You get a specific Target value when you divide the maximum Target by the Difficulty.

But when you convert the Target in to the Bits format, you're only taking the first 3 bytes of the Target and losing some accuracy.

For example:

Difficulty: 166851513282.78

Original Target: 0x696f3ffffffe0c000000000000000000000000000000000
Target -> Bits: 0x180696f4
Bits -> Target:  0x696f4000000000000000000000000000000000000000000

So when a Miner is trying to get a low enough hash value for the block, are they trying to get below the Bits value or the Target value?


Below the target value that corresponds to the bits field in the block.

  • So the Bits field is ultimately a short (and slightly less accurate) representation of the Target, just for the purposes of having a representation of the Target in the block header?
    – inersha
    Apr 14 '16 at 12:01
  • Also, I'm assuming Bits is always rounded up?
    – inersha
    Apr 14 '16 at 13:30

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