Find the highest number where:

1: all digits are different

2: the sum of its digits is a 2digit number
whose 2 digits do not appear in the number

Example:
number : 754
7 + 5 + 4 = 16 : 1 and 6 do not appear in 754

1. 2^(3^(4^(5^(6^(7^(8^(9^10)))))))

What d'heck is that, Bob

That is one way to represent powers.

Let me try it a different way leaving out the parentheses...

Yes, I knew that, BUT that has nothing to do with the problem.
The number you'll get from that will have many digits that are
the same. As example, 2^3 is NOT a number, it is an expression.
In this puzzle, you have a simple number, maximum length = 10
(since there are 10 digits and digits must be different).

Anyway, if you add all the digits in your expression:
2+3+4+5+6+7+8+9+1+0 = 45
Digits 4 and 5 (from the 45) DO appear in the expression.

The puzzle is simple; don't complicate it

Having answered the wrong question a couple of times, I think it may be 9,876,532 with digits adding up to 40 (assuming of course that I have actually interpreted the puzzle correctly this time!).

I defer to CCb's answer below.

98754210

Yepper CC ; 9+8+7+5+4+2+1+0 = 36 !

Lex, I found it somewhat difficult to word that problem;
how would you have worded it? Anybody?

Denis, I think it was fine, but it needed reading with mathematical precision, especially with a rather foggy brain on a Saturday morning!

I began by thinking that the answer was going to be a two digit number, but 1 does not say that; I then thought that the answer was going to be the sum of the other digits but 2 does not say that; however, both thoughts led to some amusing light-hearted calculations over my morning coffee...

Thanks. Was asking cause I intend to
post it at a coupla udder sites...

Cowabunga!

Denis, you might make it clear that there is one answer that satisfies both conditions. It appeared to me that you were asking for two different answers. And, by the way, the reason I answered as I did is because an answer to just condition number 1 would be too simple. I assumed you were asking a trick question, which you have to admit you often do.

Thanks Bob.

Ask a trick question, moi?