Alright, let's work through this problem step by step. First, let's recall the key rule for divisibility by 11:

A number is divisible by 11 if and only if the absolute difference between the sum of its digits in odd positions and the sum of its digits in even positions is a multiple of 11 (including 0).

Step 1: Calculate the total sum of digits #

First, let's add up all our digits: 2 + 3 + 4 + 5 + 6 = 20.

For a 5-digit number, we have 3 odd positions (1st, 3rd, 5th digits) and 2 even positions (2nd, 4th digits). Let's call the sum of digits in odd positions S₁ and the sum in even positions S₂. We know two things:

• S₁ + S₂ = 20 (since it's the total sum of all digits)
• |S₁ - S₂| = 11k where k is an integer (0, 1, -1, etc.)

Step 2: Narrow down valid values for S₁ and S₂ #

Let's think about possible values of k:

• The maximum possible sum for S₁ (3 digits) is 4 + 5 + 6 = 15, and the maximum for S₂ (2 digits) is 5 + 6 = 11. The largest possible absolute difference between them is 15 - 5 = 10, which is less than 11. So k can only be 0.
• This means S₁ = S₂ = 10 (since their sum is 20, splitting equally gives 10 each).

Step 3: Find valid digit groupings #

We need to split our digits into two groups:

1. A group of 3 digits that add up to 10 (for the odd positions)
2. A group of 2 digits that add up to 10 (for the even positions)

Let's check all possible pairs first (since there are fewer pairs):

• The only pair of digits from our set that sums to 10 is 4 and 6 (4+6=10; other pairs like 3+7 or 5+5 don't exist with our unique digits).
• The remaining digits are 2, 3, 5, which sum to 2+3+5=10—exactly what we need for the 3-digit group.

This is the only valid grouping of digits that satisfies the divisibility condition.

Step 4: Calculate the number of valid permutations #

Now let's count how many unique numbers we can form:

• For the 3 odd positions: we can permute the digits 2,3,5 in 3! = 6 different ways.
• For the 2 even positions: we can permute the digits 4,6 in 2! = 2 different ways.
• Multiply these together to get the total number of valid numbers: 6 × 2 = 12.

So the total number of 5-digit numbers formed by permuting 2,3,4,5,6 that are divisible by 11 is 12.

内容的提问来源于stack exchange，提问作者Mr.SrJenea