POTW 6 Posted

There's a typo in the pdf, in the example that is supposed to read 22^22.

Thanks for catching the error, which I have now corrected.

By inspection it has to be 2^(2^22), surely? As any computer programmer knows 2^20 is about a million, so that's 2^about 4 million. This is obviously bigger than any of the other numbers by a fairly wide margin.

I agree with Gazza: 2^(2^22) is definitely the largest of these numbers. I'd memorized the exact value of 2^20 (it's 1048576), so 2^22 would be 4194304.

Second largest is 2^(22^2), which is 2^484. Then 2^222, 22^22, 22^(2^2), 2^(2^(2^2)), 222^2, and lowest is 2222. Reasoning: The base 2 logarithm of 22 is a little less than 4.5, and the base 2 logarithm of 222 is a little less than 8.

I was able to find the full order of the numbers by comparison with powers of two:

2222 = 2 * 1111 < 2 * 2048 so 2222 < 2^12
222^2 = 2^2 * 111^2 < 2^2 * 128^2 so 222^2 < 2^16
22^22 = 2^22 * 11^22 < 2^22 * 16^22 so 22^22 < 2^110
2^222 gives a power of two immediately
22^(2^2) = 22^4 = 2^4 * 11^4< 2^4 * 16^4 so 22^2^2 2*1048 = 2^11
222^2 > 2^2 * 64^2 = 2^14
22^22 > 2^22 * 8^22 = 2^88
22^4 > 2^4 * 8^4 = 2^16

We thus have (letting A through G stand for the numbers in the order given):

A=2^11<A<2^12
2^14<B<2^16
2^88<C<2^110
2^222 = D
2^16<EF>D>C>E>H>B>A

2^(2^22) is therefore the largest.

Somehow that last part didn't format right:

2^11 < A < 2^12
2^14< B < 2^16
2^88 < C < 2^110
2^222 = D
2^16 < E < 2^20
2^484 = F
2^(2^22) = G
2 ^ 16 = H

Therefore, the order from largest to smallest is GFDCEHBA

Sean T, this is essentially the same reasoning Eric Lund used above with his note about base 2 logs.