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 ...

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 ...

Originally posted by fetofs
I'd say 3 is one of them.

Originally posted by fetofs
I'd say 3 is one of them.

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...

Originally posted by fetofs
9 as well.

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.

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.

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.

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?

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.

Originally posted by TheMaster37
148 : 37 = 38

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

Try harder Doggie 🙂

Originally posted by TheMaster37
148 : 37 = 38

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

Try harder Doggie 🙂

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.

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.