Jane is thinking of a 3-digit number. She counts on by hundreds, "100, 200, 300."

Next, she counts on by tens, and then ones.

The number she gets has the same number of tens and ones. What can the number be?

Write down all the possible numbers.

To solve this problem, we need to break down the information given step by step:

### 1. **Understanding the pattern:**
   - Jane starts with a 3-digit number, then she counts up by hundreds, tens, and ones.
   - After completing this counting, the resulting number has the same digit for both the tens and ones places.

### 2. **Constraints:**
   - The tens digit and the ones digit of the final number must be the same.
   
Let's express Jane's starting 3-digit number as \( ABC \), where:
- \( A \) is the hundreds digit,
- \( B \) is the tens digit,
- \( C \) is the ones digit.

Jane counts on by hundreds, tens, and ones, and the final number has the property that the tens and ones digits are the same. Therefore, we need to find all 3-digit numbers where the tens and ones digits are identical.

### 3. **Possible numbers:**
The tens digit and the ones digit must be the same. Therefore, we can express the number as \( Axx \), where \( x \) represents both the tens and ones digits, and \( A \) represents the hundreds digit.

The possible numbers, considering all values for \( A \) (hundreds digit) and \( x \) (tens and ones digits), are:

- \( A \) can be any digit from 1 to 9 (since it's a 3-digit number).
- \( x \) can be any digit from 0 to 9 (since the tens and ones can be any valid digit).

So, the list of all possible numbers is:

\[
100, 111, 122, 133, 144, 155, 166, 177, 188, 199,
200, 211, 222, 233, 244, 255, 266, 277, 288, 299,
300, 311, 322, 333, 344, 355, 366, 377, 388, 399,
400, 411, 422, 433, 444, 455, 466, 477, 488, 499,
500, 511, 522, 533, 544, 555, 566, 577, 588, 599,
600, 611, 622, 633, 644, 655, 666, 677, 688, 699,
700, 711, 722, 733, 744, 755, 766, 777, 788, 799,
800, 811, 822, 833, 844, 855, 866, 877, 888, 899,
900, 911, 922, 933, 944, 955, 966, 977, 988, 999.
\]

### Final Answer:
The possible numbers that Jane can think of are:

\[
100, 111, 122, 133, 144, 155, 166, 177, 188, 199,
200, 211, 222, 233, 244, 255, 266, 277, 288, 299,
300, 311, 322, 333, 344, 355, 366, 377, 388, 399,
400, 411, 422, 433, 444, 455, 466, 477, 488, 499,
500, 511, 522, 533, 544, 555, 566, 577, 588, 599,
600, 611, 622, 633, 644, 655, 666, 677, 688, 699,
700, 711, 722, 733, 744, 755, 766, 777, 788, 799,
800, 811, 822, 833, 844, 855, 866, 877, 888, 899,
900, 911, 922, 933, 944, 955, 966, 977, 988, 999.
\]

### Do you have any questions or want further clarification?

### Five additional questions:
1. How many total numbers have identical tens and ones digits?
2. What if the problem asked for different digits in the tens and ones places instead?
3. Can you find a pattern in the numbers Jane could be thinking of?
4. What would happen if the number had a different number of tens and ones?
5. How would the problem change if Jane started with a two-digit number?

### Tip:
When working with constraints in digit-based problems, writing down the form of the number (hundreds, tens, ones) can help visualize and solve the problem efficiently.

Jane is thinking of a 3-digit number. She counts on by hundreds, '100, 200, 300.' Next, she counts on by tens, and then ones. The number she gets has the same number of tens and ones. What can the number be? Write down all the possible numbers.

Explore how to find all 3-digit numbers where the tens and ones digits are identical. This problem helps students practice number patterns and place value concepts. Suitable for Grades 4-6.

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