1
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Four rows of numbers:

1234567890
5916708234
192345678
???

Fill in the last row with the appropriate digits (not necessarily 3, even though there are 3 '?'s). Please explain your answer.

Hint/clarification:

I'm in an English-speaking country.

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  • $\begingroup$ Is it a coincidence all given numbers are made out of 1-9 (each number only used once for each) the first 2 also have 0. $\endgroup$
    – Vincent
    Commented Apr 30, 2015 at 9:13

2 Answers 2

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Shouldn't the first row be this:

2345678901

?

My answer:

7625143

Reason:

The digits correspond to the standard qwerty keyboard layout (restricted to alphanumeric characters). Each digit describes the corresponding character's alphabetical (or ASCII) order within its row (modulo 10).

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  • $\begingroup$ You've got the right answer, but I'm not sure why you think the first row should be different (unless you're using a different keyboard style from me). $\endgroup$ Commented Apr 30, 2015 at 18:58
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    $\begingroup$ In the qwertyuiop and asdfghjkl rows you clearly denote the first letter with 1, the second with 2 etc., the ninth with 9 and the tenth with 0 (that is, 10 modulo 10). The upper row of the US keyboard layout contains 1234567890, the last key clearly being 0 and not 10. This is not the natural ordering. To order correctly, the key labeled 0 should receive the sequence number 1, the key labeled 1 should receive the sequence number 2 and so on. $\endgroup$
    – egmont
    Commented Apr 30, 2015 at 19:15
  • $\begingroup$ Ah, fair enough. I was imagining identifying 0 with 10 throughout. $\endgroup$ Commented Apr 30, 2015 at 19:18
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    $\begingroup$ Probably the puzzle would have been clearer with 0-based sequence numbers. Anyway, it was a nice one :) I quickly discovered a high correlation with the keyboard layout, e.g. the 345678 continuous run corresponding to fghjkl, but "i" is missing caused quite a bit of headache. I even looked at the EBCDIC code table which has similar gaps between some adjacent letters, but of course it didn't lead to a solution. $\endgroup$
    – egmont
    Commented Apr 30, 2015 at 19:24
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Based on what egmont said (credits to him for the solution), the answer is

7625143

Take a qwerty keyboard and consider the english letters and numbers. First line is 1234567890. Second line is qwertyuiop, if you sort it eiopqrtuwy. As you see, q is the 5th letter in eiopqrtuwy, w is the 9th, e is 1st and so on, we get 5916708234. If we do the same for the last line, we get 7625143. Also, there's no contradiction in the first line, as he couldn't write 12345678910 (ugly, don't you think?) Also, no need for modulo 10.

Disclaimer: I wrote this answer because felt that egmont's answer, even though correct, wasn't very easy to understand.

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  • $\begingroup$ "Disclaimer: I wrote this answer because felt that egmont's answer, even though correct, wasn't very easy to understand." - then why didn't you just edit egmont's answer? And there is need for modulo 10, precisely by your previous sentence: 0 represents 10 in 1234567890! $\endgroup$ Commented Apr 30, 2015 at 18:59

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