How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3?

1. Translate the problem requirements: We need to find all integers from 0 to 50 (including both endpoints) that leave remainder 1 when divided by 3. In other words, numbers of the form \(3\mathrm{k} + 1\) where k is a non-negative integer.
2. Identify the pattern of numbers with remainder 1: Find the first few numbers that have remainder 1 when divided by 3 to establish the arithmetic sequence pattern.
3. Determine the range boundaries: Find the smallest and largest numbers in our range (0 to 50) that satisfy the remainder condition.
4. Count using arithmetic sequence properties: Apply the pattern to count how many such numbers exist within our specified range.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this question is really asking us to find. We want integers (whole numbers) from 0 to 50 that have a remainder of 1 when divided by 3.

Think about what "remainder of 1 when divided by 3" means in everyday terms: if you divide the number by 3, you get some whole number of groups of 3, plus 1 left over.

For example: \(10 ÷ 3 = 3\) remainder \(1\) (because \(3 × 3 = 9\), and \(10 - 9 = 1\))

Mathematically, we're looking for numbers that can be written as \(3\mathrm{k} + 1\), where k is some non-negative whole number (0, 1, 2, 3, ...).

Process Skill: TRANSLATE - Converting the remainder condition into a clear mathematical pattern

2. Identify the pattern of numbers with remainder 1

Let's find the first several numbers that have remainder 1 when divided by 3 by testing some examples:

• \(0 ÷ 3 = 0\) remainder \(0\) ✗
• \(1 ÷ 3 = 0\) remainder \(1\) ✓
• \(2 ÷ 3 = 0\) remainder \(2\) ✗
• \(3 ÷ 3 = 1\) remainder \(0\) ✗
• \(4 ÷ 3 = 1\) remainder \(1\) ✓
• \(5 ÷ 3 = 1\) remainder \(2\) ✗
• \(6 ÷ 3 = 2\) remainder \(0\) ✗
• \(7 ÷ 3 = 2\) remainder \(1\) ✓
• \(8 ÷ 3 = 2\) remainder \(2\) ✗
• \(9 ÷ 3 = 3\) remainder \(0\) ✗
• \(10 ÷ 3 = 3\) remainder \(1\) ✓

The pattern is clear: 1, 4, 7, 10, 13, ...

Notice that each number is exactly 3 more than the previous one. This makes sense because adding 3 doesn't change the remainder when dividing by 3.

We have an arithmetic sequence with first term = 1 and common difference = 3.

3. Determine the range boundaries

Now we need to find which numbers in our sequence fall between 0 and 50, inclusive.

The smallest number: Looking at our pattern, the first number ≥ 0 with remainder 1 is clearly 1.

The largest number: We need the biggest number ≤ 50 that fits our pattern \(3\mathrm{k} + 1\).

Let's work backwards from 50:

• \(50 ÷ 3 = 16\) remainder \(2\) ✗
• \(49 ÷ 3 = 16\) remainder \(1\) ✓

So our sequence within the range 0 to 50 goes from 1 to 49.

Process Skill: APPLY CONSTRAINTS - Ensuring we stay within the given range of 0 to 50

4. Count using arithmetic sequence properties

We have an arithmetic sequence: 1, 4, 7, 10, 13, ..., 46, 49

To count how many terms are in this sequence, we can use the pattern we established.

Since each term is of the form \(3\mathrm{k} + 1\):

• First term: \(1 = 3(0) + 1\), so \(\mathrm{k} = 0\)
• Last term: \(49 = 3(16) + 1\), so \(\mathrm{k} = 16\)

The values of k go from 0 to 16, inclusive.

Counting: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16

That's 17 different values of k, which means 17 numbers in our sequence.

We can verify this with the arithmetic sequence formula: Number of terms = \((\text{last term} - \text{first term}) ÷ (\text{common difference}) + 1 = (49 - 1) ÷ 3 + 1 = 48 ÷ 3 + 1 = 16 + 1 = 17\)

4. Final Answer

There are 17 integers from 0 to 50, inclusive, that have a remainder of 1 when divided by 3.

The answer is C. 17

Verification: The sequence is 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49 - counting these gives us 17 numbers, confirming our answer.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding the range boundaries
Students often misinterpret "from 0 to 50, inclusive" and forget to check whether 0 and 50 themselves satisfy the remainder condition. They might assume these boundary numbers are automatically included or excluded without testing them. In this case, 0 gives remainder 0 (not 1) when divided by 3, so it shouldn't be counted, while 50 gives remainder 2 (not 1), so it also shouldn't be counted.

Faltering Point 2: Confusion about remainder notation
Some students struggle to translate "remainder of 1 when divided by 3" into the mathematical form \(3\mathrm{k} + 1\). They might try to work backwards from division results or get confused about what the remainder actually represents, leading to incorrect identification of the pattern.

Errors while executing the approach

Faltering Point 1: Incorrect identification of sequence boundaries
When finding the largest number ≤ 50 that fits the pattern \(3\mathrm{k} + 1\), students might incorrectly conclude that 50 itself works (since \(50 ÷ 3 = 16\) remainder \(2\), not 1) or make arithmetic errors when checking 49. This leads to using wrong endpoints in their counting.

Faltering Point 2: Miscounting the arithmetic sequence
When counting terms from \(\mathrm{k} = 0\) to \(\mathrm{k} = 16\), students frequently forget to include both endpoints or make off-by-one errors. They might count 16 terms instead of 17, forgetting that when you count from 0 to 16 inclusive, you have 17 different values, not 16.

Faltering Point 3: Arithmetic errors in verification
When using the arithmetic sequence formula \((\text{last} - \text{first}) ÷ (\text{common difference}) + 1\), students might forget the "+1" at the end or make calculation errors with \((49 - 1) ÷ 3\), leading to an answer of 16 instead of 17.

Errors while selecting the answer

Faltering Point 1: Selecting based on partial counting
Some students might manually list out the sequence but stop counting prematurely or miscount the numbers they've written down. For instance, they might correctly identify the sequence 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49 but then miscount and select 16 (choice B) instead of 17 (choice C).