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

Revision as of 22:17, 8 March 2013

Problem

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

Solution

Solution 1

Let's call our four consecutive integers $(n-1), n, (n+1), (n+2)$ . Notice that $(n-1)(n)(n+1)(n+2)+1=(n^2+n)^2-2(n^2+n)+1 \Rightarrow (n^2+n-1)^2$ . Thus, $\sqrt{(31)(30)(29)(28)+1} = (29^2+29-1) = \boxed{869}$ .

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 $\boxed{869}$ .

@@ Line 7: / Line 7: @@
 === 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>. Now clearly <math>868\cdot 870 + 1 = (869-1)(869+1) + 1 = 869^2 - 1^2 + 1 = 869^2</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===

1989 AIME (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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