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The numbers could be (149, 151) or (249, 251), differing by 2, and rounding to 100 and 200,respectively.

Let's denote the two numbers as [tex]\( A \) and \( B \).[/tex]

1. The difference between the two numbers is 2:

 [tex]\[ |A - B| = 2 \][/tex]

2. When each number is rounded to the nearest hundred, the difference between them is 100.

Step-by-Step Solution:

**Step 1: Understanding the rounding impact**

When a number is rounded to the nearest hundred, it falls within the range of the midpoint between two hundreds. For example:

- Any number from 50 to 149 rounds to 100.

- Any number from 150 to 249 rounds to 200.

- And so on.

**Step 2: Establishing conditions based on rounding**

We know the rounded difference is 100, so the rounded values of ( A ) and ( B ) must differ by 100. Let's denote the rounded values as  [tex]\( A_r \)[/tex] and [tex]\( B_r \):[/tex]

[tex]\[|A_r - B_r| = 100\][/tex]

**Step 3: Analyzing the ranges for [tex]\( A \) and \( B \)[/tex] **

Given the possible ranges for rounding:

• If A rounds to Ar and B rounds to Br, the actual values A and B must be within +-50 of their

respective rounded values.

Potential Cases:

[tex]1. \( A_r = 100 \) and \( B_r = 200 \):\\ - For \( A \): \( 50 \leq A < 150 \)\\ - For \( B \): \( 150 \leq B < 250 \)[/tex]

[tex]2. \( A_r = 200 \) and \( B_r = 300 \):\\ - For \( A \): \( 150 \leq A < 250 \)\\ - For \( B \): \( 250 \leq B < 350 \)[/tex]

Step 4: Solving for exact numbers**

Since the difference between ( A ) and ( B ) is 2, we test within these ranges:

Case 1: ( A ) in [50, 150) and ( B ) in [150, 250)

  - Assume ( A ) is at the upper end of its range and ( B ) just 2 units more.

[tex]\[ B = A + 2 \][/tex]

  - Since ( B ) must round to 200 and \( A \) to 100:

[tex]\[ 50 \leq A < 150 \quad \text{and} \quad 150 \leq B < 250 \][/tex]

Check a specific example:

  - Let ( A = 149 ) (rounds to 100)

  - Then ( B = 149 + 2 = 151 ) (rounds to 200)

Thus, ( A = 149 ) and ( B = 151 ) is one solution.

Case 2: ( A ) in [150, 250) and ( B ) in [250, 350)

  - Assume ( A) is at the upper end of its range and ( B \) just 2 units more.

B = A + 2

  - Since ( B ) must round to 300 and ( A ) to 200:

    [tex]\[ 150 \leq A < 250 \quad \text{and} \quad 250 \leq B < 350 \][/tex]

Check a specific example:

• Let A 249 (rounds to 200)

• Then B 249 4- 2 251 (rounds to 300)

Thus, A = 249 and B=  251 is another solution.

Conclusion

The pairs (A, B) that satisfy the conditions are:

• (149, 151)

• (249,251)

These pairs of numbers differ by 2 and, when rounded to the nearest hundred, differ by 100.