GCD is MIN of Exponents of Prime Factors, LCM is MAX of Exponents of Prime Factors. Let $a=p_1^{x_1}p_2^{x_2}\cdots\cdot p_q^{x_q}$ and $b=p_1^{y_1}p_2^{y_2}\cdots p_r^{x_r}$ http://www.cut-the-knot.org/blue/gcd_fta.shtml,  https://math.stackexchange.com/a/349867/85100 say —
Since gcd(a,b) is the largest common divisor of a and b and is divisible by any other common divisor of the two, 
GCD$(a,b)=p_1^{\min(x_1,y_1)}p_2^{\min (x_2,y_2)} \cdots p_q^{\min (x_q,y_q)}$
LCM$(a,b)= p_1^{\max(x_1,y_1)}p_2^{\max(x_2,y_2)}\cdots p_r^{\max (x_r,y_r)}$
I understand GCD(a,b) has to divide both $a,b$. Therefore the exponent of any prime factor $p_i$ in GCD has to be in both $a,b$  — therefore $p_i^{\min(x_i, y_i)}$. But I'm muddled and anxious. Why does $\min$ appear in GREATEST common divisor? Why does $\max$ appear in LOWEST common multiple?
 A: If we are several persons, the GREATEST height that we can all reach is the MINIMUM of our heights (the smallest person amongst us is the problem), the lowest door that we can all pass through is the MAXIMUM of our heights (the tallest person is the problem).
A: Hint $ $ Assume $\rm\,p\nmid a,b,\,$ and write $\ \rm(x,y) := \gcd(x,y),\ \ [x,y] := {\rm lcm}(x,y)$. By FTA (existence and uniqueness of prime factorizations) we can recursively compute gcd and lcm by "peeling off" one-prime-power at at time as follows 
$\rm\ (ap^j,bp^k) = (a,b)(p^j,p^k) = (a,b)\,p^{\large \min(j,k)}\,$ by $\,\rm p^i\mid p^j,p^k\!\!\iff i\le j,k \!\iff\! i\le \min(j,k)$
$\rm\ \, [ap^j,bp^k] =\, [a,b]\,[p^j,p^k]\, =\, [a,b]\,p^{\large \max(j,k)}$ by $\,\rm  p^j,p^k\mid p^i\!\! \iff j,k\le i\! \iff\! \max(j,k)\le i$
A: Although the GCD is the greatest common divisor of $a$ and $b$, it still has to divide both of them! If the exponent of $p_i$ was greater than $\min(x_i,y_i)$ then the GCD wouldn't divide both $a$ and $b$.
Take for instance $40 = 2^3 \cdot 5$ and $100 = 2^2 \cdot 5^2$. Then their GCD is $20 = 2^2 \cdot 5 = 2^{\min(3,2)} \cdot 5^{\min(1,2)}$.
