Fermi Questions C

[jonboyage]

Does anyone know any good ways to do factorials? I've been having problems estimating them, and memorization isn't helping.

I looked it up recently and Stirling’s Approximation seems to be good:

[Unome]

Does anyone know any good ways to do factorials? I've been having problems estimating them, and memorization isn't helping.

I looked it up recently and Stirling’s Approximation seems to be good:

That seems to be a really good approximation... but how are you supposed to use that practically?

[whythelongface]

(n/e)^n can get a bit nasty if the numbers are large but you can use the logarithms method to find that.

[shamu111]

Does anyone have a good way to calculate (1.03)^2500? There was a question involving this on our invitational test, and my partner and I were wayyy off.

I have tried using the binomial expansion, solving for the index of the largest term, and then using Stirlings to solve for the coefficients, but it seems too tedious.
I've also tried using 1.024 = 1024/100 = 2^10/10^3 and 1.05 = 21/20 = 3*7/(4*5) which aren't too bad to compute using logs to bound it. However this gives 26 and 53 respectively while the answer is 32. If log13 (which i cba to memorize) can be used then 1.04 = 104/100 = 26/25 = 2*13/5^2 can be done with logs, but this gives 42 which is pretty far off too. Averaging 42 (from 1.04) and 26 (1.024) and then shifting the answer down to account for the for the latter being closer to 1.03 seems pretty good tho.

Can someone give me a method aside from memorizing logs up to 103?

Apart from this question I also think that for smaller powers there isn't much need to memorize logs if you can remember certain exponents:
2^n = (2^10)^(n/10) = (1024)^(n/10) = (1000)^(n/10) = 10^(3n/10)
4^n = (2^2)^n = 10^(3n/5)
5^n = (10/2)^n = 10^n/2^n = 10^n/2^(3n/10) = 10^(7n/10)
7^n = 49^(n/2) = 50^(n/2) = (5*10)^(n/2) = 5^(n/2) * 10^(n/2) = 10^(7n/20) * 10^(10n/20) = 10^(17n/20)
8^n = (2^3)^n = 10^(9n/10)
e^n = (e^3)^(n/3) = (20)^(n/3) = (2*10)^(n/3) = 2^(n/3) * 10^(n/3) = 10^(n/10) * 10^(n/3) = 10^(13n/30)
pi^n = (pi^4)^(n/4) = (100)^(n/4) = (10^2)^(n/4) = 10^(n/2)

3^n = 9^(n/2) = 10^(n/2) however this one sucks and stops working after even just 3^14

[Name]

Does anyone have a good way to calculate (1.03)^2500? There was a question involving this on our invitational test, and my partner and I were wayyy off.

I have tried using the binomial expansion, solving for the index of the largest term, and then using Stirlings to solve for the coefficients, but it seems too tedious.
I've also tried using 1.024 = 1024/100 = 2^10/10^3 and 1.05 = 21/20 = 3*7/(4*5) which aren't too bad to compute using logs to bound it. However this gives 26 and 53 respectively while the answer is 32. If log13 (which i cba to memorize) can be used then 1.04 = 104/100 = 26/25 = 2*13/5^2 can be done with logs, but this gives 42 which is pretty far off too. Averaging 42 (from 1.04) and 26 (1.024) and then shifting the answer down to account for the for the latter being closer to 1.03 seems pretty good tho.

Can someone give me a method aside from memorizing logs up to 103?

Apart from this question I also think that for smaller powers there isn't much need to memorize logs if you can remember certain exponents:
2^n = (2^10)^(n/10) = (1024)^(n/10) = (1000)^(n/10) = 10^(3n/10)
4^n = (2^2)^n = 10^(3n/5)
5^n = (10/2)^n = 10^n/2^n = 10^n/2^(3n/10) = 10^(7n/10)
7^n = 49^(n/2) = 50^(n/2) = (5*10)^(n/2) = 5^(n/2) * 10^(n/2) = 10^(7n/20) * 10^(10n/20) = 10^(17n/20)
8^n = (2^3)^n = 10^(9n/10)
e^n = (e^3)^(n/3) = (20)^(n/3) = (2*10)^(n/3) = 2^(n/3) * 10^(n/3) = 10^(n/10) * 10^(n/3) = 10^(13n/30)
pi^n = (pi^4)^(n/4) = (100)^(n/4) = (10^2)^(n/4) = 10^(n/2)

3^n = 9^(n/2) = 10^(n/2) however this one sucks and stops working after even just 3^14

It seems like every decimal increase in the thousands increases the Fermi answer by about one. So 1.03^2500 is 32 and 1.031 is 33. With that find 1.04^2500 and decrease final answer by 10

[Name]

Does anyone have a good way to calculate (1.03)^2500? There was a question involving this on our invitational test, and my partner and I were wayyy off.

I have tried using the binomial expansion, solving for the index of the largest term, and then using Stirlings to solve for the coefficients, but it seems too tedious.
I've also tried using 1.024 = 1024/100 = 2^10/10^3 and 1.05 = 21/20 = 3*7/(4*5) which aren't too bad to compute using logs to bound it. However this gives 26 and 53 respectively while the answer is 32. If log13 (which i cba to memorize) can be used then 1.04 = 104/100 = 26/25 = 2*13/5^2 can be done with logs, but this gives 42 which is pretty far off too. Averaging 42 (from 1.04) and 26 (1.024) and then shifting the answer down to account for the for the latter being closer to 1.03 seems pretty good tho.

Can someone give me a method aside from memorizing logs up to 103?

Apart from this question I also think that for smaller powers there isn't much need to memorize logs if you can remember certain exponents:
2^n = (2^10)^(n/10) = (1024)^(n/10) = (1000)^(n/10) = 10^(3n/10)
4^n = (2^2)^n = 10^(3n/5)
5^n = (10/2)^n = 10^n/2^n = 10^n/2^(3n/10) = 10^(7n/10)
7^n = 49^(n/2) = 50^(n/2) = (5*10)^(n/2) = 5^(n/2) * 10^(n/2) = 10^(7n/20) * 10^(10n/20) = 10^(17n/20)
8^n = (2^3)^n = 10^(9n/10)
e^n = (e^3)^(n/3) = (20)^(n/3) = (2*10)^(n/3) = 2^(n/3) * 10^(n/3) = 10^(n/10) * 10^(n/3) = 10^(13n/30)
pi^n = (pi^4)^(n/4) = (100)^(n/4) = (10^2)^(n/4) = 10^(n/2)

3^n = 9^(n/2) = 10^(n/2) however this one sucks and stops working after even just 3^14

It seems like every decimal increase in the thousands increases the Fermi answer by about one. So 1.03^2500 is 32 and 1.031 is 33. With that find 1.04^2500 and decrease final answer by 10

Clarification that's only if the exponent is in the thousands. If it's in the hundreds then the 1 number increase would apply for the hundreds place etc

[Name]

What good resources are there besides the invite tests in the test exchange and the website

[PM2017]

What good resources are there besides the invite tests in the test exchange and the website

One good resource is this Wikipedia page: https://en.wikipedia.org/wiki/Orders_of ... _(numbers)

[Unome]

Does anyone have a good way to calculate (1.03)^2500? There was a question involving this on our invitational test, and my partner and I were wayyy off.

I have tried using the binomial expansion, solving for the index of the largest term, and then using Stirlings to solve for the coefficients, but it seems too tedious.
I've also tried using 1.024 = 1024/100 = 2^10/10^3 and 1.05 = 21/20 = 3*7/(4*5) which aren't too bad to compute using logs to bound it. However this gives 26 and 53 respectively while the answer is 32. If log13 (which i cba to memorize) can be used then 1.04 = 104/100 = 26/25 = 2*13/5^2 can be done with logs, but this gives 42 which is pretty far off too. Averaging 42 (from 1.04) and 26 (1.024) and then shifting the answer down to account for the for the latter being closer to 1.03 seems pretty good tho.

Can someone give me a method aside from memorizing logs up to 103?

Apart from this question I also think that for smaller powers there isn't much need to memorize logs if you can remember certain exponents:
2^n = (2^10)^(n/10) = (1024)^(n/10) = (1000)^(n/10) = 10^(3n/10)
4^n = (2^2)^n = 10^(3n/5)
5^n = (10/2)^n = 10^n/2^n = 10^n/2^(3n/10) = 10^(7n/10)
7^n = 49^(n/2) = 50^(n/2) = (5*10)^(n/2) = 5^(n/2) * 10^(n/2) = 10^(7n/20) * 10^(10n/20) = 10^(17n/20)
8^n = (2^3)^n = 10^(9n/10)
e^n = (e^3)^(n/3) = (20)^(n/3) = (2*10)^(n/3) = 2^(n/3) * 10^(n/3) = 10^(n/10) * 10^(n/3) = 10^(13n/30)
pi^n = (pi^4)^(n/4) = (100)^(n/4) = (10^2)^(n/4) = 10^(n/2)

3^n = 9^(n/2) = 10^(n/2) however this one sucks and stops working after even just 3^14

I made an attempt, but my heuristic was off - ended up with 28.

[PM2017]

Does anyone have a good way to calculate (1.03)^2500? There was a question involving this on our invitational test, and my partner and I were wayyy off.

I have tried using the binomial expansion, solving for the index of the largest term, and then using Stirlings to solve for the coefficients, but it seems too tedious.
I've also tried using 1.024 = 1024/100 = 2^10/10^3 and 1.05 = 21/20 = 3*7/(4*5) which aren't too bad to compute using logs to bound it. However this gives 26 and 53 respectively while the answer is 32. If log13 (which i cba to memorize) can be used then 1.04 = 104/100 = 26/25 = 2*13/5^2 can be done with logs, but this gives 42 which is pretty far off too. Averaging 42 (from 1.04) and 26 (1.024) and then shifting the answer down to account for the for the latter being closer to 1.03 seems pretty good tho.

Can someone give me a method aside from memorizing logs up to 103?

Apart from this question I also think that for smaller powers there isn't much need to memorize logs if you can remember certain exponents:
2^n = (2^10)^(n/10) = (1024)^(n/10) = (1000)^(n/10) = 10^(3n/10)
4^n = (2^2)^n = 10^(3n/5)
5^n = (10/2)^n = 10^n/2^n = 10^n/2^(3n/10) = 10^(7n/10)
7^n = 49^(n/2) = 50^(n/2) = (5*10)^(n/2) = 5^(n/2) * 10^(n/2) = 10^(7n/20) * 10^(10n/20) = 10^(17n/20)
8^n = (2^3)^n = 10^(9n/10)
e^n = (e^3)^(n/3) = (20)^(n/3) = (2*10)^(n/3) = 2^(n/3) * 10^(n/3) = 10^(n/10) * 10^(n/3) = 10^(13n/30)
pi^n = (pi^4)^(n/4) = (100)^(n/4) = (10^2)^(n/4) = 10^(n/2)

3^n = 9^(n/2) = 10^(n/2) however this one sucks and stops working after even just 3^14

IMO its easier, and more accurate to just memorize a log table, and multiply the exponent by the log if the base. (That's what you're doing here, but I feel like my way is simpler)

(ex.
for 10^A = 4^678, [A here is the fermi answer]
A = log (4^678)
A= 678 log(4)
A= 678*0.6021
A=~ 408)


I'm guessing that you already knew this, but I'm putting this up mainly for the benefit of other users.

Also, here's a link to a log table https://www.rapidtables.com/math/algebr ... Table.html