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Prime number puzzle

  1. Dec 9, 2008 #1
    What three digit prime number has all prime digits and forms primes with its first two and last two digits?
     
  2. jcsd
  3. Dec 10, 2008 #2
  4. Dec 10, 2008 #3
    It's
    373
     
  5. Dec 10, 2008 #4
    Well it can't be 523 - 52 isn't prime.

    d_leet's answer is the only one.
     
  6. Dec 10, 2008 #5
    i guess its 373 (i had to add the "i guess its" because PF told me my post has to be atleast 4 charachters long :P)
     
  7. Dec 12, 2008 #6
    Ah. I misread the OP. I thought the *sums* d1 + d2 and d2 + d3 had to be prime.

    ie 5 +2 = 7 and 2 + 3 = 5

    At least my answer can be used to form a new puzzle :p

    k
     
  8. Dec 12, 2008 #7
    A quick&dirty computer program shows:
    11 13 113
    13 31 131
    13 37 137
    17 73 173
    17 79 179
    19 97 197
    31 11 311
    31 13 313
    31 17 317
    37 73 373
    37 79 379
    41 19 419
    43 31 431
    47 79 479
    61 13 613
    61 17 617
    61 19 619
    67 73 673
    71 19 719
    79 97 797
    97 71 971
     
  9. Dec 12, 2008 #8
    Remember, each digit has to be prime too, so the only one that works is 373. :)
     
  10. Dec 14, 2008 #9
    Noisy signal has seen 373, and adriank has seen after the programming of Dodo, that it is the only solution.

    This result can be achieved by logic: We look at the four one digit primes: 2, 3,5,7. We see that 5 can not be in the second or third place, and neither can 2. Any number containing digits 5,3,7 or 2,3,7 is divisible by 3. So only 3 and 7 can be used. 7 can not be in two successive places, and neither can 3. 737 is divisible by 11, so only 373 is left!
     
    Last edited: Dec 14, 2008
  11. Dec 14, 2008 #10
    Testing each number from 100 to 999 is logic too.
     
  12. Dec 14, 2008 #11
    Perhaps robert's solution is more elegant?
     
  13. Dec 14, 2008 #12
    Certainly; if I was without a calculator or so that's probably how I would do it. (That was probably the intention of the problem anyway.)
     
  14. Dec 15, 2008 #13
    adriank: Testing each number from 100 to 999 is logic too.

    I kept thinking I should say, "By logic alone," but I thought that might sound too egotisical. (But? Oh, well!)
     
  15. Dec 15, 2008 #14
    I'm still beating myself over not even imagining that the problem was solvable by logic. :smile:
     
  16. Dec 17, 2008 #15
    Well, I might add here, in case there was a problem. If the sum of the digits is divisible by 3 the number is divisible by 3. That takes care of any three terms containing 237, or 537. Now, a number is divisible by 11 if the alternating sum of its digits is divisible by 11. Here we have 7-3+7=11.

    Of course, years ago people knew those things, but since calculators, well, I am not so sure! In checking 373, we can eliminate 2,3,5,7,11 by simple rules. (For 7 the rule is to double the last term and subtract it from the rest.) Thus 37-6 is not divisible by 7.
    (That rule is really neat because 91x7 =637, but to check for 7 it is only necessary to subtract 14 from 63, leaving 49.)

    Possible factors of 373 such as 13,17,19 then could be checked by long division or a calculator.
     
    Last edited: Dec 17, 2008
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