MHB Compute a square root of a sum of two numbers

[anemone]

Compute $\sqrt{2000(2007)(2008)(2015)+784}$ without the help of calculator.

 

[MarkFL]

My solution:

Spoiler

$$2000\cdot2015=4030028-28$$

$$2007\cdot2008=4030028+28$$

$$784=28^2$$

Hence:

$$2000\cdot2007\cdot2008\cdot2015+784=4030028^2-28^2+28^2=4030028^2$$

And so:

$$\sqrt{2000\cdot2007\cdot2008\cdot2015+784}=\sqrt{4030028^2}=4030028$$

 

[anemone]

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]

Spoiler

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

 

[anemone]

Spoiler

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)

 

[agentredlum]

Spoiler

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.

:)

 

[anemone]

Spoiler

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)