# Why was finding a hash beginning with a certain number of zeroes chosen as the proof of work?

It seems to me that the reason the proof-of-work involves finding a hash beginning with a (temporarily) fixed number of zeroes is because one can easily check whether they have solved the problem by comparing the hash they have generated with the value 2^k, where k = (256 - # of zero bits) and doing so is a relatively inexpensive operation. Also, it is easy to make the problem arbitrarily difficult up to a certain point (as well as having the difficulty increase exponentially with the number of zeros).

Are there any other reasons for why this particular problem was chosen for the proof-of-work?

• It doesn't check for a number of zero bits. It checks whether the hash, when interpreted as a 256-bit number, is lower than a target number. That target number is occasionally adjusted. Nov 14, 2015 at 22:19
• Is the target number a power of two, or can it be any 256-bit number?
– Ali
Nov 15, 2015 at 1:34
• @Ali It can't be any 256 bit number (there's a format it needs to follow, see here) but you can use very nearly any value. Nov 15, 2015 at 2:28