1998 CEMC Pascal Problems/Problem 7

Problem

The value of $490 - 491 + 492 - 493 + 494 - 495 + ... - 509 + 510$ is

$\text{ (A) }\ 500 \qquad\text{ (B) }\ -10 \qquad\text{ (C) }\ -11 \qquad\text{ (D) }\ 499 \qquad\text{ (E) }\ 510$

Solution

We can see that until we get to $510$ in the pattern, each group of two has a value of $-1$ because $n - (n + 1) = n - n - 1 = -1$ for all values of $n$ .

We have $10$ pairs because $509$ is $18$ more than $491$ , which equates to $9$ extra pairs after the first one. We can also list the numbers to see this. This means that the total value is:

$510 - 1 \times 10 = 510 - 10 = \boxed {\textbf {(A) } 500}$

~anabel.disher

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1998 CEMC Pascal Problems/Problem 7

Problem

Solution