1996 AJHSME Problems/Problem 4
Problem
Solution 1
First, notice that each number in the numerator is a multiple of , and each number in the denominator is a multiple of . This suggests that each expression can be factored. Factoring gives:
Since all the material in the parentheses is the same, the common factor in the numerator and the denominator may be cancelled, leaving , which is option .
Solution 2
There are terms in the numerator . Consider adding those terms, but in a different order. Start with the last two terms, , and then add the next two terms on the outside, , and continue. You will get pairs of numbers that add to , while the number in the middle will be alone. That number is . Adding all the numbers gives .
Similarly, the denominator has terms of . There are pairs of numbers that add up to , with the number in the center being . The total of all the numbers is .
The answer is . Eyeballing the options, the fraction is clearly under , but more than . Thus, the answer must be , or . Alternately, you can do the work by factoring out a in the numerator to give . Factoring out will give the desired answer.
Solution 3
Start by finding a pattern:
Each step doesn't seem to change the value of the fraction, so is the right answer.
See also
1996 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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