Difference between revisions of "1989 AIME Problems/Problem 1"
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== Solution == | == Solution == | ||
=== Solution 1=== | === Solution 1=== | ||
− | + | 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>. | |
=== Solution 2=== | === Solution 2=== | ||
− | Note that the four numbers to multiply are symmetric with the center at <math>29.5</math>. Multiply the symmetric pairs to get <math>31\cdot 28=868</math> and <math>30\cdot 29=870</math>. | + | Note that the four numbers to multiply are symmetric with the center at <math>29.5</math>. |
+ | Multiply the symmetric pairs to get <math>31\cdot 28=868</math> and <math>30\cdot 29=870</math>. | ||
+ | <math>\sqrt{868\cdot 870 + 1} = \sqrt{(869-1)(869+1) + 1} = \sqrt{869^2 - 1^2 + 1} = \sqrt{869^2} = \boxed{869}</math>. | ||
+ | |||
+ | === Solution 3=== | ||
+ | The last digit under the radical is <math>1</math>, so the square root must either end in <math>1</math> or <math>9</math>, since <math>x^2 = 1\pmod {10}</math> means <math>x = \pm 1</math>. Additionally, the number must be near <math>29 \cdot 30 = 870</math>, narrowing the reasonable choices to <math>869</math> and <math>871</math>. | ||
+ | |||
+ | Continuing the logic, the next-to-last digit under the radical is the same as the last digit of <math>28 \cdot 29 \cdot 3 \cdot 31</math>, which is <math>6</math>. Quick computation shows that <math>869^2</math> ends in <math>61</math>, while <math>871^2</math> ends in <math>41</math>. Thus, the answer is <math>\boxed{869}</math>. | ||
+ | |||
+ | ===Solution 4=== | ||
+ | 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=== | ||
+ | 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> | ||
+ | |||
+ | ===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> | ||
+ | |||
+ | ===Solution 7=== | ||
+ | Notice that <math>(a+1)^2 = a \cdot (a+2) +1</math>. Then we can notice that <math>30 \cdot 29 =870 </math> and that <math>31 \cdot 28 = 868</math>. Therefore, <math> \sqrt{(31)(30)(29)(28) +1} = \sqrt{(870)(868) +1} = \sqrt{(868 +1)^2} = \boxed{869}</math>. This is because we have that <math>a=868</math> as per the equation <math>(a+1)^2 = a \cdot (a+2) +1</math>. | ||
+ | |||
+ | ~qwertysri987 | ||
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 12:55, 2 July 2019
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
Solution 7
Notice that . Then we can notice that and that . Therefore, . This is because we have that as per the equation .
~qwertysri987
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.