A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3....

1. Translate the problem requirements: We have a six-digit number in the form \(1k2k24\) where \(k\) appears twice as the 2nd and 4th digits. We need to find how many single-digit values of \(k\) (0-9) make this number divisible by 3.
2. Set up the digit sum formula: Since divisibility by 3 depends on the sum of digits being divisible by 3, calculate the sum: \(1 + k + 2 + k + 2 + 4 = 9 + 2k\)
3. Apply the divisibility condition: For the number to be divisible by 3, we need \((9 + 2k)\) to be divisible by 3. Since 9 is already divisible by 3, we need \(2k\) to be divisible by 3.
4. Find valid values of k: Since \(2k\) must be divisible by 3, and \(\mathrm{gcd}(2,3) = 1\), we need \(k\) itself to be divisible by 3. Test \(k = 0, 3, 6, 9\) to verify they work.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're given. We have a six-digit number that looks like \(1k2k24\), where \(k\) is some digit that appears in both the 2nd and 4th positions.

For example, if \(k = 5\), our number would be \(152524\). If \(k = 0\), our number would be \(102024\).

The question asks: how many different values can \(k\) have so that this six-digit number is divisible by 3? Since \(k\) is a digit, it can be any whole number from 0 to 9.

Process Skill: TRANSLATE - Converting the abstract form \(1k2k24\) into concrete examples helps us visualize what we're working with.

2. Set up the digit sum formula

Now we use the key rule given in the problem: a number is divisible by 3 if and only if the sum of its digits is divisible by 3.

Let's add up all the digits in our number \(1k2k24\):
• First digit: 1
• Second digit: \(k\)
• Third digit: 2
• Fourth digit: \(k\)
• Fifth digit: 2
• Sixth digit: 4

Total sum = \(1 + k + 2 + k + 2 + 4\)

Let's simplify this step by step:
\(1 + 2 + 2 + 4 = 9\)
\(k + k = 2k\)

So our total sum = \(9 + 2k\)

3. Apply the divisibility condition

For our six-digit number to be divisible by 3, the sum \((9 + 2k)\) must be divisible by 3.

Here's the key insight: since 9 is already divisible by 3 (because \(9 ÷ 3 = 3\) exactly), we need \(2k\) to also be divisible by 3 for the entire sum to be divisible by 3.

So our question becomes: when is \(2k\) divisible by 3?

Process Skill: INFER - Recognizing that since 9 is divisible by 3, we only need to focus on when \(2k\) is divisible by 3.

4. Find valid values of k

For \(2k\) to be divisible by 3, we need to think about what values of \(k\) work.

Since 2 and 3 share no common factors (2 and 3 are coprime), the only way for \(2k\) to be divisible by 3 is if \(k\) itself is divisible by 3.

Let's check each possible value of \(k\) from 0 to 9:
• \(k = 0\): Is 0 divisible by 3? Yes (\(0 ÷ 3 = 0\)) ✓
• \(k = 1\): Is 1 divisible by 3? No ✗
• \(k = 2\): Is 2 divisible by 3? No ✗
• \(k = 3\): Is 3 divisible by 3? Yes (\(3 ÷ 3 = 1\)) ✓
• \(k = 4\): Is 4 divisible by 3? No ✗
• \(k = 5\): Is 5 divisible by 3? No ✗
• \(k = 6\): Is 6 divisible by 3? Yes (\(6 ÷ 3 = 2\)) ✓
• \(k = 7\): Is 7 divisible by 3? No ✗
• \(k = 8\): Is 8 divisible by 3? No ✗
• \(k = 9\): Is 9 divisible by 3? Yes (\(9 ÷ 3 = 3\)) ✓

Let's verify with one example: if \(k = 6\), our number is \(162624\), and the digit sum is \(9 + 2(6) = 9 + 12 = 21\). Since \(21 ÷ 3 = 7\), this works!

Process Skill: CONSIDER ALL CASES - Systematically checking each possible value of \(k\) ensures we don't miss any valid options.

5. Final Answer

We found that \(k\) can have 4 different values: 0, 3, 6, and 9.

Therefore, the answer is C. 4.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misinterpreting the notation \(1k2k24\)
Students may struggle to understand that \(k\) represents the same digit appearing in both the 2nd and 4th positions. They might think \(k\) can be different values in each position, leading them to set up the problem incorrectly as finding values for two different variables instead of one repeated digit.

Faltering Point 2: Forgetting the constraint that \(k\) must be a single digit
Students may overlook that \(k\) is explicitly described as "a digit" and therefore must be between 0 and 9 inclusive. This constraint is crucial for determining the final count of valid values.

Faltering Point 3: Misapplying the divisibility rule
Students might confuse the divisibility by 3 rule with other divisibility rules (like divisibility by 9) or incorrectly remember it as requiring the sum to equal 3 rather than be divisible by 3.

Errors while executing the approach

Faltering Point 1: Arithmetic errors when calculating the digit sum
Students may make simple addition mistakes when computing \(1 + k + 2 + k + 2 + 4 = 9 + 2k\), potentially getting the wrong constant term or coefficient for \(k\).

Faltering Point 2: Incorrectly determining when \(2k\) is divisible by 3
Students might not recognize that since \(\mathrm{gcd}(2,3) = 1\), the expression \(2k\) is divisible by 3 if and only if \(k\) is divisible by 3. They may try to solve \(2k ≡ 0 \pmod{3}\) incorrectly, perhaps thinking \(k\) must equal specific values like 1.5.

Faltering Point 3: Missing valid values of \(k\) during systematic checking
When checking \(k = 0\) through \(k = 9\), students might skip \(k = 0\) thinking it's not a valid digit for the middle positions, or make individual divisibility errors (like thinking 6 is not divisible by 3).

Errors while selecting the answer

Faltering Point 1: Counting the actual values instead of the count
Students might list the valid values (0, 3, 6, 9) but then select an answer choice that matches one of these values rather than counting that there are 4 valid values total.