2022 AMC 10A Problem 19

Step 1: build $L_{17}$. The least common multiple of $1$ to $17$ takes the highest power of each prime that is at most $17$: $16 = 2^4$, $9 = 3^2$, and $5, 7, 11, 13, 17$. So
$$L_{17} = 2^4 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17.$$

Step 2: find $h$. Multiply the equation by $L_{17}$:
$$h = \frac{L_{17}}{1} + \frac{L_{17}}{2} + \cdots + \frac{L_{17}}{17}.$$
Each term is a whole number, because every $k$ from $1$ to $17$ divides $L_{17}$.

Step 3: throw away the terms that are multiples of 17. For $k = 1, 2, \ldots, 16$, the number $k$ has no factor of $17$, so dividing $L_{17}$ by $k$ leaves the $17$ in place. Those $16$ terms are all multiples of $17$ and leave remainder $0$. Only the last term matters:
$$\frac{L_{17}}{17} = 2^4 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13.$$
This one is not a multiple of $17$, because $L_{17}$ has only one $17$.

Step 4: find its remainder when divided by 17. Multiply one factor at a time, keeping only the remainder after each step. (This is allowed: the part that is a multiple of $17$ stays a multiple of $17$ after multiplying.)

So $h$ leaves remainder $5$ when divided by $17$.

The answer is $\boxed{\textbf{(C)}\ 5}$.

When fractions are put over a common denominator $L$, the numerators are $L$ divided by each denominator. If a prime appears in $L$ exactly once, every term except the one that divides it away is still a multiple of that prime. So look for the single term that loses the prime; all the others drop out of the remainder.