In the validation of a block it is said that a miner has a valid block if the transactions can be tracked in the chain and the hash of the block header is less than the difficulty. I am having a hard time seeing the relationship in real world blocks that are being solved.

For Example from Block#496785 :

Difficulty("target") = 1,347,001,430,558.57 or in hex 000000000000000000000000000000000000000000000000000001399F8AB21E

Mined Hash = 000000000000000000cf3620d570d08d1799a1cafbbfae512fdba2124665eca0

so it seems to me the hash is now greater than the difficulty so an invalid block but this is obviously not the case.

I have heard also that the difficulty is related to the number of leading zeros after the maximum target in which case may make sense where 1399F8AB21E is a 11 byte number and the hash after the default 8 byte leading target contains 10 bytes which is less than the difficulty 11 bytes of zeros.


00000000 0000000000 cf3620d570d08d1799a1cafbbfae512fdba2124665eca0
8 bytes  difficulty             some value

Is this correct? How does this work?

1 Answer 1


Difficulty is not the same as target hash (target in short).

They are inversely related, so minimal possible difficulty (of 1) gives maximal target, by definition given as (https://en.bitcoin.it/wiki/Target):


Because Bitcoin stores the target as a floating-point type, this is truncated:


The current difficulty is calculated as maximal target divided by the current target. For example (https://en.bitcoin.it/wiki/Difficulty):

0x00000000FFFF0000000000000000000000000000000000000000000000000000 / 0x00000000000404CB000000000000000000000000000000000000000000000000 = 16307.420938523983

So in your example the Difficulty = 1,347,001,430,558.57 would give the current target of:

current_target = 0x00000000FFFF0000000000000000000000000000000000000000000000000000 / 1,347,001,430,558.57

If you calculate this (take care as one number is hexadecimal and other decimal), the comparison should give that this current_target is higher than the mined hash of:


EDIT: Pen and paper calculation gives me that the current target is around:


How I got this. 1,347,001,430,558.57 is around 1.225 * (16^10). As the target is in base 16, this means that dividing by 1 * (16^10) is like shifting the digits 10 places to the right. To get precise result after digits shifting I should divide by 1.225, but I ignored this in this pen and paper calculation.

EDIT2: To be even more precise I take 16/1.225 is around 13 (D in hex), so the target is around:


  • Why does 0x00000000FFFF0000000000000000000000000000000000000000000000000000 represent the truncated floating-point type version of 0x00000000FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF? Jun 24, 2018 at 9:58

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