A teacher's total salary after $n$ years is $\dfrac{nx(n+19)}{20}$ , where $\$x$ is his first year's salary. Find the least value of $n$ for which his total salary exceeds 100 times his first salary.

n = 45

n = 100

n = 37

n = 36

Explanation:

\tfrac{nx(n+19)}{20}>100x \Rightarrow n^2+19n-2000>0

. The positive root is

n=\tfrac{-19+\sqrt{8361}}{2}=36.2

, so the least integer is

n=37

Derived from ZIMSEC Mathematics Paper 1, June 2011, Q12

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