2017 AMC 12B Problem 19

Since $45 = 9 \cdot 5,$ find $N \bmod 5$ (the last digit) and $N \bmod 9$ (the digit sum)

The digit sum is a multiple of $9,$ so $N \equiv 0 \pmod 9;$ combine with the residue mod $5$

The last digit of $N$ is $4,$ so $N \equiv 4 \pmod 5.$ For mod $9,$ sum the digits: the numbers $1$–$9$ contribute their digits, the tens digits of $10$–$44$ and the units digits together sum to $270,$ which is a multiple of $9,$ so $N \equiv 0 \pmod 9.$ The number $N - 9$ is then a multiple of $9,$ and its last digit is $5,$ so it is a multiple of $5;$ hence $N - 9$ is a multiple of $45.$ Therefore $N \equiv 9 \pmod{45}.$

Thus, the correct answer is C.