I have to find a little riddle , but i don't find a résulat.

can you help me please ?!

http://i33.tinypic.com/jtt5ad.jpg

:)

thx you!

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- Nov 6th 2009, 01:19 PMjadee[SOLVED] calcul riddle
I have to find a little riddle , but i don't find a résulat.

can you help me please ?!

http://i33.tinypic.com/jtt5ad.jpg

:)

thx you! - Nov 6th 2009, 01:23 PMWilmer
C'est quoi la question?

- Nov 6th 2009, 01:25 PMjadee
il faut résoudre A

- Nov 6th 2009, 01:28 PMWilmer
Solve for A?!

That makes no sense. - Nov 6th 2009, 01:38 PMBacteriusReply
Tu ne peux pas trouver un résultat sans avoir une question ... quelle est-elle donc ?

You can't find a solution without having a question ... so what is it ? - Nov 6th 2009, 01:48 PMmr fantastic
- Nov 6th 2009, 01:54 PMBacteriusReply
I'll give a clue (don't have time either) :

$\displaystyle 4^3 = 8^2$

Which makes me think this expression might simplify to a constant. - Nov 6th 2009, 01:55 PMDefunkt
It does make sense, actually; it is simply a hint that tells us the result is constant:

1. $\displaystyle (8^{n+1} + 8^n)^2 = ((2^3)^{n+1} + (2^3)^n)^2 = (2^{3n+3} + 2^{3n})^2 = (2^{3n}(1 + 8))^2 = \boxed{2^{6n}\cdot 81}$

2. $\displaystyle (4^{n}-4^{n-1})^3 = ((2^2)^n - (2^2)^{n-1})^3 = (2^{2n}-2^{2n-2})^3 = (2^{2n}(1 - \frac{1}{4}))^3 = \boxed{2^{6n}(\frac{27}{64})}$

Therefore $\displaystyle A = \frac{2^{6n}\cdot 81}{2^{6n}\cdot \frac{27}{64}} = 81 \cdot \frac{64}{27} = 3 \cdot 64 = 192$

Therefore $\displaystyle \boxed{A \equiv 192}$

You can plug in any value of $\displaystyle n\in \mathbb{N}, n >1$ and you will see that this is what you'll get. - Nov 6th 2009, 01:56 PMDebsta
Yes you've got to simplify the expression and you will find all the n's evenyuallu cancel out. I got 216 as the value for A.

Start by writing 8 as 2^3 and 4 as 2^2.

Then use your index laws and factorising.

A few hints: (2^3)^(n+1) = 2^(3n+3) = 2^(3n) x 2^3.

Give it a try....I'll check in again later if you need more help. Good luck. - Nov 6th 2009, 01:59 PMDebsta