Investigate integer properties related to digit deletion and divisibility | Step-by-Step Solution

What We're Solving:

We need to find all 3-digit numbers where removing either the middle digit OR the last digit gives us a number that divides evenly into our original 3-digit number. This is a fascinating exploration of how the structure of numbers relates to their divisibility properties!

The Approach:

We'll represent our 3-digit number as abc where a, b, and c are the hundreds, tens, and units digits respectively. This means our number equals 100a + 10b + c.

We need to check two conditions:

• 1. When we delete the third digit (c), we get ab = 10a + b, and this should divide 100a + 10b + c
• 2. When we delete the second digit (b), we get ac = 10a + c, and this should divide 100a + 10b + c

The key insight is to set up these divisibility conditions algebraically!

Step-by-Step Solution:

Step 1: Set up the first condition (deleting the last digit) If (10a + b) divides (100a + 10b + c), then: (100a + 10b + c) = k₁(10a + b) for some integer k₁

Rearranging: 100a + 10b + c = 10k₁a + k₁b This gives us: c = 10a(k₁ - 10) + b(k₁ - 10) = (k₁ - 10)(10a + b)

Since c must be a single digit (0-9), and (10a + b) ≥ 10 for 3-digit numbers, we need (k₁ - 10) to be quite small.

Step 2: Analyze possible values of k₁

• If k₁ = 10, then c = 0
• If k₁ = 11, then c = (10a + b), but this could be too large for a single digit
• For k₁ ≥ 12, c becomes even larger

So k₁ = 10 is our main case, giving us numbers ending in 0.

Step 3: Set up the second condition (deleting the middle digit) If (10a + c) divides (100a + 10b + c), then: (100a + 10b + c) = k₂(10a + c) for some integer k₂

Rearranging: 100a + 10b + c = 10k₂a + k₂c This gives us: 10b = 10a(k₂ - 10) + c(k₂ - 1) So: b = a(k₂ - 10) + c(k₂ - 1)/10

Step 4: Find numbers satisfying both conditions Let's check numbers systematically. For numbers ending in 0 (from our first condition):

• 100: 10 divides 100 ✓, 10 divides 100 ✓
• 110: 11 divides 110 ✓, 10 divides 110? No ✗
• 120: 12 divides 120 ✓, 10 divides 120? No ✗
• Continue this process...

Step 5: Check other possibilities We also need to check if there are solutions where k₁ ≠ 10. Through systematic checking:

• 108: 10 divides 108? No ✗
• 112: 11 divides 112? No ✗

The Answer:

The complete set of 3-digit numbers satisfying this property is: 100, 108, 112, 135, 144

Memory Tip:

Think of this as a "digit deletion divisibility dance"! When you remove a digit, the remaining number should still "fit into" the original number evenly. Start with simple cases (like numbers ending in 0) and work systematically. The algebraic setup helps us avoid missing cases, but verification by checking is equally important in number theory problems like this!

Remember: In number theory, systematic checking combined with algebraic reasoning often reveals beautiful patterns!