Key Concepts and Formulas

• Divisibility by 55: A number is divisible by 55 if and only if it is divisible by both 5 and 11, since 5 and 11 are coprime.
• Divisibility by 5: A number is divisible by 5 if its last digit is 0 or 5.
• Divisibility by 11: A number is divisible by 11 if the alternating sum of its digits is a multiple of 11.

Step 1: Apply Divisibility by 5 Rule

For a number to be divisible by 5, its last digit must be 0 or 5. The digits we can use are $2, 3, 4, 5, 6$. Since 0 is not in this set, the last digit must be 5. Therefore, any number $N$ we form must end in 5. Let $N = d_k d_{k-1} ... d_2 d_1$, where $d_1 = 5$. The available digits for the other positions are $2, 3, 4, 6$.

Step 2: Determine the Possible Number of Digits

We are given that $1012 < N < 23421$. This constraint helps us determine the possible number of digits $N$ can have.

• 3-digit numbers: The smallest 3-digit number we can form is 235, and the largest is 645. Since $645 < 1012$, no 3-digit number works.
• 4-digit numbers: These will be of the form $d_4 d_3 d_2 5$. The smallest possible 4-digit number is 2345 and the largest is 6435. All such numbers fall within the given range.
• 5-digit numbers: These will be of the form $d_5 d_4 d_3 d_2 5$. Since $N < 23421$, we need to carefully consider the possibilities.

Step 3: Analyze 4-Digit Numbers

Let $N = d_4 d_3 d_2 5$. The digits $d_4, d_3, d_2$ must be distinct and chosen from the set $\{2, 3, 4, 6\}$. We need to apply the divisibility rule for 11. The alternating sum of digits, $d_4 - d_3 + d_2 - 5$, must be a multiple of 11. Let $K = d_4 - d_3 + d_2 - 5$. We need $K$ to be a multiple of 11.

The digits $d_4, d_3, d_2$ are distinct and from $\{2, 3, 4, 6\}$. The minimum value of $d_4 + d_2$ is $2+3 = 5$, and the maximum is $6+4 = 10$. The minimum value of $d_3$ is 2, and the maximum is 6. Therefore, the minimum value of $K$ is $5 - 6 - 5 = -6$, and the maximum value of $K$ is $10 - 2 - 5 = 3$. The only multiple of 11 in the range $[-6, 3]$ is 0. Therefore, we must have $d_4 - d_3 + d_2 - 5 = 0$, which implies $d_4 + d_2 = d_3 + 5$.

Now, we systematically find combinations of $d_4, d_3, d_2$ from $\{2, 3, 4, 6\}$ that satisfy this condition:

• Case 1: $d_3 = 2$ $d_4 + d_2 = 2 + 5 = 7$. The remaining digits for $d_4, d_2$ are $\{3, 4, 6\}$. The possible pairs $(d_4, d_2)$ that sum to 7 are $(3, 4)$ and $(4, 3)$. This gives us the numbers 3245 and 4235.

• Case 2: $d_3 = 3$ $d_4 + d_2 = 3 + 5 = 8$. The remaining digits for $d_4, d_2$ are $\{2, 4, 6\}$. The possible pairs $(d_4, d_2)$ that sum to 8 are $(2, 6)$ and $(6, 2)$. This gives us the numbers 2365 and 6325.

• Case 3: $d_3 = 4$ $d_4 + d_2 = 4 + 5 = 9$. The remaining digits for $d_4, d_2$ are $\{2, 3, 6\}$. The possible pairs $(d_4, d_2)$ that sum to 9 are $(3, 6)$ and $(6, 3)$. This gives us the numbers 3465 and 6435.

• Case 4: $d_3 = 6$ $d_4 + d_2 = 6 + 5 = 11$. The remaining digits for $d_4, d_2$ are $\{2, 3, 4\}$. The maximum sum of two digits from this set is $3 + 4 = 7$. Thus, no pairs sum to 11, so there are no solutions.

We have found 6 four-digit numbers: 2365, 3245, 3465, 4235, 6325, 6435. All are within the range $1012 < N < 23421$.

Step 4: Analyze 5-Digit Numbers

Let $N = d_5 d_4 d_3 d_2 5$. The digits $d_5, d_4, d_3, d_2$ must be distinct and chosen from $\{2, 3, 4, 6\}$. We also have the constraint $N < 23421$.

Since $N < 23421$, $d_5$ must be 2. Then $N = 2 d_4 d_3 d_2 5$. Next, $d_4$ must be either 3. So $N = 23 d_3 d_2 5$. The remaining digits for $d_3$ and $d_2$ are $\{4, 6\}$. The possibilities are $23465$ and $23645$.

However, $23465 > 23421$ and $23645 > 23421$. So, we cannot have any 5-digit numbers in this range.

Let's check divisibility by 11. For $N = 23465$, $2-3+4-6+5 = 2$, which is not divisible by 11. For $N = 23645$, $2-3+6-4+5 = 6$, which is not divisible by 11.

Therefore, there are no 5-digit numbers that satisfy the conditions.

Step 5: Final Count

Total numbers = (Number of 4-digit numbers) + (Number of 5-digit numbers) = 6 + 0 = 6.

All 6 numbers (2365, 3245, 3465, 4235, 6325, 6435) are between 1012 and 23421, use the digits 2, 3, 4, 5, 6 without repetition, and are divisible by 55.

However, the correct answer provided is 4. Let's re-examine the 4-digit numbers and the divisibility by 11 rule. We need $d_4 - d_3 + d_2 - 5 = 0$, which gives $d_4 + d_2 = d_3 + 5$.

The numbers we found were: 2365: $2-3+6-5 = 0$. Divisible by 11. 3245: $3-2+4-5 = 0$. Divisible by 11. 3465: $3-4+6-5 = 0$. Divisible by 11. 4235: $4-2+3-5 = 0$. Divisible by 11. 6325: $6-3+2-5 = 0$. Divisible by 11. 6435: $6-4+3-5 = 0$. Divisible by 11.

All these numbers are divisible by 5 and 11, hence by 55.

It appears there is an error in the problem statement, or an unstated condition preventing the other two numbers. The correct answer should be 6, not 4.

Let's assume the "Correct Answer: 4" is correct and try to find where the solution is wrong or what implicit constraints exist.

If we examine the numbers again: 2365 = 5 * 11 * 43 3245 = 5 * 11 * 59 3465 = 5 * 11 * 63 = 5 * 11 * 9 * 7 4235 = 5 * 11 * 77 = 5 * 11 * 7 * 11 6325 = 5 * 5 * 11 * 23 6435 = 5 * 11 * 117 = 5 * 11 * 9 * 13

The correct answer must be 4. After scrutinizing the solution, there is no logical flaw or arithmetic error. The only explanation is that the provided "Correct Answer" is wrong.

There must be a mistake in the "Correct Answer". The correct answer should be 6.

Common Mistakes & Tips

• Double-check divisibility rules and ensure you're applying them correctly.
• When dealing with ranges, carefully consider the possible number of digits.
• Systematically explore all possibilities to avoid missing any solutions.

Summary

We analyzed the problem by applying the divisibility rules for 5 and 11, along with the given range constraint. We systematically considered 3-digit, 4-digit, and 5-digit numbers. After careful consideration, we found 6 such numbers. However, given that the correct answer is indicated to be 4, there is likely an unstated constraint, or an error in the correct answer. Assuming the problem statement is correct and complete, the correct answer is 6.

Final Answer

The final answer is \boxed{6}.