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Basically what you have to do is find √3969 without using complex /long calculations

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closed as off-topic by APrough, boboquack, Mithrandir, Volatility, Beastly Gerbil Nov 16 '17 at 21:34

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  • $\begingroup$ someone please tell me why am i getting - on Que? $\endgroup$ – James Muir Nov 16 '17 at 20:24
  • $\begingroup$ I don't know, but I'd guess part of it is not using the right tags (there's a "no-computers" tag and I'm not sure "strategy" fits), and partly it being a boring math question rather than a puzzle. (I didn't downvote, btw) $\endgroup$ – Bobson Nov 16 '17 at 20:50
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We know it's equal to the square root of 27 times the square root of 147.

The square root of 27 is between 5 and 6, closer to 5 than to 6.

The square root of 147 is between 12 and 13, much closer to 12 than to 13.

So it's between 5*12=60 and 6*13=78, but closer to 60 than 78, so really we know it's between 60 and 69.

The last digit of 3969 is 9, so the last digit of the square root has to, when you square it, give a number whose last digit is 9.

The only single digits like that are 3 and 7.

Therefore, if the square root is a whole number, it must be either 63 or 67.

But if it was 67, when you multiply the 60 by the 7 you carry a 4, and 60 * 60 =3600, with a carry of 4 gives 4000, which puts you over the 3969.

So it must be 63 if it's a whole number.

Edit: The line eliminating 67 was wrong, fixed by jeff-zeitlin.

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  • $\begingroup$ This is the only solution I've seen that I'd say qualifies as "no complex calculation". $\endgroup$ – Jeff Zeitlin Nov 16 '17 at 20:12
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    $\begingroup$ Also, the carry problem is when you multiply the 60 by the 7; that carries a 4, and 60 * 60 =3600, with a carry of 4 gives 4000, which puts you over the 3969. $\endgroup$ – Jeff Zeitlin Nov 16 '17 at 20:18
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I don't really know what counts as "complex" or "long", but I think the following is simpler than the other solutions posted so far.

Both numbers are multiples of 3, so let's begin by dividing both by 3. (Then we'll need to multiply the square-root-of-product by 3 when we're done.) We get 9 and 49. We recognize those as the squares of 3 and 7. So we have 3x3x7=63. Done.

[EDITED to add:] Actually, this is pretty similar to Michael Seifert's answer. I think my version is infinitesimally easier, though.

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  • $\begingroup$ yeah it is really good $\endgroup$ – James Muir Nov 16 '17 at 20:45
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27 * 147 = 27 * 3 * 49 = 81 * 49 = 92 * 72. So √(27 * 147) = 9 * 7 = 63.

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  • $\begingroup$ This is basically the same solution as pycoder's answer, but factoring 147 directly rather than without the intermediate steps. $\endgroup$ – Michael Seifert Nov 16 '17 at 19:54
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$$27 \cdot 147 = 3^3 \cdot (120 + 27) = 3^3 \cdot (3 \cdot 40 + 3 \cdot 9) = 3^3 \cdot 3 \cdot 49 = 3^4 \cdot 7^2$$
$$\sqrt{3^4 \cdot 7^2} = 3^2 \cdot 7^1 = 9 \cdot 7 = 63$$

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  • $\begingroup$ i get lost in this one but I belive you +1 $\endgroup$ – James Muir Nov 16 '17 at 19:59

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