# Transfer 1A3C from hexadecimal in octal number system

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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:

1∙163+10∙162+3∙161+12∙160 = 1∙4096+10∙256+3∙16+12∙1 = 4096+2560+48+12 = 671610

got It: 1A3C16 =671610

Translate the number 671610 в octal like this:

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

 6716 8 -6712 839 8 4 -832 104 8 7 -104 13 8 0 -8 1 5

the result of the conversion was:

671610 = 150748
the Final answer: 1A3C16 = 150748

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

let's do a direct translation from hexadecimal to binary like this:

1A3C16 = 1 A 3 C = 1(=0001) A(=1010) 3(=0011) C(=1100) = 11010001111002

the Final answer: 1A3C16 = 11010001111002

Fill in the number with missing zeros on the left

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

0011010001111002 = 001 101 000 111 100 = 001(=1) 101(=5) 000(=0) 111(=7) 100(=4) = 150748

the Final answer: 1A3C16 = 150748