Question

Define $L_n$ as the least common multiple of all the integers from $1$ to $n$ inclusive. There is a unique integer $h$ such that $\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} \cdots + \dfrac{1}{17} = \dfrac{h}{L_{17}}$ What is the remainder when $h$ is divided by $17?$

$1$

$3$

$5$

$7$

$9$

Solution:

Multiplying the harmonic sum by $L_{17}$ , we get $h=\sum_{i=1}^{17}\frac{L_{17}}{i}.$

For $1\le i\le16$ , the term $\frac{L_{17}}{i}$ is still divisible by $17$ , so these terms contribute $0\pmod{17}$ .

Thus $h\equiv \frac{L_{17}}{17}\pmod{17}.$ The least common multiple $L_{17}$ contains the prime-power factors $16,9,5,7,11,13,17$ , so $\frac{L_{17}}{17}\equiv 16\cdot9\cdot5\cdot7\cdot11\cdot13\pmod{17}.$

Reducing modulo $17$ , this is $(-1)\cdot9\cdot5\cdot7\cdot11\cdot13\equiv5\pmod{17}.$

Thus, C is the correct answer.

2022 AMC 10A Problem 19

Solution:

Problem 19 in Other Years