Thread: Barcode proof problem

A particular identification code (ID) has a 9-digit code with check vector c = [9, 8, 7, 6, 5, 4, 3, 2, 1]. A valid ID v has c · v = 0 in Z11.
A common error when entering an ID on machine-readable forms is to enter 0
instead of 9, or x + 1 instead of x for 0 ≤ x ≤ 8. (So a 0 gets entered as 1, or 1 as 2, for example.)
Prove that a single error of this type will always be detected. (That is, prove that if a single error of this type is made, and the resulting incorrect ID is v, then c · v is not equals to 0 in Z11.)

Thanks for all your help.(in adv)

Originally Posted by platon

A particular identification code (ID) has a 9-digit code with check vector c = [9, 8, 7, 6, 5, 4, 3, 2, 1]. A valid ID v has c · v = 0 in Z11.
A common error when entering an ID on machine-readable forms is to enter 0
instead of 9, or x + 1 instead of x for 0 ≤ x ≤ 8. (So a 0 gets entered as 1, or 1 as 2, for example.)
Prove that a single error of this type will always be detected. (That is, prove that if a single error of this type is made, and the resulting incorrect ID is v, then c · v is not equals to 0 in Z11.)

Thanks for all your help.(in adv)

Looks pretty straight forward. Adding any integer 9 or less to a multiple of 11 cannot give a multiple of 11.

Originally Posted by HallsofIvy

Looks pretty straight forward. Adding any integer 9 or less to a multiple of 11 cannot give a multiple of 11.

thanks, i understand that if the wrong ID is entered(1 wrong entry),
c.v will always give 1-9 mod 11?
but is there like a standard working to it, derivation based on algebra?
I am pretty stuck on the workings although i get the concept.