21.3. The numbers $2^{n}$ and $5^{n}$ start with the digit $a$. What is $a$?
21.3. Answer: 3. By the condition $a \cdot 10^{p}<2^{n}<(a+1) 10^{p}$ and $a \cdot 10^{q}<5^{n}<$ $<(a+1) 10^{q}$. Therefore, $a^{2} 10^{p+q}<10^{n}<(a+1)^{2} 10^{p+q}$, i.e., $a^{2}<10^{n-p-q}<$ $<(a+1)^{2}$. At the same time, $(a+1)^{2} \leqslant 100$. Thus, $a^{2}<10<(a+1)^{2}$, i.e., $a=3$. The number 3 is the leading digit of the numbers $2^{5}$ and $5^{5}$.