Let $a$ and $b$ be positive integers for which $45 a+b=2021$. The minimum possible value of $a+b$ is
(A) 44
(B) 82
(C) 85
(D) 86
(E) 130
Solution 1

Since $a$ is a positive integer, $45 a$ is a positive integer.

Since $b$ is a positive integer, $45 a$ is less than 2021 .

The largest multiple of 45 less than 2021 is $45 \times 44=1980$. (Note that $45 \cdot 45=2025$ which is greater than 2021.)

If $a=44$, then $b=2021-45 \cdot 44=41$.

Here, $a+b=44+41=85$.

If $a$ is decreased by 1 , the value of $45 a+b$ is decreased by 45 and so $b$ must be increased by 45 to maintain the same value of $45 a+b$, which increases the value of $a+b$ by $-1+45=44$.

Therefore, if $a44$, then $45 a>2021$ which makes $b$ negative, which is not possible.

Therefore, the minimum possible value of $a+b$ is 85 .

## Solution 2

We re-write $45 a+b=2021$ as $44 a+(a+b)=2021$.

Since $a$ and $b$ are positive integers, $44 a$ and $a+b$ are positive integers.

In particular, this tells us that $44 a$, which is a multiple of 44 , is less than 2021.

Since the sum of $44 a$ and $a+b$ is constant, to minimize $a+b$, we can try to maximize $44 a$.

Since $44 \cdot 45=1980$ and $44 \cdot 46=2024$, the largest multiple of 44 less than 2021 is 1980 .

This means that $a+b \geq 2021-1980=41$.

However, $a+b$ cannot equal 41 since we would need $44 a=1980$ and so $a=45$ (making $b=-4$ ) to make this possible.

The next multiple of 44 less than 1980 is $44 \cdot 44=1936$.

If $a=44$, then $a+b=2021-44 a=85$.

If $a=44$ and $a+b=85$, then $b=41$ which is possible.

Since $a+b=41$ is not possible and 85 is the next smallest possible value for $a+b$, then the minimum possible value for $a+b$ is 85 .

ANswer: (C)