2022 AMC 10A Problem 19

Taking the highest power of each prime up to $17$,
$$L_{17} = 2^4 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17.$$
Multiplying the equation by $L_{17}$ gives
$$h = \frac{L_{17}}{1} + \frac{L_{17}}{2} + \cdots + \frac{L_{17}}{17},$$
and every term is an integer because each $k \le 17$ divides $L_{17}$.

Reduce modulo $17$. For $k \ne 17$, the factor $17$ of $L_{17}$ survives division by $k$, so $\frac{L_{17}}{k} \equiv 0 \pmod{17}$. Only the $k = 17$ term matters:
$$h \equiv \frac{L_{17}}{17} = 2^4 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \pmod{17}.$$
Multiply step by step, reducing as we go: $16 \equiv -1$; $-1 \cdot 9 = -9 \equiv 8$; $8 \cdot 5 = 40 \equiv 6$; $6 \cdot 7 = 42 \equiv 8$; $8 \cdot 11 = 88 \equiv 3$; $3 \cdot 13 = 39 \equiv 5$.

So $h \equiv 5 \pmod{17}$. The answer is $\boxed{\textbf{(C)}\ 5}$.

When a sum of fractions is put over a common denominator $L$, each numerator is $L/k$. Modulo a prime $p$ that divides $L$ exactly once, every $L/k$ with $p \nmid k$ is still a multiple of $p$, and only the $k = p$ term survives. Look for the one term that loses the prime; everything else vanishes.

• Trying to compute the harmonic sum or L_17 exactly, or forgetting that 17 divides L_17 only once (so L_17/17 is not a multiple of 17).
• Arithmetic slips in the mod $17$ product; reduce after every multiplication and use $16 \equiv -1$ to keep numbers small.