[SOLVED] calcul riddle

jadee · #1 $Old$ November 6th, 2009, 03:19 PM

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

can you help me please ?!

thx you!

Wilmer · #2 $Old$ November 6th, 2009, 03:23 PM

C'est quoi la question?

jadee · #3 $Old$ November 6th, 2009, 03:25 PM

il faut résoudre A

Wilmer · #4 $Old$ November 6th, 2009, 03:28 PM

Solve for A?!
That makes no sense.

Bacterius · #5 $Old$ November 6th, 2009, 03:38 PM

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 ?

mr fantastic · #6 $Old$ November 6th, 2009, 03:48 PM

Quote:

Originally Posted by Bacterius $View Post$

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 ?

The question probably is to simplify the given expression. I don't have time but someone else might chip in.

Bacterius · #7 $Old$ November 6th, 2009, 03:54 PM

I'll give a clue (don't have time either) :

$4^3 = 8^2$

Which makes me think this expression might simplify to a constant.

Defunkt · #8 $Old$ November 6th, 2009, 03:55 PM

Quote:

Originally Posted by Wilmer $View Post$

Solve for A?!
That makes no sense.

It does make sense, actually; it is simply a hint that tells us the result is constant:

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

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

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

Therefore $\boxed{A \equiv 192}$

You can plug in any value of $n\in \mathbb{N}, n >1$ and you will see that this is what you'll get.

Debsta · #9 $Old$ November 6th, 2009, 03:56 PM

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.

Debsta · #10 $Old$ November 6th, 2009, 03:59 PM

Quote:

Originally Posted by Debsta $View Post$

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.

Sorry - did the calculation at the end wrong. The answer is 192.