This transfer is possible in two ways: direct transfer and using the decimal system.
first, let\'s make a direct transfer.
let\'s translate to decimal like this:
4∙84+3∙83+6∙82+4∙81+3∙80 = 4∙4096+3∙512+6∙64+4∙8+3∙1 = 16384+1536+384+32+3 = 1833910
got It: 436438 =1833910
Translate the number 1833910 в hexadecimal like this:
the Integer part of the number is divided by the base of the new number system:
18339 | 16 | | | |
-18336 | 1146 | 16 | | |
3 | -1136 | 71 | 16 | |
| A | -64 | 4 | |
| | 7 | | |
|
the result of the conversion was:
1833910 = 47A316
the Final answer: 436438 = 47A316
now let\'s make the transfer using the decimal system.
let\'s do a direct translation from octal to binary like this:
436438 = 4 3 6 4 3 = 4(=100) 3(=011) 6(=110) 4(=100) 3(=011) = 1000111101000112
the Final answer: 436438 = 1000111101000112
Fill in the number with missing zeros on the left
let\'s do a direct translation from binary to hexadecimal like this:
01000111101000112 = 0100 0111 1010 0011 = 0100(=4) 0111(=7) 1010(=A) 0011(=3) = 47A316
the Final answer: 01000111101000118 = 47A316