Spring break has ended. It is a great sadness. But that means that POTW can now return! Life is full of tradeoffs.
More like this
I have now posted the final POTW for the semester. That's right! After this, there will no more POTWs until September. So go have a look, and tell me what you think.
The fifth Problem of the Week has now been posted. This one is probably my favorite of the term. I think it's fairly challenging. It will have to hold you for a while, though, since POTW will be taking next week off.
My trip to New York was a lot of fun. Some friends from Kentucky were visiting me this weekend, and that was fun too. But in all the chaos Sunday Chess Problem ended up taking the week off. Sorry about that! It will return next week.
The second POTW has now been posted, along with one possible set of solutions for the first problem. This week's problem has a similar flavor to last week's, so if you liked that one you'll like this one.
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.