How many positive integers less than 500 have a remainder of 1 when divided by 7 and a remainder of...

1. Translate the problem requirements: We need positive integers less than 500 that satisfy two remainder conditions simultaneously: when divided by 7, remainder is 1 (numbers of form \(7k+1\)), and when divided by 3, remainder is 2 (numbers of form \(3j+2\)).
2. Find the pattern for numbers satisfying both conditions: Use the Chinese Remainder Theorem concept to find what form these numbers take by testing small examples and identifying the repeating pattern.
3. Determine the range and interval: Once we know the general form of numbers satisfying both conditions, identify the first such number and the interval between consecutive numbers.
4. Count occurrences within the limit: Calculate how many numbers of this form exist that are positive and less than 500.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're looking for in everyday language. We need positive integers (whole numbers greater than 0) that are less than 500, and these numbers must satisfy two specific conditions at the same time:

Condition 1: When we divide the number by 7, we get a remainder of 1
This means our number looks like: \(7 \times (\text{some whole number}) + 1\)

Condition 2: When we divide the number by 3, we get a remainder of 2
This means our number looks like: \(3 \times (\text{some whole number}) + 2\)

Let's see some examples of each type:

• Numbers with remainder 1 when divided by 7: 1, 8, 15, 22, 29, 36, 43, 50...
• Numbers with remainder 2 when divided by 3: 2, 5, 8, 11, 14, 17, 20, 23...

We need to find numbers that appear in BOTH lists. Looking at our examples, we can see that 8 appears in both lists!

Process Skill: TRANSLATE - Converting the remainder conditions into concrete number patterns

2. Find the pattern for numbers satisfying both conditions

Now let's systematically find more numbers that satisfy both conditions. We'll check each number from our first list to see if it also satisfies the second condition:

• \(1 \div 3 = 0\) remainder \(1\) ❌ (we need remainder 2)
• \(8 \div 3 = 2\) remainder \(2\) ✓ (this works!)
• \(15 \div 3 = 5\) remainder \(0\) ❌ (we need remainder 2)
• \(22 \div 3 = 7\) remainder \(1\) ❌ (we need remainder 2)
• \(29 \div 3 = 9\) remainder \(2\) ✓ (this works!)
• \(36 \div 3 = 12\) remainder \(0\) ❌ (we need remainder 2)
• \(43 \div 3 = 14\) remainder \(1\) ❌ (we need remainder 2)
• \(50 \div 3 = 16\) remainder \(2\) ✓ (this works!)

So far we have: 8, 29, 50...

Let's look at the differences: \(29 - 8 = 21\), and \(50 - 29 = 21\)

The pattern shows that once we find one number that works, the next one is always 21 more!

This makes sense because we need the pattern to repeat for both divisors (3 and 7), so we need a common cycle. The numbers repeat every \(3 \times 7 = 21\) integers.

Process Skill: INFER - Recognizing that the difference pattern reveals the repeating cycle

3. Determine the range and interval

From our pattern discovery, we now know:

• First number that works: 8
• Interval between consecutive numbers: 21
• General form: 8, \(8+21\), \(8+2\times 21\), \(8+3\times 21\), ... which is \(8 + 21k\) where \(k = 0, 1, 2, 3, ...\)

So our numbers are: 8, 29, 50, 71, 92, 113, 134, 155, 176, 197, 218, 239, 260, 281, 302, 323, 344, 365, 386, 407, 428, 449, 470, 491...

4. Count occurrences within the limit

We need to find how many numbers of the form \(8 + 21k\) are positive and less than 500.

Since we start with 8 (when \(k = 0\)), all our numbers are automatically positive.

For the upper limit, we need: \(8 + 21k < 500\)

Solving: \(21k < 500 - 8\)

\(21k < 492\)

\(k < 492 \div 21\)

\(k < 23.43...\)

Since k must be a whole number, the largest value of k we can use is 23.

This means k can be: 0, 1, 2, 3, 4, ..., 23

Counting from 0 to 23, that's 24 different values of k.

Let's verify our largest number: \(8 + 21 \times 23 = 8 + 483 = 491 < 500\) ✓

And check the next one would be too big: \(8 + 21 \times 24 = 8 + 504 = 512 > 500\) ❌

Process Skill: APPLY CONSTRAINTS - Ensuring we count only numbers within the specified range

Final Answer

There are 24 positive integers less than 500 that have a remainder of 1 when divided by 7 and a remainder of 2 when divided by 3.

The answer is C. 24

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding simultaneous conditions
Students often try to solve each remainder condition separately and then somehow combine the results, rather than recognizing that they need to find numbers that satisfy BOTH conditions simultaneously. They might count all numbers with remainder 1 when divided by 7, then count all numbers with remainder 2 when divided by 3, and incorrectly try to add or manipulate these counts.

Faltering Point 2: Missing the Chinese Remainder Theorem pattern
Many students don't recognize that this is a Chinese Remainder Theorem problem where the solution will repeat every \(\text{LCM}(7,3) = 21\) integers. Instead, they might try to manually check every single number from 1 to 500, leading to a very time-consuming and error-prone approach.

Faltering Point 3: Confusing remainder notation
Students sometimes misinterpret "remainder of 1 when divided by 7" as meaning the number equals 1 or 7, rather than understanding it means numbers of the form \(7k + 1\) where k is a non-negative integer.

Errors while executing the approach

Faltering Point 1: Arithmetic errors in finding the pattern
When manually checking which numbers from the first sequence (8, 15, 22, 29...) also satisfy the second condition, students often make division errors. For example, they might incorrectly calculate \(22 \div 3\) or \(29 \div 3\), leading them to identify the wrong starting number or miss the correct pattern interval.

Faltering Point 2: Incorrect interval calculation
Even if students recognize the need for a repeating pattern, they might incorrectly calculate the interval. Some might use \(7 + 3 = 10\) instead of \(7 \times 3 = 21\), or make other errors in determining how often the pattern repeats.

Faltering Point 3: Boundary condition errors
When solving the inequality \(8 + 21k < 500\), students commonly make errors such as: forgetting to subtract 8 first (solving \(21k < 500\) instead of \(21k < 492\)), making division errors when calculating \(492 \div 21\), or incorrectly handling the inequality direction.

Errors while selecting the answer

Faltering Point 1: Off-by-one counting errors
After correctly finding that k ranges from 0 to 23, students often miscount the total number of values. They might count 23 values instead of 24, forgetting to include \(k = 0\), or they might count 25 values by incorrectly including both endpoints in a different way.

Faltering Point 2: Using k value instead of count
Some students correctly determine that the largest value of k is 23, but then select 23 as their final answer instead of recognizing that they need to count how many different values of k are possible (which is 24, since k goes from 0 to 23 inclusive).