What is the smallest 5-digit number divisible by 11 and containing each of the digits 2, 3, 4, 5, 6?

We have the digits 2, 3, 4, 5, 6. Using these digits, we need to form the smallest 5-digit number divisible by 11.

If we had had the (smallest) number as abcde, this would have satisfied the criteria:

(a + c + e) = (b + d) (mod 11)

Adding these digits, 2 + 3 + 4 + 5 + 6 = 20, so this means that (a + c + e) and (b + d) must be either equal or the difference must come out to be 11 or a multiple of 11. Since, the numbers are small, we can say that the difference won’t come out to be greater multiples of 11.

Observe that from the given digits, (a + c + e) = (b + d) = 10

With the given digits, b + d = 10 is possible.

So, if (b + d) = 10, then (a + c + e) = 10

Such that, (a + c + e) = (b + d) (mod 11)

⇒ 10 = 10 (mod 11)

From the given digits, {b, d} = {4, 6}

⇒ {a, c, e} = {2, 3, 5}

⇒ The number is abcde, that is, 24365, which is the smallest possible 5-digit number containing each of the digit 2, 3, 4, 5, 6.

Thus, the number is 24365.