# Math Help - Factors Of 3-digit Numbers

1. ## Factors Of 3-digit Numbers

Hi all!

I have this question:

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.

So far I have worked out that 1 is special of course, so is 3 (factor test sum of digits will always be the same) and 9 as well as 111,222,333....999 and any number greater than 1000 as it will only divide into the 3 digit number 000.

I am told that there are two more such numbers n under 100. How do you find them? And please explan your reasoning. Thanks!

2. Originally Posted by Infinate Possibilities
Hi all!

I have this question:

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.

So far I have worked out that 1 is special of course, so is 3 (factor test sum of digits will always be the same) and 9 as well as 111,222,333....999 and any number greater than 1000 as it will only divide into the 3 digit number 000.

I am told that there are two more such numbers n under 100. How do you find them? And please explan your reasoning. Thanks!
Hello,

interesting problem.
First idea: The digits must be manifolds of your divisor.
So 2 divides all numbers build of 2, 4, 6 or 8. (248, 862, 628, ...)
So 5 divides all numbers build of 5 or 0: 550, 505, (maybe even 055).
So 4 divides all numbers build of 4 or 8. (484, 444, 848, ...)

There should be more numbers of the described properties, but I don't know how to get them. Good luck!

Greetings

EB

3. We say that n is not special as long as we can find one example of it not being special.

Thus for 5: See (455)

5 divides 455
But it does not divide 554 therefore 5 is not special.

Do you kind of see what im after?

4. Now 4 is not special either:

4(312)

4 divides 312
But not 123.

5. Interesting problem.
Maybe you can do this:
If ABC is a number then it by definition of digits (base 10) is expressable as $ABC=100x+10y+z$
Thus, $BCA=100y+10z+x$
and $CAB=100z+10x+y$.
Now $n$ divides these, thus (if and only if)
$100x+10y+z=nk$
$100y+10z+x=nk$
$100z+10x+y=nk$ For some $k$.

$111(x+y+z)=3nk$.
Since $\gcd(111,3)=1$ Euclid's Lemmas tells us that $3|(x+y+z)$. Thus, the sum of the digits must be divisible by 3.
Further information is that $x,y,z$ are integers between 1 and 9 inclusive. Thus, you can go through all possibilities.

6. Ok i'll thy that, thanks a lot. I'll post when I get some answers.

7. I'm trying to do it, but their are too many numbers. As a lot of numbers add to 9 in their digits and divide 3. Any more ideas?

8. So far I have found 1 more : 27

But tell me why 27 can be n. I can't just say I did guess + check or something like that.

Also there is still 1 more n under 100.