Difference between revisions of "1989 AIME Problems/Problem 1"
Lilcritters (talk | contribs) (→Solution 6) |
Mathlete2017 (talk | contribs) (→Solution 5) |
||
Line 19: | Line 19: | ||
Similar to Solution 1 above, call the consecutive integers <math>\left(n-\frac{3}{2}\right), \left(n-\frac{1}{2}\right), \left(n+\frac{1}{2}\right), \left(n+\frac{3}{2}\right)</math> to make use of symmetry. Note that <math>n</math> itself is not an integer - in this case, <math>n = 29.5</math>. The expression becomes <math>\sqrt{\left(n-\frac{3}{2}\right)\left(n + \frac{3}{2}\right)\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) + 1}</math>. Distributing each pair of difference of squares first, and then distributing the two resulting quadratics and adding the constant, gives <math>\sqrt{n^4 - \frac{5}{2}n^2 + \frac{25}{16}}</math>. The inside is a perfect square trinomial, since <math>b^2 = 4ac</math>. It's equal to <math>\sqrt{\left(n^2 - \frac{5}{4}\right)^2}</math>, which simplifies to <math>n^2 - \frac{5}{4}</math>. You can plug in the value of <math>n</math> from there, or further simplify to <math>\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) - 1</math>, which is easier to compute. Either way, plugging in <math>n=29.5</math> gives <math>\boxed{869}</math>. | Similar to Solution 1 above, call the consecutive integers <math>\left(n-\frac{3}{2}\right), \left(n-\frac{1}{2}\right), \left(n+\frac{1}{2}\right), \left(n+\frac{3}{2}\right)</math> to make use of symmetry. Note that <math>n</math> itself is not an integer - in this case, <math>n = 29.5</math>. The expression becomes <math>\sqrt{\left(n-\frac{3}{2}\right)\left(n + \frac{3}{2}\right)\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) + 1}</math>. Distributing each pair of difference of squares first, and then distributing the two resulting quadratics and adding the constant, gives <math>\sqrt{n^4 - \frac{5}{2}n^2 + \frac{25}{16}}</math>. The inside is a perfect square trinomial, since <math>b^2 = 4ac</math>. It's equal to <math>\sqrt{\left(n^2 - \frac{5}{4}\right)^2}</math>, which simplifies to <math>n^2 - \frac{5}{4}</math>. You can plug in the value of <math>n</math> from there, or further simplify to <math>\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) - 1</math>, which is easier to compute. Either way, plugging in <math>n=29.5</math> gives <math>\boxed{869}</math>. | ||
− | ==Solution 5== | + | ===Solution 5=== |
Note that <math>a(a+1)(a+2)(a+3)+1=(a^2+3a+1)^2</math>. | Note that <math>a(a+1)(a+2)(a+3)+1=(a^2+3a+1)^2</math>. | ||
So, our answer is just <math>28^2+3\cdot 28+1=\boxed{869}</math> | So, our answer is just <math>28^2+3\cdot 28+1=\boxed{869}</math> | ||
− | ===Solution | + | ===Solution 6=== |
Multiplying <math>(31)(30)(29)(28)</math> gives us <math>755160</math>. Adding <math>1</math> to this gives <math>755161</math>. Now we must choose a number squared that is equal to <math>755161</math>. Taking the square root of this gives <math>\boxed{869}</math> | Multiplying <math>(31)(30)(29)(28)</math> gives us <math>755160</math>. Adding <math>1</math> to this gives <math>755161</math>. Now we must choose a number squared that is equal to <math>755161</math>. Taking the square root of this gives <math>\boxed{869}</math> | ||
Revision as of 22:08, 12 December 2018
Contents
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 .
Solution 5
Note that . So, our answer is just
Solution 6
Multiplying gives us . Adding to this gives . Now we must choose a number squared that is equal to . Taking the square root of this 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.