# BIP39 Manual Phrase Calculations - How are Multiple checksums valid?

Need some help understanding the math regarding why multiple checksums work for mnemonic phrase generation (BIP39).

Let's assume a 12 word passphrase. If we divide the 2048 wordlist into groups of 16 ... exactly 1 word out of 16 word "block" will be a valid checksum for the 11 words selected.

With a 24 word passphrase, 1 word out of every 256 would be a correct checksum.

When generating a phrase by hand... I know that the ENT / 32 bits of the sha-256 hash are appended to the entropy to generate the checksum word.. but this generates one specific word.

So I guess my long-winded question is ... what is the math behind other checksum values being valid? I guess my real question is how is the ENT + CS validated as legitimate?

See this example:

Entropy (128 bits): 11010011 01100100 00000010 01011110 01010011 11101100 01010011 01101110 01101010 01111000 11010010 11011000 10111010 00100011 11101100 11110010

SHA-256 Hash of entropy = 14 c5 8b c9 05 11 5e 08 27 49 61 1e 48 d6 04 c0 2a 70 8c 39 ad 6c dc 0c 91 2f 70 62 c3 24 71 23

First 4 bits of SHA-256 Hash = 1 (hex) Binary = 0001

Generated Recovery Phrase: square cactus nurse pond share rescue prepare bottom suffer speed will tomorrow

another valid phrase (with same entropy but different checksum): square cactus nurse pond share rescue prepare bottom suffer speed will account

Account = 13 (on the BIPS39 wordlist) Subtract 1 since word index starts at 0 = 12 12 to binary = 1100 Hex = C

C (hex) != 1 (hex)

another valid phrase (with same entropy but different checksum): square cactus nurse pond share rescue prepare bottom suffer speed will acoustic

Acoustic = 17 (on the BIPS39 wordlist) Subtract 1 since word index starts at 0 = 16 16 to Binary: 10000 Hex = 10

10 (hex) != 1 (hex)

I guess I'm also confused how these checksums are valid when their values do not equal the first 4 bits of the SHA-256 Hash of ENT? I'm guessing it has to do with how the checksum is validated (and pointing back to my original question)?

I believe the checked answer is incorrect. However, so are some assumptions in the question. First off, since there are 2048 words, each word encodes eleven bits (not 8). And since `128/11 = 11 remainder 7`, the entropy itself encodes the first 11 words and part of the 12th word. Conveniently the 4 bit checksum appended to the remaining 7 bits makes exactly 11 to encode the final word ("tomorrow").