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. - Nelissen Grade advocaten