Difference between revisions of "1989 AIME Problems/Problem 1"
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Notice <math>{31*28 = 868}</math> and <math>{30*29 =870}</math>. So now our expression is <math>\sqrt{(870)(868) + 1}</math>. Setting 870 equal to <math>y</math>, we get <math>\sqrt{(y-1)^{2}}</math> which then equals <math>{(y-1)}</math>. So since <math>{y = 870}</math>, <math>{y-1}=869</math>, our answer is <math>\boxed{869}</math>. | Notice <math>{31*28 = 868}</math> and <math>{30*29 =870}</math>. So now our expression is <math>\sqrt{(870)(868) + 1}</math>. Setting 870 equal to <math>y</math>, we get <math>\sqrt{(y-1)^{2}}</math> which then equals <math>{(y-1)}</math>. So since <math>{y = 870}</math>, <math>{y-1}=869</math>, our answer is <math>\boxed{869}</math>. | ||
Revision as of 10:47, 23 April 2017
Contents
[hide]Problem
Compute .
Solution
Solution 1
Notice and . So now our expression is . Setting 870 equal to , we get which then equals . So since , , our answer is .
Solution 2
Note that the four numbers to multiply are symmetric with the center at . Multiply the symmetric pairs to get and . .
Solution 3
The last digit under the radical is , so the square root must either end in or , since means . Additionally, the number must be near , narrowing the reasonable choices to and .
Continuing the logic, the next-to-last digit under the radical is the same as the last digit of , which is . Quick computation shows that ends in , while ends in . Thus, the answer is .
Solution 4
Similar to Solution 1 above, call the consecutive integers to make use of symmetry. Note that itself is not an integer - in this case, . The expression becomes . Distributing each pair of difference of squares first, and then distributing the two resulting quadratics and adding the constant, gives . The inside is a perfect square trinomial, since . It's equal to , which simplifies to . You can plug in the value of from there, or further simplify to , which is easier to compute. Either way, plugging in gives .
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.