Let $a_n$ be the integer obtained by writing all the integers from $1$ to $n$ from left to right. For example, $a_3 = 123$ and $a_{11} = 1234567891011$. Compute the remainder when $a_{44}$ is divided by $45$.
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{align*}
a_{44}&=44+43\cdot 10^2 + 42 \cdot 10^4+41\cdot 10^6+\cdots+10\cdot10^{68}\\
&\qquad+9\cdot 10^{70}+8\cdot 10^{71}+\cdots + 1\cdot 10^{78} \\ &\equiv 44+43+42+\cdots+1\pmod{9},
\end{align*}since $10^n\equiv 1^n\equiv 1\pmod{9}$ 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 $\boxed{9}$.
9
Level 5
Number Theory
Problem: Let $a_n$ be the integer obtained by writing all the integers from $1$ to $n$ from left to right. For example, $a_3 = 123$ and $a_{11} = 1234567891011$. Compute the remainder when $a_{44}$ is divided by $45$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the remainder of $a_{44}$ when divided by $45$, we first need to determine the value of $a_{44}$. This involves concatenating all integers from $1$ to $44$. 

Next, we can analyze the structure of $a_{44}$. The total number of digits contributed by each range of numbers is as follows:

- Numbers from $1$ to $9$ contribute $9$ digits (each number has $1$ digit).
- Numbers from $10$ to $44$ contribute $35 \times 2 = 70$ digits (each number has $2$ digits).

Thus, the total number of digits in $a_{44}$ is $9 + 70 = 79$ digits.

To compute $a_{44} \mod 45$, we can use the property that $45 = 5 \times 9$. Therefore, we will find $a_{44} \mod 5$ and $a_{44} \mod 9$ separately, and then combine the results using the Chinese Remainder Theorem.

First, we calculate $a_{44} \mod 5$. The last digit of $a_{44}$ is the last digit of $44$, which is $4$. Therefore, $a_{44} \equiv 4 \mod 5$.

Next, we calculate $a_{44} \mod 9$. To do this, we can find the sum of the digits of $a_{44}$, as a number is congruent to the sum of its digits modulo $9$.

The sum of the digits from $1$ to $9$ is:
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45.

The sum of the digits from $10$ to $44$ can be calculated as follows:
- For the tens place (from $10$ to $19$), we have $1$ appearing $10$ times, contributing $10$.
- For the units place (from $10$ to $19$), the digits are $0$ to $9$, contributing $45$.
- For the tens place (from $20$ to $29$), we have $2$ appearing $10$ times, contributing $20$.
- For the units place (from $20$ to $29$), the digits are $0$ to $9$, contributing $45$.
- For the tens place (from $30$ to $39$), we have $3$ appearing $10$ times, contributing $30$.
- For the units place (from $30$ to $39$), the digits are $0$ to $9$, contributing $45$.
- For the tens place (from $40$ to $44$), we have $4$ appearing $5$ times, contributing $20$.
- For the units place (from $40$ to $44$), the digits are $0$, $1$, $2$, $3$, and $4$, contributing $10$.

Now, we sum these contributions:
From $1$ to $9$: $45$.
From $10$ to $19$: $10 + 45 = 55$.
From $20$ to $29$: $20 + 45 = 65$.
From $30$ to $39$: $30 + 45 = 75$.
From $40$ to $44$: $20 + 10 = 30$.

Total sum of digits:
$45 + 55 + 65 + 75 + 30 = 270$.

Now, we find $270 \mod 9$. The sum of the digits of $270$ is $2 + 7 + 0 = 9$, which gives $270 \equiv 0 \mod 9$.

Now we have:
1. $a_{44} \equiv 4 \mod 5$.
2. $a_{44} \equiv 0 \mod 9$.

Using the Chinese Remainder Theorem, we need to find a number $x$ such that:
- $x \equiv 4 \mod 5$
- $x \equiv 0 \mod 9$

The numbers that satisfy $x \equiv 0 \mod 9$ are $0, 9, 18, 27, 36, 45, \ldots$. Among these, we check which one satisfies $x \equiv 4 \mod 5$:
- $0 \mod 5 \equiv 0$
- $9 \mod 5 \equiv 4$ (this works)
- $18 \mod 5 \equiv 3$
- $27 \mod 5 \equiv 2$
- $36 \mod 5 \equiv 1$
- $45 \mod 5 \equiv 0$

Thus, the solution is $x = 9$.

Therefore, the final answer is $9$.<|END_OF_LATENT|> The final answer is $9$.