2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 = ... ?

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2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 = ... ?

by Fractal » Sun Jul 03, 2011 4:59 am
2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 = ... ?

Is there a simple method to calculate a multiplication such the one stated above?

Thx,
fractal

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by winniethepooh » Sun Jul 03, 2011 5:42 am
This question will never be asked at the Gmat, and incase it is asked, then they will ask it is near which power of 10 or similar questions!
So, don't worry!

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by Fractal » Sun Jul 03, 2011 6:14 am
You are right! The question asks how near the result is to wich power of 10.

so how are you handling such questions?

Thx a lot!

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by goalevan » Sun Jul 03, 2011 2:32 pm
Here's a quick way to calculate utilizing difference of squares formula:

2 * 3 * 5 * 7 * 11 * 13 * 17 * 19

(2 * 5) * (3 * 7) * (11 * 19) * (13 * 17)

(10 * 21) * (15 - 4)(15 + 4) * (15 - 2)(15 + 2)

(210) * (15^2 - 4^2) * (15^2 - 2^2)

210 * 209 * 221

The answer will be slightly above 200^3, or (2*10*10)^3 = 8 * 10^6. This brings the answer very close to 10^7.

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by kevincanspain » Sun Jul 03, 2011 3:04 pm
Fractal wrote:2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 = ... ?

Is there a simple method to calculate a multiplication such the one stated above?

Thx,
fractal
Also, you can regroup and approximate!

5 * 19 approx 10^2
2*3*17 approx 10^2
7*13*11 approx 10^3

total 10^7 approx
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