View source for 1989 AIME Problems/Problem 1

== Problem ==
Compute <math>\sqrt{(31)(30)(29)(28)+1}</math>.

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

~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:
<cmath>\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*}</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>
~Novus677 ~MRENTHUSIASM

== Solution 4 (Symmetry with Generalization) ==
Similar to Solution 1 above, call the consecutive integers <math>\left(n-\frac{3}{2}\right), \left(n-\frac{1}{2}\right), \left(n+\frac{1}{2}\right), \left(n+\frac{3}{2}\right)</math> to make use of symmetry.  Note that <math>n</math> itself is not an integer - in this case, <math>n = 29.5</math>.  The expression becomes <math>\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}</math>.  Distributing each pair of difference of squares first, and then distributing the two resulting quadratics and adding the constant, gives <math>\sqrt{n^4 - \frac{5}{2}n^2 + \frac{25}{16}}</math>.  The inside is a perfect square trinomial, since <math>b^2 = 4ac</math>.  It's equal to <math>\sqrt{\left(n^2 - \frac{5}{4}\right)^2}</math>, which simplifies to <math>n^2 - \frac{5}{4}</math>.  You can plug in the value of <math>n</math> from there, or further simplify to <math>\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) - 1</math>, which is easier to compute.  Either way, plugging in <math>n=29.5</math> gives <math>\boxed{869}</math>.

== Solution 5 (Prime Factorization) ==
We have <math>(31)(30)(29)(28)+1=755161.</math> Since the alternating sum of the digits <math>7-5+5-1+6-1=11</math> is divisible by <math>11,</math> we conclude that <math>755161</math> is divisible by <math>11.</math>

We evaluate the original expression by prime factorization:
<cmath>\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*}</cmath>
~Vrjmath ~MRENTHUSIASM

== Solution 6 (Observation) ==
The last digit under the radical is <math>1</math>, so the square root must either end in <math>1</math> or <math>9</math>, since <math>x^2  = 1\pmod {10}</math> means <math>x = \pm 1</math>.  Additionally, the number must be near <math>29 \cdot 30 = 870</math>, narrowing the reasonable choices to <math>869</math> and <math>871</math>.

Continuing the logic, the next-to-last digit under the radical is the same as the last digit of <math>28 \cdot 29 \cdot 3 \cdot 31</math>, which is <math>6</math>.  Quick computation shows that <math>869^2</math> ends in <math>61</math>, while <math>871^2</math> ends in <math>41</math>.  Thus, the answer is <math>\boxed{869}</math>.

==Solution 7 (Brute force)==
<math>31 \times 30 \times 29 \times 28</math> is <math>755160</math>. <math>755160 + 1=755161</math>, and <math>\sqrt{755161}=869</math>. So the answer is <math>\boxed{869}</math>.

~[[shunyipanda]]

== See also ==
{{AIME box|year=1989|before=First Question|num-a=2}}

[[Category:Intermediate Algebra Problems]]
{{MAA Notice}}