Reversing digits, then adding and finding divisible integers of result.

[VinnyCee]

1. The problem statement, all variables and given/known data

If you find the sum of any two digit number and the number formed by reversing its digits, the resulting number is always divisible by which three positive integers?


2. Relevant equations

None.


3. The attempt at a solution

\left(10\ x\ +\ y\right)\ +\ \left(10\ y\ +\ x\right)\ =\ 11\ x\ +\ 11\ y

Where x is the tens place and y is the ones place. This is divisible by 11, and 1 (as are all integers) and x + y.

\therefore\ x\ +\ y\ ,11\ ,1\ |\ 11\ x\ +\ 11\ y

[DaveC426913]

Was there a question in there? Did you determine the 3 numbers?

[Dick]

1. The problem statement, all variables and given/known data

If you find the sum of any two digit number and the number formed by reversing its digits, the resulting number is always divisible by which three positive integers?


2. Relevant equations

None.


3. The attempt at a solution

\left(10\ x\ +\ y\right)\ +\ \left(10\ y\ +\ x\right)\ =\ 11\ x\ +\ 11\ y

Where x is the tens place and y is the ones place. This is divisible by 11, and 1 (as are all integers) and x + y.

\therefore\ x\ +\ y\ ,11\ ,1\ |\ 11\ x\ +\ 11\ y

The number is pretty likely to be divisible by itself as well as 1 and 11, isn't it? Kind of a trick question.

[DaveC426913]

The number is pretty likely to be divisible by itself as well as 1 and 11, isn't it? Kind of a trick question.

Well, if we include one and itself, we are up to 4 numbers (11 and x+y).

So, either they're being generous, asking only for 3 of 4, or they don't count 1 and itself, which would mean we're still missing one.

[Dick]

Well, if we include one and itself, we are up to 4 numbers (11 and x+y).

So, either they're being generous, asking only for 3 of 4, or they don't count 1 and itself, which would mean we're still missing one.

Well, 11+11=22. The only divisors are 1,2,11 and 22. I count four. So that's the max for the minimum number of divisors. Don't think any can be missing. 1 divides everything. So that's probably the one I would throw out. But every number divides itself as well. Tough call.

[DaveC426913]

Well, 11+11=22. The only divisors are 1,2,11 and 22. I count four. So that's the min. Don't think any can be missing. 1 divides everything. So that's probably the one I would throw out. But every number divides itself as well. Tough call.

22??

22 isn't one of them.


34,43 = 77, divisible by 1,7,11,77
27,72 = 99, divisible by 1,9,11,99
etc.

[Dick]

22??

22 isn't one of them.


34,43 = 77, divisible by 1,7,11,77
27,72 = 99, divisible by 1,9,11,99
etc.

11 isn't a two digit number? This is getting silly.

[DaveC426913]

11 isn't a two digit number? This is getting silly.

Oh sorry. I thought you were claiming 22 was a common divisor. As in: any 2 digit number and the sum of its reversed version would be divisible by 22. My error.:blushing:

[I like Serena]

Well, if we include one and itself, we are up to 4 numbers (11 and x+y).

So, either they're being generous, asking only for 3 of 4, or they don't count 1 and itself, which would mean we're still missing one.

Yes, there are 4 positive numbers, 2 fixed, and 2 depending on the choice of digits.
However, these numbers do not have to be distinct.

Note that with the choice "10", the resulting number is 11, and we only get 1 and 11 as divisors.
In all other cases, there are 3 distinct numbers guaranteed: itself, 11, and 1.

So either we get one number for free, or the problem is broken.

[Dick]

Yes, there are 4 positive numbers, 2 fixed, and 2 depending on the choice of digits.
However, these numbers do not have to be distinct.

Note that with the choice "10", the resulting number is 11, and we only get 1 and 11 as divisors.
In all other cases, there are 3 distinct numbers guaranteed: itself, 11, and 1.

So either we get one number for free, or the problem is broken.

Ha! That's good. Attention to detail is even more important to exposing lame problem statements than I thought.