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Find solution, how work algoritem for last, control number in 8 places barcode. Here is some results:

  1. Here we got last number 8Example 1
  2. Here we got last number 7: Example 2
  3. Here we got last number 5 Example3

What is last number of this 2 Barcode?

  1. Barcode 1

and this

  1. Barcode 2

Try multiply some numbers with 3

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closed as off-topic by Deusovi, Mike Earnest, AJL, Spencerkatty, Will Mar 6 '16 at 3:41

  • This question does not appear to be about creation and solving of puzzles, within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ Finding industry barcode algorithms is not a puzzle. $\endgroup$ – Deusovi Mar 5 '16 at 22:36
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    $\begingroup$ I'm voting to close this question as off-topic because it's not a puzzle. $\endgroup$ – Deusovi Mar 5 '16 at 22:43
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    $\begingroup$ I have a feeling that if the answer wasn't publicly available information, the three examples given would not have been enough to figure it out. $\endgroup$ – f'' Mar 5 '16 at 23:16
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The barcode checksum is derived through the following process:

  1. Add the digits at odd positions, then multiply by 3.

  2. Add the digits at even positions to your answer.

  3. 10 minus your result, modulo 10, is the check digit.

So for the first one: $ 10 -( ((5+1+4+8)\times3 + 1 + 8 + 3) \mod 10)=7$

And for the second: $10- (((5+0+2+4)\times 3 + 7 + 1 + 3 )\mod 10) = 6$

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  • $\begingroup$ You have right, but only first one you write 1 not 0. $\endgroup$ – neiiic Mar 5 '16 at 22:42
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Pretty sure its:

1. 5

and

2. 0

Just by looking at previous numbers.

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  • $\begingroup$ @Deusovi why did I get a downvote for answering? $\endgroup$ – Z. Dailey Mar 5 '16 at 22:37
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    $\begingroup$ Because this answer is wrong. $\endgroup$ – Deusovi Mar 5 '16 at 22:38
  • $\begingroup$ Someone needs their 886609 546***. Sheesh. $\endgroup$ – Z. Dailey Mar 5 '16 at 23:04
  • $\begingroup$ You need to provide justification for why you think the answer is correct. "Just by looking at previous numbers" is not helpful. $\endgroup$ – f'' Mar 7 '16 at 5:19
  • $\begingroup$ Yah. I didn't assume it was necessary after it was pointed out to be wrong and the right answer was already posted. I looked at the section of the barcode over the blank space and attempted to match to a known number. $\endgroup$ – Z. Dailey Mar 7 '16 at 5:24

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