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Question is just out of interest.

If we take md5 hash from all integers from 1 to n, what is the first n, where its md5 collides with some previous md5?

Has anyone ever calculated and proved the smallest integer, where the collision occurs?

If not, is there considerations about md5 pseudo-randomness, which make probability of colliding md5-s higher for consecutive integers than for cryptographically random strings (birthday attack -> 2^64)?

This, for example, might be useful to know, when trying to hide real sequential order numbers.

Rauli Rajande
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MD5 has a 128-bit output, so it's expected that the first collision is around $2^{64}$. This is because of the Birthday-Problem in probability theory.

So you should be safe to take orders up to $$2^{64} = 18,446,744,073,709,551,616$$

But that shouldn't be a problem for you, because even though this number might not seem like a large number it actually is huge:

An Example:

If every human ($ \approx 7,000,000,000$ people) would place an order every second, then it would take $${2^{64} \over 7,000,000,000 * 3600 * 24 * 365} \approx 83.5$$ years.

So you can be extremely sure that you won't have a collision of hashed order-numbers.

AleksanderCH
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