2009/10 AHSMC Part 1 Problem 3

Problem 3 of 16

Ace calculates the average of all the integers from $1$ to $100$ . Bea calculates the average of all the integers from $1001$ to $1100$ and subtracts $1000$ . Cec calculates the average of all the integers from $1000001$ to $1000100$ and subtracts $1000000$ . The largest answer is given by

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Solution

The calculation is $\frac{(n+1)+(n+2)+\cdots+(n+100)}{100} - n = \frac{100n+(1+2+\cdots+100)}{100} - n = \frac{1+2+\cdots+100}{100},$ with $n = 0$ for Ace, $n = 1000$ for Bea and $n = 1000000$ for Cec. The answer is (E).

University of Alberta

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