HEX TO DECIMAL TO OCTAL TO BINARY
0hex |
= |
= |
0oct |
0 |
0 |
0 |
0 |
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1hex |
= |
= |
1oct |
0 |
0 |
0 |
1 |
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2hex |
= |
= |
2oct |
0 |
0 |
1 |
0 |
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3hex |
= |
= |
3oct |
0 |
0 |
1 |
1 |
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4hex |
= |
= |
4oct |
0 |
1 |
0 |
0 |
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5hex |
= |
= |
5oct |
0 |
1 |
0 |
1 |
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6hex |
= |
= |
6oct |
0 |
1 |
1 |
0 |
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7hex |
= |
= |
7oct |
0 |
1 |
1 |
1 |
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8hex |
= |
= |
10oct |
1 |
0 |
0 |
0 |
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9hex |
= |
= |
11oct |
1 |
0 |
0 |
1 |
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Ahex |
= |
= |
12oct |
1 |
0 |
1 |
0 |
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Bhex |
= |
= |
13oct |
1 |
0 |
1 |
1 |
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Chex |
= |
= |
14oct |
1 |
1 |
0 |
0 |
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Dhex |
= |
= |
15oct |
1 |
1 |
0 |
1 |
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Ehex |
= |
= |
16oct |
1 |
1 |
1 |
0 |
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Fhex |
= |
= |
17oct |
1 |
1 |
1 |
1 |
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The hexadecimal term 2AF316 is equal to a sum of (200016 + A0016 + F016 + 316).
By decomposing the numeral into a series of place value terms -- converting each term to decimal -- one can further write:
(2 × 163) + (A × 162) + (F × 161) + (3 × 160),
(2 × 4096) + (10 × 256) + (15 × 16) + (3 × 1), or 10995 in decimal format
The octal term 112 is equal to the sum of (1008 + 108 + 28) which leads us to
(1 x 82) + (1 x 81) + (2 x 80),
(1 x 64) + (1 x 8) + (2 x 1), or 74 in decimal format.