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Compute a square root of a sum of two numbers

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,678
Compute $\sqrt{2000(2007)(2008)(2015)+784}$ without the help of calculator.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
My solution:

\(\displaystyle 2000\cdot2015=4030028-28\)

\(\displaystyle 2007\cdot2008=4030028+28\)

\(\displaystyle 784=28^2\)

Hence:

\(\displaystyle 2000\cdot2007\cdot2008\cdot2015+784=4030028^2-28^2+28^2=4030028^2\)

And so:

\(\displaystyle \sqrt{2000\cdot2007\cdot2008\cdot2015+784}=\sqrt{4030028^2}=4030028\)
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,678
This is probably the most genius way to collect the four numbers $2000$, $2007$, $2008$ and $2015$ in such a manner so that their product takes the form $(a+b)(a-b)$ and what's more, $b^2=784$!

Bravo, MarkFL!(Clapping)(Sun)
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
Letting 2000 = a
we have 2000 * 2007 *2008 * 2015 + 784
a(a+7)(a+8)(a+15) + 784
= a(a+15) (a+7) (a+8) + 784
= (a^2+15a) (a^2 + 15a + 56) + 784
= $(a^2 + 15 a + 28 - 28 ) (a^2 + 15a + 28 + 28) + 28^2$
= $(a^2 + 15 a + 28)^2 - 28^2 + 28^2$
= $(a^2 + 15 a + 28)^2$
hence square root is $a^2 + 15a + 28$ or 4000000 + 30000 + 28
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,678
Letting 2000 = a
we have 2000 * 2007 *2008 * 2015 + 784
a(a+7)(a+8)(a+15) + 784
= a(a+15) (a+7) (a+8) + 784
= (a^2+15a) (a^2 + 15a + 56) + 784
= $(a^2 + 15 a + 28 - 28 ) (a^2 + 15a + 28 + 28) + 28^2$
= $(a^2 + 15 a + 28)^2 - 28^2 + 28^2$
= $(a^2 + 15 a + 28)^2$
hence square root is $a^2 + 15a + 28$ or 4000000 + 30000 + 28
Hey kaliprasad, thanks for participating and your method is good and I'm particularly very happy to see you finally picking up on LaTeX!(Tongueout)(Sun)
 

agentmulder

Active member
Feb 9, 2012
33
Almost identical to MarkFL's , i just factored 16 to make the multiplications a bit smaller.

$ \sqrt{(2000)(2007)(2008)(2015)+ 784} \ = $

$ \sqrt{16 [ (\frac{2000}{4})(2007) ( \frac{2008}{4} ) (2015) + \frac{16 \cdot 7^2}{16} ]} \ = $

$ 4 \sqrt{(500)(2015)(502)(2007) \ + \ 7^2} \ = $

$ 4 \sqrt{(1007507 -7)(1007507 + 7) + 7^2} \ = $

$ 4 \sqrt{(1007507)^2} \ = \ 4030028 $

Admittedly , had i not seen MarkFL's method i probably would not have discovered it on my own.

:)

 
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anemone

MHB POTW Director
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Feb 14, 2012
3,678
Almost identical to MarkFL's , i just factored 16 to make the multiplications a bit smaller.

$ \sqrt{(2000)(2007)(2008)(2015)+ 784} \ = $

$ \sqrt{16 [ (\frac{2000}{4})(2007) ( \frac{2008}{4} ) (2015) + \frac{16 \cdot 7^2}{16} ]} \ = $

$ 4 \sqrt{(500)(2015)(502)(2007) \ + \ 7^2} \ = $

$ 4 \sqrt{(1007507 -7)(1007507 + 7) + 7^2} \ = $

$ 4 \sqrt{(1007507)^2} \ = \ 4030028 $

Admittedly , had i not seen MarkFL's method i probably would not have discovered it on my own.

:)

Nice one, agentmulder!:)(Sun)