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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.

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    $\begingroup$ another one from Brilliant.org $\endgroup$
    – lab bhattacharjee
    May 19, 2013 at 4:50

4 Answers 4

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\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}

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    $\begingroup$ For a symmetric (though fraction-riddled) version, use $x=29.5$. Conjugate polynomials $(x\pm \frac{1}{2})$ and $(x \pm \frac{3}{2})$ give differences of squares, which are convenient to expand. Ultimately, though, you get $\frac{1}{4}(59^2-5)$, which isn't quite as immediate a computation as $30^2-30+1$. $\endgroup$
    – Blue
    May 19, 2013 at 0:16
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    $\begingroup$ Wow, this answer was posted only 55 seconds after the question! $\endgroup$
    – Jonas Meyer
    May 19, 2013 at 0:33
  • $\begingroup$ Can you tell me how you can see the solution, or the way to solve this problem right away? $\endgroup$
    – Jeff
    May 19, 2013 at 0:35
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    $\begingroup$ @jeff More practice of problem-solving $\endgroup$
    – user67258
    May 19, 2013 at 0:41
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    $\begingroup$ Danny, posting one line and then editing the rest is somewhat frowned upon here. It contributes to the Fastest Gun In The West problem which prefers speed over quality. True, there are badges for getting the first answer right, but I don't think anyone really appreciates this behavior here. Finish your answer, then submit it. $\endgroup$
    – Asaf Karagila ♦
    May 19, 2013 at 0:51
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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.

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  • $\begingroup$ Nice interpretation! $\endgroup$
    – Calvin Lin
    Jun 15, 2013 at 0:59
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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 $869$.

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  • $\begingroup$ Using $(a+b)(a-b)=a^2-b^2$, the inside of the square root is $(29.5^2-0.5^2)(29.5^2-1.5^2)+1$, and it is easy to see that this is less than $29.5^4$, so the square root is less that $29.5^2$. $\endgroup$
    – Jonas Meyer
    May 19, 2013 at 4:29
  • $\begingroup$ @JonasMeyer: right you are-I meant $30$ not $29$, so you should be on the low side $\endgroup$
    – Ross Millikan
    May 19, 2013 at 4:36
  • $\begingroup$ Thank you. I still don't find that particularly convincing, but I think I might understand it now. $\endgroup$
    – Jonas Meyer
    May 19, 2013 at 4:41
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Hint: Use $(x)(x+1)(x+2)(x+3)+1 = (x)(x+3)(x+2)(x+1)+1 =(x^2+3x)(x^2+3x+2)+1= (x^2+3x)^2+2(x^2+3x)+1=(x^2+3x+1)^2$

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