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
Exponent factoring
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Source: Beat The GMAT — Data Sufficiency |
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logitech
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if it is wrong, I will not post the solutioncramya 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
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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logitech
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ltg356,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
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]
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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cramya
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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!
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!
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logitech
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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 ?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!
When are you taking the test by the way ?
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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cramya
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Lets say 9 billion,7 trillion and 25 gazillion was given how many zeros would u take off, seriously?Let's call 3^p = X and lets get rid of three zeros at the end
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)
