We seek 3-digit numbers divisible by 9 and ending in 3. A number divisible by 9 has its digit sum divisible by 9. Also, the number must end in 3, so let it be of the form \( 100a + 10b + 3 \), where \( a \in \{1,2,\dots,9\} \), \( b \in \{0,1,\dots,9\} \), and the number is divisible by 9.

["Finding All 3-Digit Numbers Divisible by 9 Ending in 3 — Step-by-Step Guide", "Numbers divisible by 9 have a special property: the sum of their digits is also divisible by 9. If we restrict our search to 3-digit numbers ending in 3, we can determine exactly how many such numbers exist and list them.", "This article explains how to find all 3-digit numbers divisible by 9 that end in the digit 3, using logic based on number properties and digit analysis.", "---", "### What is a 3-Digit Number Ending in 3?", "A 3-digit number ending in 3 takes the general form:", "[\nN = 100a + 10b + 3\n]", "where:\n- ( a ) is the hundreds digit, so ( a \in {1, 2, 3, \dots, 9} ) (since 000–099 is not 3-digit),\n- ( b ) is the tens digit, so ( b \in {0, 1, 2, \dots, 9} ).", "Thus, ( N ) ranges from 103 to 993 (e.g., 103, 123, ..., 993). All such numbers end in 3.", "---", "### Apply the Divisibility Rule for 9", "A number is divisible by 9 if and only if the sum of its digits is divisible by 9.", "Let’s express the digit sum for ( N = 100a + 10b + 3 ):", "[\n\ ext{Sum of digits} = a + b + 3\n]", "We want:", "[\na + b + 3 \equiv 0 \pmod{9}\n]", "This simplifies to:", "[\na + b \equiv 6 \pmod{9}\n]", "So, the goal is to find all digit pairs ( (a,b) ) with ( a \in {1,\dots,9} ), ( b \in {0,\dots,9} ), such that:", "[\na + b = 6 \quad \ ext{or} \quad a + b = 15\n]", "Why 6 and 15?\nBecause ( a + b + 3 ) must be divisible by 9, and the smallest possible sum is ( 1 + 0 + 3 = 4 ), the next possible multiple of 9 is 9 and then 18. So:", "- ( a + b + 3 = 9 \Rightarrow a + b = 6 )\n- ( a + b + 3 = 18 \Rightarrow a + b = 15 )\n- ( a + b + 3 = 27 ) → too large (max is ( 9+9+3=21 ))", "So only two cases:\nCase 1: ( a + b = 6 )\nCase 2: ( a + b = 15 )", "---", "### Find All Valid Pairs", "#### Case 1: ( a + b = 6 ), ( a \in [1,9] ), ( b \in [0,9] )", "Possible values:\n- ( a = 1 \Rightarrow b = 5 )\n- ( a = 2 \Rightarrow b = 4 )\n- ( a = 3 \Rightarrow b = 3 )\n- ( a = 4 \Rightarrow b = 2 )\n- ( a = 5 \Rightarrow b = 1 )\n- ( a = 6 \Rightarrow b = 0 )", "So pairs:\n(1,5), (2,4), (3,3), (4,2), (5,1), (6,0)\n→ Corresponding numbers:\n153, 243, 333, 423, 513, 603", "#### Case 2: ( a + b = 15 ), ( a \in [1,9] ), ( b \in [0,9] )", "Possible values:\n- ( a = 6 \Rightarrow b = 9 )\n- ( a = 7 \Rightarrow b = 8 )\n- ( a = 8 \Rightarrow b = 7 )\n- ( a = 9 \Rightarrow b = 6 )", "Note: ( a = 5 \Rightarrow b = 10 ) → invalid (b must be ≤ 9)\nSo valid pairs: (6,9), (7,8), (8,7), (9,6)\n→ Corresponding numbers:\n693, 783, 873, 963", "---", "### Complete List of Valid Numbers", "Combine both cases:", "| Number | Digit Sum | Digit Sum ÷ 9 |\n|--------|-----------|----------------|\n| 153 | 1+5+3 = 9 | ✅ |\n| 243 | 2+4+3 = 9 | ✅ |\n| 333 | 3+3+3 = 9 | ✅ |\n| 423 | 4+2+3 = 9 | ✅ |\n| 513 | 5+1+3 = 9 | ✅ |\n| 603 | 6+0+3 = 9 | ✅ |\n| 693 | 6+9+3 = 18| ✅ |\n| 783 | 7+8+3 = 18| ✅ |\n| 873 | 8+7+3 = 18| ✅ |\n| 963 | 9+6+3 = 18| ✅ |", "---", "### Summary", "There are exactly 10 three-digit positive integers divisible by 9 that end in the digit 3:", "[\n\boxed{153,\ 243,\ 333,\ 423,\ 513,\ 603,\ 693,\ 783,\ 873,\ 963}\n]", "These numbers satisfy the dual condition: ending in 3 and divisible by 9, verified by checking both divisibility rules — digit sum divisible by 9 and ending digit compatible with the sum.", "---", "### Final Notes", "- This method efficiently narrows the search by combining number theory with digit constraints.\n- No brute-force checking is needed — modular logic applies.\n- Useful for math learners, programmers solving number puzzles, or anyone exploring divisibility patterns.", "Start exploring this pattern: look for 3-digit numbers ( <em> </em> 3 ) where digit sum is 9 or 18 — perfect candidates!", "---", "Keywords: 3-digit numbers divisible by 9 ending in 3, numbers ending in 3 divisible by 9, digit sum divisibility rule, find divisible by 9 3-digit ending 3, modular arithmetic digit sum, 153 243 333 423 513 603 693 783 873 963."]