jmc

We will use the Chinese Remainder Theorem as follows: We will compute the remainders when $a_{44}$ is divided by $5$ and $9$. The remainder when $a_{44}$ is divided by 45 will be the residue (mod 45) which leaves the same remainders when divided by 5 and 9 as $a_{44}$ does. Since $a_{44}$ ends in $4$, it gives a remainder of $4$ when divided by $5$.

For the remainder when $a_{44}$ is divided by 9, note that $$\begin{array}{rl}{a}_{44}& =44+43\cdot 10^2+42\cdot 10^4+41\cdot 10^6+\cdots +10\cdot 10^{68}\\ & +9\cdot 10^{70}+8\cdot 10^{71}+\cdots +1\cdot 10^{78}\\ & \equiv 44+43+42+\cdots +1(\mathrm{mod}9),\end{array}$$since $10^n\equiv 1^n\equiv 1(\mathrm{mod}9)$ for all nonnegative integers $n$. In words, this calculation shows that we can sum groups of digits in any way we choose to check for divisibility by 9. For example, 1233 is divisible by 9 since $12+33=45$ is divisible by 9. This is a generalization of the rule that a number is divisible by 9 if and only if the sum of its digits is divisible by 9. Getting back to the problem at hand, we sum $44+43+\cdots +1$ using the formula $1+2+\cdots +n=n(n+1)/2$ to find that $a_{44}$ is divisible by 9.

We are looking for a multiple of $9$ that gives a remainder of $4$ when divided by $5$. Nine satisfies this condition, so the remainder when $a_{44}$ is divided by 45 is $\boxed{9}$.