# Math Help - [SOLVED] calcul riddle

1. ## [SOLVED] calcul riddle

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

can you help me please ?!

thx you!

2. C'est quoi la question?

3. il faut résoudre A

4. Solve for A?!
That makes no sense.

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 ?

6. Originally Posted by Bacterius
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.

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.

8. Originally Posted by Wilmer
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.

9. 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.

10. Originally Posted by Debsta
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.