I am getting B) Whats the OA? If its right I will post the solution

Please post OA'S using spoiler function when posting the question if possible

cramya wrote:I am getting B) Whats the OA? If its right I will post the solution

Please post OA'S using spoiler function when posting the question if possible

if it is wrong, I will not post the solution

ltg356 wrote:Is 3^(p-2) > 1000
1) 3^(p+1) < 54000
2) 3^p < 3^(p-1) + 2000


What's the most efficient way to approach this problem?
Thanks

ltg356,

Since you are asking the most efficient way, I assume that you already solved this question and you are trying to find a faster/better solution.

I really do not trade my solutions for official answers ( it is just funny ) , bit please keep in mind to put the OA with spoiler function in the future.

Going back to question:

Is 3^(p-2) > 1000 ?

You can rephrase this to : Is 3^p > 9000 ?

Here is the trick..do not carry on this calculation and try to find P , leave it as it is because both statement are using the same expression. Let's call 3^p = X and lets get rid of three zeros at the end

so Is X > 9

1) 3^(p+1) < 54000

means : X < 54/3

X<18 INSUF - something less than 18 does not need to be bigger than 9


2) 3^p < 3^(p-1) + 2000

X<X/3 + 2

2x/3 < 2

X < 3

Tatamm!! Sufficient!

Hence, B

I hope you find this method useful.


[/b]

Logitech,
If what you (speaking in general) have said is correct then there is a point explaining it. If you are wrong to begin with then its best to observe and make notes of other's solutions and make make sure u dont commit the same mistake again



By the way nice approach on making 3 ^ p =x. I did it just by using 3 ^ p ;like I said not all minds are alike.

Good luck!

cramya wrote:Logitech,
If what you (speaking in general) have said is correct then there is a point explaining it. If you are wrong to begin with then its best to observe and make notes of other's solutions and make make sure u dont commit the same mistake again



By the way nice approach on making 3 ^ p =x. I did it just by using 3 ^ p ;like I said not all minds are alike.

Good luck!

I am just teasing you Cramya, I really enjoy having you here and learning from you. But we have to make this place a little bit fun right ?

When are you taking the test by the way ?

Let's call 3^p = X and lets get rid of three zeros at the end

Lets say 9 billion,7 trillion and 25 gazillion was given how many zeros would u take off, seriously?

Jus kidding!

I am explaining my solution in detail since we have had one more vote for B.

To find 3 ^ p * 3 ^ -2 > 1000 i.e 3 ^ p / 9 > 1000 or 3 ^ p > 9000
( a ^ (b+c) = a ^ b * a ^ c where a=3 b=p c=-2 here AND a ^ -m = 1 /a ^ m 3 ^ -2 = 1 / 3 ^ 2 i.e 1/9)


Stmt I)

3^(p+1) < 54000
3 ^ p . 3 ^1 < 54000
3^p < 18000

Could be greater than 9000 or less dont know INSUFF

Stmt II)

3 ^ p < 3^(p-1) + 2000 ( a ^ (b+c) = a ^ b * a ^ c where a=3 b=p c=-1 here and a ^ -m = 1 /a ^ m 3 ^ -1 = 1 / 3 ^ 1 i.e 1/3)

3^p < 3 ^ p /3 + 2000
3 ^ p < 3 ^p + 6000 / 3
3 ^ (p+1) < 3^p + 6000
3 ^ (p+1) - 3 ^ p < 6000
3 ^P 3 ^ 1 - 3 6 P < 6000
3 ^ p * (3-1) < 6000
3 ^ p < 3000

SUFF

B)

sorry mates, I meant to tell you it is B.
My inefficient method involves calculating the roots of 3... too much room for error

My answer was B- I used Cramya's method. DEFINITELY more efficient than calculating roots of 3.

Horray consensus![/i]