### Brain Teasers

# 2^1100???

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)?

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

### Hint

The numbers are too big to deal with on a normal graphics calculator scale, think of it in terms of powers of ten.### 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.

Hide Hint Show Hint Hide Answer Show Answer

## What Next?

**Solve a Similar Brain Teaser...**

Or, get a random brain teaser.

If you become a registered user you can vote on this brain teaser, keep track of

which ones you have seen, and even make your own.

## Comments

Way hard, but cool.

Yay! First one!

Yay! First one!

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? --)

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? --)

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.

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

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.

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.

Wow, that's awesome. I never would have thought to do it like that.

Yeah

3^700 = (3^7)^100 = 2187^100

2^1100 = (2^11)^100 = 2048^100

(2187/204^100 = 1.06787109375^100 = 711.0220569369852982035305300795

2^1100 = (2^11)^100 = 2048^100

(2187/204^100 = 1.06787109375^100 = 711.0220569369852982035305300795

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

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

Obviously my calculator has less significant digits...

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

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

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

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.

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.

I of course meant 3^70 is slightly more than 2^110, not the other way around.

Arrgh. One more time: 3^70 is slightly less than twice 2^110.

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.)

## Follow Braingle!