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Home » This Unbelievable TETSUWAN ATOM Feature Will Blow Your Mind—You Won’t Believe #7! - Get link 4share

This Unbelievable TETSUWAN ATOM Feature Will Blow Your Mind—You Won’t Believe #7! - Get link 4share

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 How many positive 3-digit numbers are divisible by 9 and end with the digit 3?

📌 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.

📌 Let \( N = 100a + 10b + 3 \). We require \( N \equiv 0 \pmod{9} \). Since \( 100 \equiv 1 \pmod{9} \), \( 10 \equiv 1 \pmod{9} \), so:

📌 N \equiv a + b + 3 \pmod{9}

📌 a + b + 3 \equiv 0 \pmod{9} \quad \Rightarrow \quad a + b \equiv 6 \pmod{9}

📌 So \( a + b = 6 \) or \( a + b = 15 \) (since \( a \geq 1 \), \( b \leq 9 \), maximum \( a + b = 9 + 9 = 18 \), but we'll check both).

📌 First, count solutions to \( a + b = 6 \), \( a \geq 1 \), \( b \geq 0 \):

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