Transfer 110001101.1100100111 from binary in octal number system

This transfer is possible in two ways: direct transfer and using the decimal system.

first, let\'s make a direct transfer.

Fill in the number with missing zeros on the right

let\'s make a direct translation from binary to post-binary like this:

110001101.110010011100₂ = 110 001 101. 110 010 011 100 = 110₍₌₆₎ 001₍₌₁₎ 101₍₌₅₎. 110₍₌₆₎ 010₍₌₂₎ 011₍₌₃₎ 100₍₌₄₎ = 615.6234₈

the Final answer: 110001101.1100100111₂ = 615.6234₈

now let\'s make the transfer using the decimal system.

let\'s translate to decimal like this:

1∙2⁸+1∙2⁷+0∙2⁶+0∙2⁵+0∙2⁴+1∙2³+1∙2²+0∙2¹+1∙2⁰+1∙2^-1+1∙2^-2+0∙2^-3+0∙2^-4+1∙2^-5+0∙2^-6+0∙2^-7+1∙2^-8+1∙2^-9+1∙2^-10+0∙2^-11+0∙2^-12 = 1∙256+1∙128+0∙64+0∙32+0∙16+1∙8+1∙4+0∙2+1∙1+1∙0.5+1∙0.25+0∙0.125+0∙0.0625+1∙0.03125+0∙0.015625+0∙0.0078125+1∙0.00390625+1∙0.001953125+1∙0.0009765625+0∙0.00048828125+0∙0.000244140625 = 256+128+0+0+0+8+4+0+1+0.5+0.25+0+0+0.03125+0+0+0.00390625+0.001953125+0.0009765625+0+0 = 397.7880859375₁₀

got It: 110001101.110010011100₂ =397.7880859375₁₀

Translate the number 397.7880859375₁₀ в octal like this:

the Integer part of the number is divided by the base of the new number system:

397	8
-392	49	8
5	-48	6
	1

the Fractional part of the number is multiplied by the base of the new number system:


	0.	7880859375*8
	6	.30469*8
	2	.4375*8
	3	.5*8
	4	.0*8

the result of the conversion was:

397.7880859375₁₀ = 615.6234₈

the Final answer: 110001101.1100100111₂ = 615.6234₈

0
(	)	^	←
C	±	÷	X
7	8	9	-
4	5	6	+
1	2	3	=
0		.	=