1989 AIME Problems/Problem 1

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 $30\cdot 29=870$ . Now clearly $868\cdot 870 + 1 = (869-1)(869+1) + 1 = 869^2 - 1^2 + 1 = 869^2$ .

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 $(n-\frac{3}{2}), (n-\frac{1}{2}), (n+\frac{1}{2}), (n+\frac{3}{2})$ to make use of symmetry. Note that $n$ itself is not an integer - in this case, $n = 29.5$ . The expression becomes $\sqrt{(n-\frac{3}{2})(n + \frac{3}{2})(n - \frac{1}{2})(n + \frac{1}{2}) + 1}$ . 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 $\sqrt{(n^2 - \frac{5}{4})^2}$ , which simplifies to $n^2 - \frac{5}{4}$ . You can plug in the value of $n$ from there, or further simplify to $(n - \frac{1}{2})(n + \frac{1}{2} - 1)$ , which is easier to compute. Either way, plugging in $n=29.5$ gives $\boxed{869}$ .

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

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