- 22 Feb '06 14:07isnt 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. - 22 Feb '06 23:55 / 2 edits1, 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. - 23 Feb '06 00:59 / 1 edit

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.*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.

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? - 23 Feb '06 01:40 / 1 edit

I believe your interpretation of the stipulation is indeed what the original poster intended. Thanks for clearing that up.*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?

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. - 23 Feb '06 07:26

148 : 37 = 38*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.

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

Try harder Doggie - 23 Feb '06 07:46

You have a minor problem. You did not change the position of the "1".*Originally posted by TheMaster37***148 : 37 = 38**

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

Try harder Doggie

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. - 23 Feb '06 11:45 / 1 edit

My bad BigProblem, I do apologise :p*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.

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? - 23 Feb '06 15:24

cant wait to see that*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.**

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.