🔢 Digit Counting

Problems involving counting numbers based on their digits, divisibility rules, and base representations.

🎯 Recognition Cues

  • Keywords: “digits”, “divisible by”, “base”, “leading zeros”, “palindrome”
  • Setup: Numbers with specific digit properties
  • Constraints: Digit sums, divisibility, base conversions

📋 Solution Template

  1. Identify the digit property:

    • Divisibility: Use divisibility rules
    • Digit sum: Use digit counting techniques
    • Base representation: Use base conversion methods

  2. Apply the appropriate method:

    • Divisibility: Use modular arithmetic
    • Digit sum: Use generating functions or casework
    • Base representation: Use place value system

  3. Check for special cases:

    • Leading zeros: May or may not be allowed
    • Base constraints: Different bases have different rules
    • Range constraints: May limit the counting

💡 Micro-Examples

Divisibility by 3

  • Problem: How many 3-digit numbers are divisible by 3?
  • Solution: Use the fact that a number is divisible by 3 if and only if the sum of its digits is divisible by 3.

Digit Sum

  • Problem: How many 3-digit numbers have digit sum 15?
  • Solution: Use casework on the hundreds digit: 6, 7, 8, 9 are possible.

Base Representation

  • Problem: How many 3-digit numbers in base 5 are there?
  • Solution: First digit: 1-4 (4 choices), other digits: 0-4 (5 choices each). Total: $4 \cdot 5^2 = 100$.

⚠️ Common Pitfalls & Variants

Pitfall: Forgetting about leading zeros

  • Wrong: “3-digit numbers” = Always 3 digits
  • Right: Check if leading zeros are allowed or not.

Pitfall: Confusing base representations

  • Wrong: “3-digit number in base 5” = Same as base 10
  • Right: Different bases have different digit ranges and place values.

Pitfall: Misapplying divisibility rules

  • Wrong: “Divisible by 3” = Last digit must be 3, 6, or 9
  • Right: Sum of digits must be divisible by 3.

🏆 AMC-Style Worked Example

Problem: How many 4-digit numbers have exactly 2 even digits?

Solution:

  • Step 1: Choose 2 positions for even digits: $\binom{4}{2} = 6$ ways
  • Step 2: Choose even digits for those positions: $5^2 = 25$ ways (0,2,4,6,8)
  • Step 3: Choose odd digits for remaining positions: $5^2 = 25$ ways (1,3,5,7,9)
  • Step 4: Total: $6 \cdot 25 \cdot 25 = 3750$ ways

Key insight: Use combinations to choose positions, then fill each position independently!

🔗 Related Topics

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