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 Powerful Powers Pileup! (Posted on 2005-04-03)
Given a list of positive integers a, b, c, ... z, you can calculate the superpower a^b^c^...^z. [Note that this is a^(b^(c^(...(y^z)...))).]

What's the largest/least superpower value you can get with the list 2, 3, 4, ... n?

 See The Solution Submitted by Old Original Oskar! Rating: 3.0000 (1 votes)

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 A try Comment 3 of 3 |

Here, I do not include power's parenthesis, but I'll consider them infra.

In a crescent series 2, 3, .., a, a+1,...n, a superpower with base a and the other terms ordained as crescent powers, is always higher than a superpower with base (a+1) and the other terms ordained also as crescent powers.

You can substitute for easiness the expression 2^3^...^(a-1) = @ and the value (a+1)^(a+2)^...^n = s. To compare       (a^@^(a +1)^s) with ((a+1)^@^a^s)) note that s is the same for both, so it doesn't count  to determine the higher term.

But a^@^(a +1) is higher than (a+1)^@^a. If we call p the exponent of the second expression, the exponent of the first can be expressed as p^(a-1). But a^p^(a-1) is always superior to (a+1)^p for a, p>3

It means that setting powers in crescent order, the superpower with minor base is always higher (excluding 2^3 and 3^2 as has been pointed).

If instead of changing the base, we change the order of the powers, the conclusion is the same, because to calculate any exponent of the chain, we have to take it as a base, so we find ourselves in the above case. For each base, setting exponents in crescent order will give higher numbers.

To conclude that the order 2^3^4^...^n give the highest superpower.

 Posted by armando on 2005-04-10 11:05:52

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