- #1

leetchaos

- 2

- 0

Key: 1134487964316

Lock1: 23144767481431

Lock2: 43148651671461

Lock3: 43517649819891

Lock4: 76487995462131

Lock5: 47687982498191

Lock6: 46149897275131

Lock7: 46768491794850

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- Thread starter leetchaos
- Start date

- #1

leetchaos

- 2

- 0

Key: 1134487964316

Lock1: 23144767481431

Lock2: 43148651671461

Lock3: 43517649819891

Lock4: 76487995462131

Lock5: 47687982498191

Lock6: 46149897275131

Lock7: 46768491794850

- #2

Soca fo so

- 51

- 0

Okay so we've got a key with 13 digits and 7 locks with 14 digits each, so I'm guessing we need to find the missing digit in the key for each lock.

One way you could do this is to add the digits in the lock and find the remainder MOD 10, and do the same for the key and have these remainders be the same i.e. have the sums be equivalent modulo 10.

So for lock 1 the sum of the digits is 55 = 5 mod 10, and for the key the sum is 57 + x (where x is the missing digit) which must be equivalent to 5 mod 10, which would give x = 8. The problem then is where to insert this digit into the key. One, completely arbitrary, way could be to insert the digit in place 1 for lock 1, place 2 for lock 2, and so on. So for lock 1 the key would be 81134487964316.

Alternatively instead of the straight sum of the digits you could do the weighted sum, then the missing digit would have to be inserted at place 1.

So for lock 1 the sum would be 1(2) + 2(3) + 3(1) + 4(4) + ... + 14(1) = 415 = 5 mod 10, and x + 491 for the key, giving x = 4. So the key in this case would be 41134487964316.

One way you could do this is to add the digits in the lock and find the remainder MOD 10, and do the same for the key and have these remainders be the same i.e. have the sums be equivalent modulo 10.

So for lock 1 the sum of the digits is 55 = 5 mod 10, and for the key the sum is 57 + x (where x is the missing digit) which must be equivalent to 5 mod 10, which would give x = 8. The problem then is where to insert this digit into the key. One, completely arbitrary, way could be to insert the digit in place 1 for lock 1, place 2 for lock 2, and so on. So for lock 1 the key would be 81134487964316.

Alternatively instead of the straight sum of the digits you could do the weighted sum, then the missing digit would have to be inserted at place 1.

So for lock 1 the sum would be 1(2) + 2(3) + 3(1) + 4(4) + ... + 14(1) = 415 = 5 mod 10, and x + 491 for the key, giving x = 4. So the key in this case would be 41134487964316.

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