2^1100???
Math brain teasers require computations to solve.
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)?
HintThe numbers are too big to deal with on a normal graphics calculator scale, think of it in terms of powers of ten.
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Answer
3^700 is greater, by approximately 710 times (711.0220569 to be exact).
To start, one must realize these numbers are too big to deal with by hand. So, one must use exponents and the "log" function on a calculator.
2^1100=10^a
2^x=10^1
log(10)/log(2)=x
3.321928095=x
1100/x=331.1329952
2^3.321928095=10^1
2^(331.132995*3.321928095)=10^(331.132995*1)
2^1100=10^331.132995
Now that we know what 2^1100 equals, apply the same concept to 3^700.
3^700=10^b
3^y=10^1
log(10)/log(3)=y
2.095903274=y
3^2.095903274=10^1
3^(333.9848783*2.095903274)=10^(333.9848783*1)
3^700=10^333.9848783
So, now we have 2^1100 and 3^700 in terms of "ten to the power of".
2^1100=10^331.1379952
3^700=10^333.9848783
Obviously, 3^700 is greater. To figure out how much greater, requires a little more work, but not much.
First, subtract the exponents.
333.9848783-331.1379952
This equals 2.851883073
Then, take 10^2.851883073
This equals 711.0220569, which is around 710.
Thus, 3^700 is approximately 710 times greater than 2^1100. Congratulations, you passed your professor's exam.
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Comments
monkeygrl99   
Jul 23, 2006
| Way hard, but cool.
Yay! First one! |
stil 
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? --) |
HiImDavid
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
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. |
HiImDavid
Jul 26, 2006
| Wow, that's awesome. I never would have thought to do it like that. |
monkeygrl99
Jul 27, 2006
| Yeah |
cnmne
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 |
leftclick
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  |
leftclick
Jul 29, 2007
| Obviously my calculator has less significant digits...  |
kwelchans
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 |
kwelchans
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
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
Dec 30, 2008
| I of course meant 3^70 is slightly more than 2^110, not the other way around. |
javaguru
Dec 30, 2008
| Arrgh. One more time: 3^70 is slightly less than twice 2^110.  |
javaguru
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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