How many 5-digit

We know the rule of divisibility by 11. A number abcde is divisible by 11, if the difference of the sum of digits at odd places and even places is a multiple of 11.

That is, if (a + c + e) – (b + d) = 11z, where z = integers

We have the digits 2, 3, 4, 5, 6. If using these we can estimate a possible number divisible by 11, then our work would become easier.

The 5-digit number abcde would obviously follow the criteria to be divisible by 11.

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

Adding digits 2, 3, 4, 5, 6, we get

2 + 3 + 4 + 5 + 6 = 20, which means (a + c + e) and (b + d) must be either equal or the difference must come out to be equal to 11 or a multiple of 11. Noticing the numbers, it is not possible to obtain a result equal to 11 or multiple of 11.

So, we can say that (a + c + e) = (b + d) = 10

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

⇒ {a, c, e} = {2, 3, 5}

Using this result, we can take a number of combinations which will satisfy the question.

For instance, if we take a = 2, b = 4, c = 3, d = 6, e = 5,

We get the number, 24365.

Now, if we take a = 3, b = 4, c = 2, d = 6, e = 5,

Then, we get 34265.

Like that, using permutation and combination, there are 2 orders possible for 4 and 6.

∵ There are two numbers in the set {b, d}, so arrangement can be done in 2! ways.

⇒ Arrangement can be done in (2 × 1) ways.

⇒ Arrangement can be done in 2 ways. …(i)

Similarly, there are 6 ways possible for 2, 3 and 5.

∵ There are 3 numbers in the set {a, c, e}, so arrangement can be done in 3! ways.

⇒ Arrangement can be done in (3 × 2 × 1) ways.

⇒ Arrangement can be done in 6 ways. …(ii)

Using result (i) and (ii), we get

2 × 6 = 12 ways to form a number using digits 2, 3, 4, 5, 6, which is divisible by 11.

Hence, the answer is 12.