BdMO Online Forum

Problem 1:
Find a three digit number so that when its digits are arranged in reverse order and added with the original number, the result is a three digit number with all of its digits being equal. In case of two digit numbers, here is an example: $23+32=55 $

Totaly 6 ta ans hoy.

jkisor wrote:Totaly 6 ta ans hoy.

Which are $111,222,333,444,123,234,345$

oops !! if I am not miscounting, there are $7$

nafistiham wrote:

jkisor wrote:Totaly 6 ta ans hoy.

Which are $111,222,333,444,123,234,345$

oops !! if I am not miscounting, there are $7$

Wait a minute bro. The last three can be reversed an they will also fulfill the condition,i.e. $321,234,543$ are also answers. And $135,531,147,741,246,642$ also fulfill the condition.
Check my idea please.

Actually there are twenty of them. 111,123, 135, 147, 210, 222, 234, 246, 321, 333, 345, 420, 432, 444, 531, 543, 630, 642, 741, 840. Now prove it without calculating

Let the number be $100x+10y+z$.
Here $x+z=2y$. $2y$ have to be an one digit even number which is either 8,6,4 or 2.
For these we will get total $(8+6+4+2)=20$ solutions.

111,
210,
123,
222,
420,
321,
333,
531,
135,
432,
234,
630,
444,
840,
642,
246,
543,
345,
741,
147,
yes only these are the correct numbers...............

123+321=444