Permutation and tenths place

As said previously, $abcde$ is a multiple of 11 iff $a - b + c - d + e$ is itself a multiple of $11$ .

Now with $\{a,b,c,d,e\} = \{1,3,4,5,6\}$ , we see that $a - b + c - d + e = a + b + c + d + e - 2(b + d) = 19 - 2(b+d)$ .

Also, $4 \leq b + d \leq 11$ . But $2(b+d) = 8 \mod 11$ , which implies $b + d = 4 \mod 11$ . These two constraints force $b+d = 4$ , i.e. $\{b,d\} = \{1,3\}$

We seek to minimize the number $abcde$ , and this forces the pair $b,d$ and the three numbers $a,c,e$ to be (separately) placed in ascending order. I.e. $abcde = 41536$ .

When a number $abcde$ is divisible by $11$ , it has the property that $e-d+c-b+a$ is also divisible by $11$ .

$1 \pm 3 \pm 4 \pm 5 \pm 6$ always has odd parity, so therefore it must add to $11$ or $-11$ (it cannot reach to $33$ or $-33$ ). The only way that this can happen is if we have $4+5+6-1-3=11$ . Finally, to make the number the smallest, we arrange the numbers to get $41536$ .

If abcde is the five digit number then as it is divisible by 11 (a+c+e) - (b+d) = 0 , 11 or -11. So ( a+c+e ) = 15 and (b+d) = 4 . As abcde is the smallest as well then (a = 4, c =5, e = 6) and (b = 1 , d = 3). Tenths place = 3