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

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m (→‎Solution 3 (Symmetry with Generalization))
 
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More generally, we can prove that one more than the product of four consecutive integers must be a perfect square:
 
More generally, we can prove that one more than the product of four consecutive integers must be a perfect square:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
(a+3)(a+2)(a+1)(a)+1 &= \left[(a+3)(a)\right]\left[(a+2)(a+1)\right]+1 \\
+
(a+3)(a+2)(a+1)(a)+1 &= [(a+3)(a)][(a+2)(a+1)]+1 \\
−
&= \left[a^2+3a\right]\left[a^2+3a+2\right]+1 \\
+
&= [a^2+3a][a^2+3a+2]+1 \\
−
&= \left[a^2+3a\right]^2+2\left[a^2+3a\right]+1 \\
+
&= [a^2+3a]^2+2[a^2+3a]+1 \\
−
&= \left[a^2+3a+1\right]^2.
+
&= [a^2+3a+1]^2.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 
At <math>a=28,</math> we have <cmath>\sqrt{(a+3)(a+2)(a+1)(a)+1}=a^2+3a+1=\boxed{869}.</cmath>
 
At <math>a=28,</math> we have <cmath>\sqrt{(a+3)(a+2)(a+1)(a)+1}=a^2+3a+1=\boxed{869}.</cmath>

Latest revision as of 23:12, 7 October 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)

More generally, we can prove 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 &= [(a+3)(a)][(a+2)(a+1)]+1 \\ &= [a^2+3a][a^2+3a+2]+1 \\ &= [a^2+3a]^2+2[a^2+3a]+1 \\ &= [a^2+3a+1]^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 4 (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 5 (Prime Factorization)

We have $(31)(30)(29)(28)+1=755161.$ Since the alternating sum of the digits $7-5+5-1+6-1=11$ is divisible by $11,$ we conclude that $755161$ is divisible by $11.$

We evaluate the original expression by prime factorization: \begin{align*} \sqrt{(31)(30)(29)(28)+1}&=\sqrt{755161} \\ &=\sqrt{11\cdot68651} \\ &=\sqrt{11^2\cdot6241} \\ &=\sqrt{11^2\cdot79^2} \\ &=11\cdot79 \\ &=\boxed{869}. \end{align*} ~Vrjmath (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 6 (Observation)

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 $\boxed{869}$.

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

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