This transfer is possible in two ways: direct transfer and using the decimal system.
First we will perform the translation through the decimal system
let\'s translate to decimal like this:
11∙162+12∙161+1∙160+2∙16-1+3∙16-2 = 11∙256+12∙16+1∙1+2∙0.0625+3∙0.00390625 = 2816+192+1+0.125+0.01171875 = 3009.1367187510
got It: BC1.2316 =3009.1367187510
Translate the number 3009.1367187510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
3009 | 8 | | | |
-3008 | 376 | 8 | | |
1 | -376 | 47 | 8 | |
| 0 | -40 | 5 | |
| | 7 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 13671875*8 |
1 | .09375*8 |
0 | .75*8 |
6 | .0*8 |
the result of the conversion was:
3009.1367187510 = 5701.1068
answer: BC1.2316 = 5701.1068
Now we will perform a direct translation.
let\'s do a direct translation from hexadecimal to binary like this:
BC1.2316 = B C 1. 2 3 = B(=1011) C(=1100) 1(=0001). 2(=0010) 3(=0011) = 101111000001.001000112
answer: BC1.2316 = 101111000001.001000112
Fill in the number with missing zeros on the right
let\'s make a direct translation from binary to post-binary like this:
101111000001.0010001102 = 101 111 000 001. 001 000 110 = 101(=5) 111(=7) 000(=0) 001(=1). 001(=1) 000(=0) 110(=6) = 5701.1068
answer: BC1.2316 = 5701.1068