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More ways to get Braingle...

2^1100???

Math brain teasers require computations to solve.

 

Puzzle ID:#31996
Fun:*** (2.31)
Difficulty:*** (2.97)
Category:Math
Submitted By:HiImDavidAus***!!!

 

 

 



Your professor prepared a rigorous 150-question final exam to be completed in under 30 minutes, but unfortunately, he seems to have misplaced it, and can only remember one problem. So, he has decided to waive your final exam, if you can answer the one question he can remember from his original exam.

Which is greater, 2^1100 or 3^700? Approximately how many times greater is it (to the nearest 10)?

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Comments

monkeygrl99Aus*
Jul 23, 2006

Way hard, but cool.

Yay! First one!
stil*us*
Jul 23, 2006

You take the long road and I'll take the short road and I'll get an answer afore ye...
How about -
3=2^x
x=log(3)/log(2)
x=0.477121/0.301030
x=1.58496
3=2^1.58496
Then,
3^700=(2^1.58496)^700=2^1109.47
The second value is 2^9.47, i.e. 709 point something, times the first. (Keep in mind that this teaser's talk of 711 point whatever being exact isn't absolutely true. It's simply more precise and carries a lot more decimal places than one would find in a common carry-it-to-class log chart. -- Do they still print such things? --)
HiImDavidAus*
Jul 23, 2006

I am not sure, I had this problem on my final exam for my AP Calculus class, and I thought it was a good problem. I used a graphic calculator to solve it, so I didn't need a log chart. Your approach works though, because to the closest ten 3^700 is still 710 greater than 2^1100.
galaxynova
Jul 25, 2006

okay, apparently it wasn't created for someone not even in hs yet...i'll rate it when i actually understand the concept.
stil*us*
Jul 26, 2006

Also-
3^700=
(2*1.5)^700=
2^700*1.5^700=
2^700*(1.5^2)^350=
2^700*2.25^350=
2^700*(2*1.125)^350=
2^700*2^350*1.125^350=
2^1050*1.125^350=
2^1050*(1.125^7)^50=
2^1050*(2.280697345733642578125)^50=
2^1050*(2*1.1403486728668212890625)^50=
2^1100*1.1403486728668212890625^50=
This proves precise figures; my calculator renders the second factor as 711.0220569369852982035305300795.
HiImDavidAus*
Jul 26, 2006

Wow, that's awesome. I never would have thought to do it like that.
monkeygrl99Aus*
Jul 27, 2006

Yeah
cnmne*us*
Aug 03, 2006

3^700 = (3^7)^100 = 2187^100
2^1100 = (2^11)^100 = 2048^100
(2187/204^100 = 1.06787109375^100 = 711.0220569369852982035305300795
leftclickAau*
Jul 29, 2007

3^700 / 2^1100 =
3^700 / (2^(1100/700))^700 =
(3 / (2^(1100/700))) ^ 700 =
1.009425144 ^ 700 =
711.022056937

I like cnme's method as well, the rest are a bit convoluted imho
leftclickAau*
Jul 29, 2007

Obviously my calculator has less significant digits...
kwelchansAus*
Apr 17, 2008

Here's how I did it:
a^b=e^(b*ln(a))
So:
2^1100=e^(1100*ln(2)) and
3^700=e^(700*ln(3))
1100*ln(2)=762.46
700*ln(3)=769.03
e^769.03>e^762.46, so
3^700>2^1100
(3^700)/(2^1100)=
(e^769.03)/(e^762.46)=
e^(769.03-762.46)=
e^6.57=
713.37, which is approximately 710
kwelchansAus*
Apr 17, 2008

By the way, in order to complete a 150-question exam in 30 minutes, you get 12 seconds per question. Tough exam!
javaguru*us*
Dec 30, 2008

I did it the same way as cnme, except with some shortcut approximations. In my head in less than 30 seconds I had the rough answer that 3^700 was somewhat less than 1000 times as large.

How in my head?

I know that 2^11 is 2048 (any computer programmer should know this). I also know that 3^6 = 729, so 2187 is easy to calculate as 3^7 from there. So within 5 seconds I know that 3^700 is bigger.

But how much bigger?

The difference between 2187 and 2048 is around 140. 14^2 = 198, so 14/205 is somewhat less than 1/14 which makes 3^7 somewhat less than (15/14)*2^11.

This is convenient because a rough approximation of the 5*r root of 2 is (r*7+1)/(r*7) (i.e. 2^(1/5) ~ 8/7; 2^(1/10) ~ 15/14, 2^(1/15) ~ = 22/21, ...) which makes 15/14 a good approximation of the 10th root of 2.

So now I know that 1 + 14/205 is probably a slight underestimate of the 10th root of 2. This means that 2^110 is a little less than 2 times larger than 3^70. A little less than 2^10 (to get from 2^110 to 2^1100, etc.) is somewhat less than 1000.

Since the teaser only needed the answer within an order of magnitude, 1000 was close enough. I was estimating that the actual answer was between 800 and 900, so I wasn't too far off.
javaguru*us*
Dec 30, 2008

I of course meant 3^70 is slightly more than 2^110, not the other way around.
javaguru*us*
Dec 30, 2008

Arrgh. One more time: 3^70 is slightly less than twice 2^110.
javaguru*us*
Dec 30, 2008

Sheesh. And 14^2 is 196, not 198. (I actually made the same mistake when I did it in my head, but it's insignificant for the accuracy I was shooting for.)



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