Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1989 AIME Problems/Problem 1"

(→‎Solution 1)
m (→‎Solution)
Line 18: Line 18:
 
===Solution 4===
 
===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>.
 
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===
 +
Multiplying <math>(31)(30)(29)(28)</math> gives us <math>755160</math>. Adding <math>1</math> to this gives <math>755161</math>. Through guess and check, we get <math>869</math>.
  
 
== See also ==
 
== See also ==

Revision as of 14:51, 14 July 2017

Problem

Compute $\sqrt{(31)(30)(29)(28)+1}$.

Solution

Solution 1

Notice ${31*28 = 868}$ and ${30*29 =870}$. So now our expression is $\sqrt{(870)(868) + 1}$. Setting 870 equal to $y$, we get $\sqrt{(y-1)^{2}}$ which then equals ${(y-1)}$. So since ${y = 870}$, ${y-1}=869$, our answer is $\boxed{869}$.

Solution 2

Note that the four numbers to multiply are symmetric with the center at $29.5$. Multiply the symmetric pairs to get $31\cdot 28=868$ and $30\cdot 29=870$. $\sqrt{868\cdot 870 + 1} = \sqrt{(869-1)(869+1) + 1} = \sqrt{869^2 - 1^2 + 1} = \sqrt{869^2} = \boxed{869}$.

Solution 3

The last digit under the radical is $1$, so the square root must either end in $1$ or $9$, since $x^2 = 1\pmod {10}$ means $x = \pm 1$. Additionally, the number must be near $29 \cdot 30 = 870$, narrowing the reasonable choices to $869$ and $871$.

Continuing the logic, the next-to-last digit under the radical is the same as the last digit of $28 \cdot 29 \cdot 3 \cdot 31$, which is $6$. Quick computation shows that $869^2$ ends in $61$, while $871^2$ ends in $41$. Thus, the answer is $\boxed{869}$.

Solution 4

Similar to Solution 1 above, call the consecutive integers $\left(n-\frac{3}{2}\right), \left(n-\frac{1}{2}\right), \left(n+\frac{1}{2}\right), \left(n+\frac{3}{2}\right)$ to make use of symmetry. Note that $n$ itself is not an integer - in this case, $n = 29.5$. The expression becomes $\sqrt{\left(n-\frac{3}{2}\right)\left(n + \frac{3}{2}\right)\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) + 1}$. Distributing each pair of difference of squares first, and then distributing the two resulting quadratics and adding the constant, gives $\sqrt{n^4 - \frac{5}{2}n^2 + \frac{25}{16}}$. The inside is a perfect square trinomial, since $b^2 = 4ac$. It's equal to $\sqrt{\left(n^2 - \frac{5}{4}\right)^2}$, which simplifies to $n^2 - \frac{5}{4}$. You can plug in the value of $n$ from there, or further simplify to $\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) - 1$, which is easier to compute. Either way, plugging in $n=29.5$ gives $\boxed{869}$.

Solution 5

Multiplying $(31)(30)(29)(28)$ gives us $755160$. Adding $1$ to this gives $755161$. Through guess and check, we get $869$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_1&oldid=86374"