Problem 359

Solution 1

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

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

Therefore, if $a<44$, the value of $a+b$ is always greater than 85.

If $a>44$, then $45a>2021$ which makes $b$ negative, which is not possible.

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

Solution 2

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

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

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

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

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 $44a = 1980$ 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 $a+b = 2021 - 44a = 85$.

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.