
6. (i) Consider two positive integers $a$ and $b$ which are such that $\boldsymbol{a}^{\boldsymbol{a}} \boldsymbol{b}^{\boldsymbol{b}}$ is divisible by 2000 . What is the least possible value of the product $a b$ ?

(ii) Consider two positive integers $a$ and $b$ which are such that $\boldsymbol{a}^{\boldsymbol{b}} \boldsymbol{b}^{\boldsymbol{a}}$ is divisible by 2000 . What is the least possible value of the product $a b$ ?


Solution: We have $2000=2^{4} 5^{3}$.

(i) Since 2000 divides $a^{a} b^{b}$, it follows that 2 divides $a$ or $b$ and similarly 5 divides $a$ or $b$. In any case 10 divides $a b$. Thus the least possible value of $a b$ for which $2000 \mid a^{a} b^{b}$ must be a multiple of 10 . Since 2000 divides $10^{10} 1^{1}$, we can take $a=10, b=1$ to get the least value of $a b$ equal to 10 .

(ii) As in (i) we conclude that 10 divides $a b$. Thus the least value of $a b$ for which 2000 $\mid a^{b} b^{a}$ is again a multiple of 10 . If $a b=10$, then the possibilities are $(a, b)=$ $(1,10),(2,5),(5,2),(10,1)$. But in all these cases it is easy to verify that 2000 does not divide $a^{b} b^{a}$. The next multiple of 10 is 20 . In this case we can take $(a, b)=(4,5)$ and verify that 2000 divides $4^{5} 5^{4}$. Thus the least value here is 20 .
