# Difference between revisions of "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 $\boxed{869}$.