | -9314 | 4657 | 2 | | | | | | | | | | | | |
| | | 0 | -4656 | 2328 | 2 | | | | | | | | | | | |
| | | | 1 | -2328 | 1164 | 2 | | | | | | | | | | |
| | | | | 0 | -1164 | 582 | 2 | | | | | | | | | |
| | | | | | 0 | -582 | 291 | 2 | | | | | | | | |
| | | | | | | 0 | -290 | 145 | 2 | | | | | | | |
| | | | | | | | 1 | -144 | 72 | 2 | | | | | | |
| | | | | | | | | 1 | -72 | 36 | 2 | | | | | |
| | | | | | | | | | 0 | -36 | 18 | 2 | | | | |
| | | | | | | | | | | 0 | -18 | 9 | 2 | | | |
| | | | | | | | | | | | 0 | -8 | 4 | 2 | | |
| | | | | | | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | | | | 0 | | |
|
the result of the conversion was:
7451710 = 100100011000101012
answer: 1231516 = 100100011000101012