Problem 11.7. Natural numbers $a$ and $b$ are such that $a^{a}$ is divisible by $b^{b}$, but $a$ is not divisible by $b$. Find the smallest possible value of the number $a+b$, given that the number $b$ is coprime with 210.
Answer: 374.

Solution. Obviously, $b \neq 1$. Let $p$ be a prime divisor of the number $b$; then $p \geqslant 11$, since $b$ is coprime with $210=2 \cdot 3 \cdot 5 \cdot 7$. Since $a^{a}$ is divisible by $b^{b}$, which is divisible by $p$, then $a$ is also divisible by $p$. From this, it immediately follows that the number $b$ is not a prime.

We will prove that $a+b \geqslant 253+121=374$. If $b$ can be represented as the product of at least three prime factors (not necessarily distinct), then it is not less than $11^{3}>374$, and thus $a+b$ is greater than 374. Now let $b$ be representable as exactly two prime factors. If $b=q r$ for distinct prime numbers $q$ and $r$, then by the above, $a$ is divisible by both $q$ and $r$, but then it is divisible by $b$, which is impossible.

The only case left to consider is when $b=s^{2}$ for some prime $s \geqslant 11$. The number $a$ is divisible by $s$; let $a=s k$ for some natural number $k$. The number $k$ is coprime with $s$, since $a$ is not divisible by $b$.

Since $a^{a}=(s k)^{s k}=s^{s k} \cdot k^{s k}$ is divisible by $b^{b}=\left(s^{2}\right)^{s^{2}}=s^{2 s^{2}}$, then $s^{s k}$ is divisible by $s^{2 s^{2}}$, i.e., $s k \geqslant 2 s^{2}$, and $k \geqslant 2 s$. The number $k$ is coprime with $s$, so it is not equal to $2 s$, i.e., $k \geqslant 2 s+1$.

Then $a+b=s k+s^{2} \geqslant s(2 s+1)+s^{2} \geqslant 11 \cdot 23+11^{2}=253+121=374$.

It is easy to verify that for $a=253=11 \cdot 23$ and $b=121=11^{2}$ all conditions are satisfied: $a^{a}=11^{253} \cdot 22^{253}$ is divisible by $b^{b}=11^{242}$, $a$ is not divisible by $b$, and the number $b$ is coprime with 210.