2002 AMC 12B Problem 4

Since $\dfrac12+\dfrac13+\dfrac17=\dfrac{41}{42},$ the sum $\dfrac{41}{42}+\dfrac1n$ lies strictly between $0$ and $2,$ so it must equal $1.$ Then $\dfrac1n=\dfrac1{42},$ giving $n=42.$ Now $2,$ $3,$ $6,$ and $7$ all divide $42,$ but $n\gt84$ is false. The untrue statement is $n\gt84.$ Thus, the correct answer is E.