arithematic problem

[ms.cupcake]

6^3 x 6^4 x 6^5 x....6^(p-1) x 6^p = (1/6)^-900. what is p? help me please. im stuck since (1/6)^-900 is at infinity. thanks for advance.

 

[soroban]

Hello, ms.cupcake!

Solve for $\displaystyle p:\;6^3\cdot6^4\cdot6^5\:\cdots\,6^p \:=\: \left(\tfrac{1}{6}\right)^{-900}$

I'm stuck since (1/6)^-900 is at infinity. .What does that mean?

Note that: .$\displaystyle \left(\tfrac{1}{6}\right)^{-900} \:=\:\left(6^{-1}\right)^{-900} \:=\:6^{900}$


We have: .$\displaystyle 6^{3+4+5+\cdots+p} \:=\:6^{900}$

Hence: .$\displaystyle 3+4+5+\cdots+p \:=\:900$


Formula: .$\displaystyle 1+2+3+\cdots+p \:=\:\dfrac{p(p+1)}{2}$
The left side is: .$\displaystyle \dfrac{p(p+1)}{2} - 1 -2$


The equation becomes: .$\displaystyle \dfrac{p(p+1)}{2} - 3 \:=\:900$

. . $\displaystyle \dfrac{p(p+1)}{2} \:=\:903 \quad\Rightarrow\quad p(p+1) \:=\:1806$

. . $\displaystyle p^2 + p \:=\:1806 \quad\Rightarrow\quad p^2 + p - 1806 \:=\:0$

. . $\displaystyle (p-42)(p+43) \:=\:0 \quad\Rightarrow\quad p \:=\:42,\:\color{red}{\rlap{///}}\text{-}43$


Answer: .$\displaystyle p \;=\;42$

 

[HallsofIvy]

6^3 x 6^4 x 6^5 x....6^(p-1) x 6^p = (1/6)^-900. what is p? help me please. im stuck since (1/6)^-900 is at infinity. thanks for advance.

Your statement that "(1/6)^-900 is at infinity" suggests that you think the left side is an infinite product. That is not the case- has exactly p- 2 terms in it.

 

[ms.cupcake]

haha alright, I just realize that you add 1 and 2 on the right side, it will be exactly equal to 900. I thought it should be 3+4+5+... and so it will be equal to 900. thanks for your help.

 

[HallsofIvy]

So you want to find p such that 3+ 4+ 5+ ...+ p= 900. It will help to know that 1+ 2+ 3+ ...+ p= p(p+1)/2

 

[ms.cupcake]

well,i had try that 3+4+5+... =900 but it was in decimal so 1+2+3+.. will get perfect ans p=24 and it will be right.