Solution to Problem 11

Let k be the number of the page that was counted twice. Then, 0<k<n+1, and $1+2+\ldots+n+k$ is between $1+2+\ldots+n$ and $1+2+\ldots+n+(n+1)$. In other words,

$$\frac{n(n+1)}{2}<1986<\frac{(n+1)(n+2)}{2},$$

i.e.,

n(n+1)<3972<(n+1)(n+2).

By trial and error (clearly, n is a little larger than 60) we find that n=62. Thus

$$k=1986-\frac{62\cdot 63}{2}=1986-1953=33.$$

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Sat Nov 11 13:04:39 CST 2000