How many integers between 324{,}700 and 458{,}600 have tens digit 2 and units digit 1?

1. Translate the problem requirements: We need to count how many integers in the range from 324,700 to 458,600 end in "21" (tens digit 2, units digit 1). This means we're looking for numbers like 324,721, 324,821, etc.
2. Identify the pattern structure: Every integer ending in 21 can be written as \(100\mathrm{k} + 21\) for some integer k. We need to find which values of k give us numbers in our target range.
3. Convert range boundaries to the pattern format: Transform the given boundaries into the form \(100\mathrm{k} + 21\) to find the minimum and maximum values of k that keep us within the range.
4. Count the valid integers: Calculate how many integers k exist between our minimum and maximum values, which directly gives us our answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding exactly what we're looking for. We want to count integers that have:

• Tens digit = 2
• Units digit = 1

This means we're looking for numbers that end in "21". For example:

• 324,721 ✓ (ends in 21)
• 324,831 ✗ (ends in 31, not 21)
• 458,521 ✓ (ends in 21)

So we're counting all numbers between 324,700 and 458,600 that end in exactly "21".

Process Skill: TRANSLATE - Converting the digit requirements into a concrete pattern we can work with

2. Identify the pattern structure

Now let's think about what numbers ending in 21 look like. Any number ending in 21 can be written as some whole number times 100, plus 21.

For example:

• 324,721 = \(3,247 \times 100 + 21\)
• 324,821 = \(3,248 \times 100 + 21\)
• 324,921 = \(3,249 \times 100 + 21\)

In general, every number ending in 21 has the form: \(100\mathrm{k} + 21\), where k is some integer.

So instead of counting all the numbers ending in 21, we can count how many different values of k work for our range.

3. Convert range boundaries to the pattern format

We need our numbers to be between 324,700 and 458,600. Let's figure out what values of k give us numbers in this range.

For the lower bound:

We need: \(100\mathrm{k} + 21 \geq 324,700\)

Subtracting 21: \(100\mathrm{k} \geq 324,679\)

Dividing by 100: \(\mathrm{k} \geq 3,246.79\)

Since k must be an integer, the smallest value is k = 3,247.

Let's check: \(100(3,247) + 21 = 324,721\) ✓ (this is in our range)

For the upper bound:

We need: \(100\mathrm{k} + 21 \leq 458,600\)

Subtracting 21: \(100\mathrm{k} \leq 458,579\)

Dividing by 100: \(\mathrm{k} \leq 4,585.79\)

Since k must be an integer, the largest value is k = 4,585.

Let's check: \(100(4,585) + 21 = 458,521\) ✓ (this is in our range)

4. Count the valid integers

Now we just need to count how many integers there are from k = 3,247 to k = 4,585, inclusive.

The number of integers from 3,247 to 4,585 (inclusive) is:

\(4,585 - 3,247 + 1 = 1,339\)

(We add 1 because we're counting both endpoints)

Let's verify with a simple example: counting from 5 to 8 gives us 5, 6, 7, 8 = four numbers, and \(8 - 5 + 1 = 4\) ✓

4. Final Answer

There are 1,339 integers between 324,700 and 458,600 that have tens digit 2 and units digit 1.

This matches answer choice A: 1,339.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the digit position requirements

Students often confuse which digit is the tens digit and which is the units digit. In a number like 324,721, the units digit is 1 (rightmost) and the tens digit is 2 (second from right). Some students might incorrectly think we're looking for numbers ending in "12" instead of "21".

2. Misunderstanding the range boundaries

Students may incorrectly interpret "between 324,700 and 458,600" as excluding one or both endpoints, or they might think both boundary numbers must also end in "21". The solution correctly identifies that we need numbers greater than or equal to 324,700 and less than or equal to 458,600, regardless of whether the boundaries themselves end in "21".

3. Attempting to list individual numbers instead of recognizing the pattern

Some students might try to manually count numbers ending in "21" by going through ranges like 324,721, 324,821, 324,921, etc., rather than recognizing that these numbers follow the pattern \(100\mathrm{k} + 21\). This approach becomes impractical and error-prone for such a large range.

Errors while executing the approach

1. Arithmetic errors when solving inequalities

When solving \(100\mathrm{k} + 21 \geq 324,700\), students might make calculation errors: incorrectly subtracting \(324,700 - 21 = 324,679\) or incorrectly dividing \(324,679 ÷ 100 = 3,246.79\). Similar errors can occur with the upper bound calculation.

2. Incorrect rounding of the boundary k values

After getting \(\mathrm{k} \geq 3,246.79\), students might incorrectly round down to k = 3,246 instead of rounding up to k = 3,247. Similarly, for \(\mathrm{k} \leq 4,585.79\), they might round up to k = 4,586 instead of rounding down to k = 4,585. This directly affects the final count.

3. Forgetting to add 1 when counting consecutive integers

When counting integers from 3,247 to 4,585 inclusive, students often calculate \(4,585 - 3,247 = 1,338\) and forget to add 1. The correct count of consecutive integers from a to b (inclusive) is always \(\mathrm{b} - \mathrm{a} + 1\), not just \(\mathrm{b} - \mathrm{a}\).

Errors while selecting the answer

No likely faltering points - the calculation directly gives 1,339, which matches answer choice A exactly.