(HMMT 2000 Guts Round #38) Find the largest number you can write with three 3's and three 8's, using only $+,-,/,\cdot$ and exponentiation.
Claim 1: if $x>y>e$ are real numbers, then $x^y < y^x$.
Assume that expressions without parentheses are evaluated using regular BEDMAS rules.
I initially guessed that the answer is $3^{3^{3^{8^{8^8}}}},$ though I'm not sure how to prove this. From the problem conditions, there are 6 numbers and we can put 5 operations between them. The key to showing the desired result will be that adding in the other arithmetic operations doesn't result in a large enough number. This is fairly intuitive; if one had $a-b$, where b is positive, one could replace that with $a+b$ to get a larger expression.
To conclude, without parentheses, one clearly cannot use the subtraction operator. One also cannot use the division operator, since it can similarly be replaced by multiplication without decreasing the value of the expression. Finally, note that one can assume WLOG that addition is not used. Indeed, we've assumed that we can only have multiplication and division operations. since if a and b exceed 2 then $a+b < a\cdot b$
Finally, we may replace $a\cdot b$ with $a^b$ whenever a and b are integers exceeding 2 to get a larger expression. Hence we may indeed assume only the exponentiation operator is used. And the maximum value is the claimed value. Indeed, assume a $3$, say b, follows an 8, say c, in the sequence of exponents. We may assume the 3 is preceded by an 8, since there must be an 8 followed by a 3 in the sequence of exponents between b and c. Then we claim that swapping the 3 and the 8 will result in a larger overall value for the expression. Indeed, by assumption, we have an expression of the form $8^{3^a}$, where $a$ is a sequence of exponents containing $3$. And the swap results in an expression of the form $3^{8^a}$.
Edit: It turns out that one can just use logarithms:
$8^{3^a} < 3^{8^a}\Leftrightarrow \ln 8 \cdot 3^a < 8^a \cdot \ln 3.$ $\ln 8 / \ln 3 < (8/3)^a,$ which clearly holds for any positive integer $a$.
As $\frac {\ln(8)} {\ln (3)} \leq \frac 8 3$ we see that $\ln(3)\cdot 8^a \geq \ln(8) \cdot 3^a$ for all $a\geq 1$. Hence $3^{8^a}\geq 8^{3^a}$ for all such $a$. If any nested exponential was of the form $x^{\dots^{8^{3^a}}}$, then $x^{\dots^{3^{8^ a}}}$ would be strictly larger by monotonicity of the exponential function, hence the largest nested exponential is necessarily of the form $3^{3^{3^{8^{8^8}}}}$. Along with the analysis ruling out other operations, this completes the proof.