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What is the unsigned range of a bytes32 type in Solidity?

I'm wondering the total number of unique expressible combinations. I'm assuming it is 2^8^32.

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    You are correct bytes32 have 32 bytes you have 256^32 = (2^8)^32 possible combinations. – Ismael Oct 8 '17 at 17:16
  • @Ismael Can you add this as an answer so I can mark it as answered? – Alec Kriebel Oct 9 '17 at 16:40
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You are correct bytes32 has 32 bytes, each byte has 256 combination then you will have 256×256×..{32 veces}..×256 = 256^32 = (2^8)^32 possible combinations total.

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