This One To Itself And That One To Itself

Whew, hard problem. But, finally, I solved it (I think...)

Anyways, this problem sorta requires more common sense than it does mathematics skills...

First, obviously, either $a$ or $b$ or both are multiples of 2 and 5, as a number can only have a trailing zero in base 10 if and only if the number is a multiple of 2 and 5. Then, next, we derive that $a^ab^b$ must have either 98 5's and an abundance of 2's, or 98 2's and an abundance of 5's in its factorization.

Now, consider a number $x^x$ . Now, if $x^x$ is a multiple of any number, say 5, then, $x$ must be of the form $5^nk^{5^nk}$ for some integer $k$ , and for some power of 5 $n$ . So, $x^x$ must have $n \cdot 5^nk$ 5's in its factorization. Now, this shows that $a^ab^b$ cannot have exactly 98 5's in its prime factorization, as the number of 5's in the prime factorization of the above number must be a multiple of 5, and 98 is not a multiple of 5.

Here we have the guessing part. First, we assume that $a$ is a factor of 5, and in order to minimize $a$ , $a$ must be of the form $25k$ for integers $k \geq 2$ , since if $a$ is not a multiple of 25, then either $b$ also has to be a multiple of 5, (which is not possible since the number of 2's in our factorization cannot be a multiple of 5) or a has to be at least $\geq105$ , which is too large for minimization. So, our possible values for $a$ are 50 and 75.

Now, for the values of $b$ , we have to assume values of $a$ first. Assume that $a = 50$ , then, $b^b$ must have exactly 48 2's in its prime factorization, which is not possible ( I leave the calculations to the reader, dear reader, I believe in you!) So, $a$ must be equal to $\boxed{75}$ . Then, $b$ must have exactly 98 2's in its factorization. Since $98 = 2 \cdot 7 \cdot 7$ in its factorization, $b$ must be of the form $2^1\cdot k$ for some integer k, as if not, $b$ must be of the form $2^7\cdot k$ , which is too large for minimization. Since if $b$ takes in the form $2^1\cdot k$ , then $b$ must have exactly $1 \cdot 2^1\cdot k$ 2's in its factorization. Equating this to 98, we get $k = 49$ , and $b = 2^1\cdot k = \boxed {98}$ .

Therefore, $75 + 98 = \boxed {173}$ .

P.S. I know there are many gaps in my solution, so please clarify any mistakes you may find. Better still, please write a more appropriate, understandable solution.

Let ${ a }_{ 2 }$ be the maximum integer such that ${ 2 }^{ { a }_{ 2 } }|a$ . Define ${ a }_{ 5 }$ , ${ b }_{ 2 }$ , ${ b }_{ 5 }$ . Our task translates into the following :

Find $a, b$ such that min ${ \{ a }_{ 5 }a + { b }_{ 5 }b, { a }_{ 2 }a + { b }_{ 2 }b\} = 98$ and $ab$ is minimal. Since $5|{ a }_{ 5 }a + { b }_{ 5 }b$ , we have ${ a }_{ 5 }a + { b }_{ 5 }b > 98$ and $\min { \{ a }_{ 5 }a + { b }_{ 5 }b, { a }_{ 2 }a + { b }_{ 2 }b\} = { a }_{ 2 }a + { b }_{ 2 }b = 98.$ Note that if $5|gcd(a, b)$ , then ${ a }_{ 2 }a + { b }_{ 2 }b \neq 98$ , contradiction.

Without loss of generality , suppose that ${ a }_{ 5 } \ge 1$ and ${ b }_{ 5 } = 0$ . Let $a={ 2 }^{ { a }_{ 2 } }{ 5 }^{ { a }_{ 5 } }x$ and ${ 2 }^{ { a }_{ 2 } }y$ , $gcd(2, x) = gcd(5, x)= gcd(2, y) = 1$ . Then ${ a }_{ 5 }a={ a }_{ 5 }({ 2 }^{ { a }_{ 2 } }{ 5 }^{ { a }_{ 5 } }x) > 98$ and ${ a }_{ 2 }a={ a }_{ 2 }({ 2 }^{ { a }_{ 2 } }{ 5 }^{ { a }_{ 5 } }x) \le 98$ . So ${ a }_{ 5 } > { a }_{ 2 }$ . We consider the following cases:

$(a)$ ${ a }_{ 2 } = 0$ . Then ${ b }_{ 2 }({ 2 }^{ { b }_{ 2 } }y) = 98$ ...

I will write rest of the solution later.