1. Non-Sub Recs: 0
    Joined
    08 Apr '05
    Moves
    455
    22 Feb '06 13:131 edit
    Find all positive integers n (n will be called special)
    such that if n divides a three digit number say ABC where A = hundreds, B = tens, C = ones. Then n also divides BCA and CAB.

    It's trickier than you think ...
  2. Joined
    01 May '05
    Moves
    390
    22 Feb '06 13:16
    Originally posted by phgao
    Find all positive integers n (n will be called special)
    such that if n divides a three digit number say ABC where A = hundreds, B = tens, C = ones. Then n also divides BCA and CAB.

    It's trickier than you think ...
    I'd say 3 is one of them.
  3. Non-Sub Recs: 0
    Joined
    08 Apr '05
    Moves
    455
    22 Feb '06 13:21
    Originally posted by fetofs
    I'd say 3 is one of them.
    Correct as if the ABC divides 3, then moving around the digits the sum of ABC is still the same.

    There are more...
  4. Non-Sub Recs: 0
    Joined
    08 Apr '05
    Moves
    455
    22 Feb '06 13:211 edit
    Originally posted by fetofs
    I'd say 3 is one of them.
  5. Joined
    01 May '05
    Moves
    390
    22 Feb '06 13:38
    Originally posted by phgao
    Correct as if the ABC divides 3, then moving around the digits the sum of ABC is still the same.

    There are more...
    9 as well.
  6. Non-Sub Recs: 0
    Joined
    08 Apr '05
    Moves
    455
    22 Feb '06 13:45
    Originally posted by fetofs
    9 as well.
    yea keep going...
  7. Joined
    29 Apr '05
    Moves
    827
    22 Feb '06 14:07
    isnt 1 also a solution? that would make it 1,3,9 so far which is 3^0, 3^1 and 3^2.....maybe 3^3 (=27) also works then?

    432:27 = 16
    324:27 = 12
    243:27 = 9

    621:27= 23
    216:27= 8
    162:27= 6

    you meant it like that right?

    for 3^4=81 i can find only pairs of two 3-digit-numbers though. i guess thats because 81 is a too large number compared to 3-digit numbers.
  8. SubscriberBigDoggProblem
    The Advanced Mind
    bigdogghouse.com/RHP
    Joined
    26 Nov '04
    Moves
    116791
    22 Feb '06 23:552 edits
    1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 18, 21, 24, 27, 37, 54, 74, 111, 148, 185, 222, 259, 296, 333, 444, 555, 666, 777, 888, 999

    Next Problem!

    Edit: The stipulation doesn't rule out the possibility that A=B=C and thus n=2 is a solution because it goes into 222 evenly, and switching digits has no effect.

    Edit2: Even disallowing A=B=C still leaves n=2 as a solution, because it goes into 246, 462, and 624 evenly.
  9. Standard memberXanthosNZ
    Cancerous Bus Crash
    p^2.sin(phi)
    Joined
    06 Sep '04
    Moves
    25076
    23 Feb '06 00:591 edit
    Originally posted by BigDoggProblem
    1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 18, 21, 24, 27, 37, 54, 74, 111, 148, 185, 222, 259, 296, 333, 444, 555, 666, 777, 888, 999

    Next Problem!

    Edit: The stipulation doesn't rule out the possibility that A=B=C and thus n=2 is a solution because it goes into 222 evenly, and switching digits has no effect.

    Edit2: Even disallowing A=B=C still leaves n=2 as a solution, because it goes into 246, 462, and 624 evenly.
    I believe the problem asks for numbers n such that if n is a factor of ABC then it will also be a factor of BCA and CAB.

    The reason we can show that 3 satisfies this is because of the division check where if A+B+C = 3 or a number divisible by 3 then ABC is divisible by 3. The rearrangements have the same sum so this works. 9 works by the same principle. 1 works because it divides all numbers.

    2 doesn't work. A counterexample is 136. 2 is a factor of this number but not 361 or 613.

    EDIT: I'm writing some code that will find all such numbers however I need a clarification, what happens with numbers with a zero in them? In one rearrangement they will have a leading zero. Is this a problem?
  10. SubscriberBigDoggProblem
    The Advanced Mind
    bigdogghouse.com/RHP
    Joined
    26 Nov '04
    Moves
    116791
    23 Feb '06 01:401 edit
    Originally posted by XanthosNZ
    I believe the problem asks for numbers n such that if n is a factor of ABC then it will also be a factor of BCA and CAB.

    The reason we can show that 3 satisfies this is because of the division check where if A+B+C = 3 or a number divisible by 3 then ABC is divisible by 3. The rearrangements have the same sum so this works. 9 works by the same principle. rs with a zero in them? In one rearrangement they will have a leading zero. Is this a problem?
    I believe your interpretation of the stipulation is indeed what the original poster intended. Thanks for clearing that up.

    My new adjusted list is:

    1, 3, 9, 27, 37, 111, 222, 333, 444, 555, 666, 777, 888, 999

    Edit: In my code, I ruled out any three digit numbers that included leading zeros.
  11. Standard memberTheMaster37
    Kupikupopo!
    Out of my mind
    Joined
    25 Oct '02
    Moves
    20443
    23 Feb '06 07:26
    Originally posted by BigDoggProblem
    I believe your interpretation of the stipulation is indeed what the original poster intended. Thanks for clearing that up.

    My new adjusted list is:

    1, 3, 9, 27, 37, 111, 222, 333, 444, 555, 666, 777, 888, 999

    Edit: In my code, I ruled out any three digit numbers that included leading zeros.
    148 : 37 = 38

    Though 184 isn't a multiple of 37 (185 is).

    Try harder Doggie 🙂
  12. Standard memberXanthosNZ
    Cancerous Bus Crash
    p^2.sin(phi)
    Joined
    06 Sep '04
    Moves
    25076
    23 Feb '06 07:40
    Originally posted by TheMaster37
    148 : 37 = 38

    Though 184 isn't a multiple of 37 (185 is).

    Try harder Doggie 🙂
    But 184 isn't one of the arrangements that 37 must divide (those are 481 and 814 both of which are multiples of 37).
  13. SubscriberBigDoggProblem
    The Advanced Mind
    bigdogghouse.com/RHP
    Joined
    26 Nov '04
    Moves
    116791
    23 Feb '06 07:46
    Originally posted by TheMaster37
    148 : 37 = 38

    Though 184 isn't a multiple of 37 (185 is).

    Try harder Doggie 🙂
    You have a minor problem. You did not change the position of the "1".

    Please remember, when dealing with me, you are no longer TheMaster, but the student. Call me 'doggie' one more time and I will crudely disembowel you before all the forum audience.
  14. Standard memberTheMaster37
    Kupikupopo!
    Out of my mind
    Joined
    25 Oct '02
    Moves
    20443
    23 Feb '06 11:451 edit
    Originally posted by BigDoggProblem
    You have a minor problem. You did not change the position of the "1".

    Please remember, when dealing with me, you are no longer TheMaster, but the student. Call me 'doggie' one more time and I will crudely disembowel you before all the forum audience.
    My bad BigProblem, I do apologise :p

    As long as my nickname is TheMaster, I will remain TheMaster. That nickname, however, in no way represents my level of expertise; I am entitled to my share of flaws and mistakes.

    I do find it hilarious that you of all people would name anyone a student.

    EDIT; Might be an idea to prove that your claim is correct. Are those the only numbers?
  15. Joined
    29 Apr '05
    Moves
    827
    23 Feb '06 15:24
    Originally posted by BigDoggProblem
    Please remember, when dealing with me, you are no longer TheMaster, but the student. Call me 'doggie' one more time and I will crudely disembowel you before all the forum audience.
    cant wait to see that 😀

    So 1,3,9,27 fit into a system (3^x)...so do 111, 222 etc.....but what about 37? thats a funny number there, but it surely works also.
Back to Top