# Difference between revisions of "1985 AJHSME Problems/Problem 2"

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==Solution== | ==Solution== | ||

− | We could just add them all together. But what would the point | + | We could just add them all together. But what would be the point of doing that? So we find a slicker way. |

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+ | We find a simpler problem in this problem, and simplify -> <math>90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9</math> | ||

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+ | We know <math>90 \times 10</math>, that's easy - <math>900</math>. So how do we find <math>1 + 2 + ... + 8 + 9</math>? | ||

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+ | We rearrange the numbers to make <math>(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5</math>. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. <math>4 \times 10 + 5 = 45</math>. Adding that on to 900 makes 945. | ||

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+ | 945 is <math>\boxed{\text{B}}</math> | ||

==See Also== | ==See Also== | ||

[[1985 AJHSME Problems]] | [[1985 AJHSME Problems]] |

## Revision as of 22:12, 12 January 2009

## Problem

## Solution

We could just add them all together. But what would be the point of doing that? So we find a slicker way.

We find a simpler problem in this problem, and simplify ->

We know , that's easy - . So how do we find ?

We rearrange the numbers to make . You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. . Adding that on to 900 makes 945.

945 is