How find the number of zeros at the end of the sum $4^{5^6}+6^{5^4}$? The problem is to find the number of zeros at the end of the sum $4^{5^6}+6^{5^4}$.
I tried $2^{2 \cdot 5^6}+3^{5^4} \cdot 2^{5^4}=  2^{5^4}  \cdot (  2^{2 \cdot 5^6  -5^4}+    3^{5^4}  )$.

Use the following lemma, which can be proven by induction 'on $b$':
For any odd prime $p$ and $a,b \in \mathbb{N}_{>0}$ such that $p^a \mid\mid x-y$ and $p^b \mid\mid k$:
  $p^{a+b} \mid\mid x^k-y^k$.
For this question we don't need the "strictly divides" in the conclusion, but in general this lemma is very useful.
$4^{5^6}+6^{5^4} = (4^{5^2})^{5^4}+6^{5^4} = (4^{5^2})^{5^4}-(-6)^{5^4}$ and $5 \mid 4^{5^2}-(-6)$ by Fermat's little theorem.
Therefore by the lemma $5^5 \mid 4^{5^6}+6^{5^4}$.
Combining with what you already have gives the number of zeros.
