divisibility by 11

let the number be $\overline{abcde}$

$\overline{abc}$ is divisible by 11

which means $\overline{de}$ is divisible by 11

which means d=e

the same way a=b

now the number is $\overline{aabdd}$

and since $\overline{aab}$ is divisible by 11

then b = 0

now the number is $\overline{aa0dd}$

since $\overline{a0d}$ is divisible by 11

this means a+d=11

to get the smallest number we set a = 2 and d = 9

so the solution is 22099

A number is divisible by eleven if the digits, alternately added and subtracted, is divisible by eleven. For instance, 649 is divisible by eleven because $6-4+9=11$ is.

The required number is of the form $\overline{abcde}$ . The given conditions require that $a-b+c,\ b-c+d,\ c-d+e,\ a-b+c-d+e$ are multiples of 11. This must also be true for $(a-b+c)+(c-d+e)-(a-b+c-d+e)=c,$ from which follows $c = 0$ . Now it is easy to see that $a-b$ and $e-d$ must be multiples of eleven, so that $a=b, d=e.$ Finally, $b-c+d$ must be a multiple of eleven, which gives us $b+d=11$ .

To find the smallest solution, we choose $b = 2, d = 9$ . This gives the final answer $22099$ .

Let the number be $\overline{ABCDE}$ .

You know that $A - B + C \equiv 0 \text{ (mod 11)}$ $-B +C - D \equiv 0 \text{ (mod 11)}$ $C - D + E\equiv 0 \text{ (mod 11)}$ $A - B + C - D + E \equiv 0 \text{ (mod 11)}$

Subtracting the first, second, and second, third relations, you get that $A + D \equiv 0$ and $B + E \equiv 0$ .

Thus, the least number possible for $A$ is $2$ , which implies that $D = 9$ . Then we want the least number for $B$ , so we try $B = 2$ which implies $E = 9$ . Then based on our origin relations, we get that $C = 0$ . Thus our number is $\boxed{22099}$ .

$\textbf{Note:}$ If some of the cases had not worked, we would have been forced to try more.