Calculating $\sqrt{28\cdot 29 \cdot 30\cdot 31+1}$ Is it possible to calculate $\sqrt{28 \cdot 29 \cdot 30 \cdot 31 +1}$ without any kind of electronic aid?
I tried to factor it using equations like $(x+y)^2=x^2+2xy+y^2$ but it didn't work.
 A: If you are willing to rely on the problem setter to make sure it is a natural, it has to  be close to $29.5^2=841+29+.25=870.25$  The one's digit of the stuff under the square root sign is $1$, so it is either $869$ or $871$.  You can either calculate and check, or note that two of the factors are below $30$ and only one above, which should convince you it is $869$.
A: \begin{align}
&\text{Let }x=30
\\ \\
\therefore&\ \ \ \ \ \sqrt{(x-2)(x-1)x(x+1)+1}
\\ \\
&=\sqrt{[(x-2)(x+1)[(x-1)x]+1}
\\ \\
&=\sqrt{(x^2-x-2)((x^2-x)+1}
\\ \\
&=\sqrt{(x^2-x)^2-2(x^2-x)+1}
\\ \\
&=\sqrt{(x^2-x-1)^2}
\\
&=x^2-x-1
\\
&=30^2-30-1
\\
&=\boxed{869}
\end{align}
A: One notes in base 28, that the numbers are 1.0, 1.1, 1.2, 1.3.  The product of the inner and outer pairs are 1.3.0 and 1.3.2.  Adding one to this product gives $1.3.1^2$, whence the square root is 1.3.1.  
Evaluating this in decimal gives 869.
When this series starts with 35, the result is the cube 1331.
A: Hint: Use $(x)(x+1)(x+2)(x+3)+1 = (x)(x+3)(x+2)(x+1)+1 =(x^2+3x)(x^2+3x+2)+1= (x^2+3x)^2+2(x^2+3x)+1=(x^2+3x+1)^2$
