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

Revision as of 13:02, 28 June 2021

Problem

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

Solution 1 (Symmetry)

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 2 (Symmetry)

Notice that $(a+1)^2 = a \cdot (a+2) +1$ . Then we can notice that $30 \cdot 29 =870$ and that $31 \cdot 28 = 868$ . Therefore, $\sqrt{(31)(30)(29)(28) +1} = \sqrt{(870)(868) +1} = \sqrt{(868 +1)^2} = \boxed{869}$ . This is because we have that $a=868$ as per the equation $(a+1)^2 = a \cdot (a+2) +1$ .

~qwertysri987

Solution 3 (Symmetry with Generalization)

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 4 (Symmetry with Generalization)

More generally, we can show that one more than the product of four consecutive integers must be a perfect square: $\begin{align*} (a+3)(a+2)(a+1)(a)+1 &= \left[(a+3)(a)\right]\left[(a+2)(a+1)\right]+1 \\ &= \left[a^2+3a\right]\left[a^2+3a+2\right]+1 \\ &= \left[\left(a^2+3a+1\right)-1\right]\left[\left(a^2+3a+1\right)+1\right]+1 \\ &= \left[\left(a^2+3a+1\right)^2-1^2\right]+1 \\ &= \left(a^2+3a+1\right)^2. \end{align*}$ At $a=28,$ we have $\sqrt{(a+3)(a+2)(a+1)(a)+1}=a^2+3a+1=\boxed{869}.$ ~Novus677 (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 5 (Prime Factorizations)

Multiplying $(31)(30)(29)(28)$ gives us $755160$ . Adding $1$ to this gives $755161$ . Now we must choose a number squared that is equal to $755161$ . Taking the square root of this gives $\boxed{869}$

Solution 6 (Observations)

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 $\boxed{869}$ .

@@ Line 16: / Line 16: @@
 == Solution 4 (Symmetry with Generalization) ==
-More generally, we can prove that one more than the product of four consecutive integers is always a perfect square:
+More generally, we can show that one more than the product of four consecutive integers must be a perfect square:
 <cmath>\begin{align*}
 (a+3)(a+2)(a+1)(a)+1 &= \left[(a+3)(a)\right]\left[(a+2)(a+1)\right]+1 \\

1989 AIME (Problems • Answer Key • Resources)
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