1. **State the problem:** We need to find the digit $x$ in the base 5 number $34x1_5$ such that the number is divisible by 31 in decimal. 2. **Convert the base 5 number to decimal:** The number $34x1_5$ can be expanded as: $$3 \times 5^3 + 4 \times 5^2 + x \times 5^1 + 1 \times 5^0$$ which simplifies to: $$3 \times 125 + 4 \times 25 + x \times 5 + 1$$ $$= 375 + 100 + 5x + 1 = 476 + 5x$$ 3. **Set divisibility condition:** The decimal number $476 + 5x$ must be divisible by 31. So, $$31 \mid (476 + 5x)$$ which means there exists an integer $k$ such that: $$476 + 5x = 31k$$ 4. **Find $x$ by checking values:** Since $x$ is a digit in base 5, $x \in \{0,1,2,3,4\}$. We test each: - For $x=0$: $476 + 5 \times 0 = 476$, check $476 \div 31 = 15.3548...$ not integer. - For $x=1$: $476 + 5 = 481$, check $481 \div 31 = 15.5161...$ not integer. - For $x=2$: $476 + 10 = 486$, check $486 \div 31 = 15.6774...$ not integer. - For $x=3$: $476 + 15 = 491$, check $491 \div 31 = 15.8387...$ not integer. - For $x=4$: $476 + 20 = 496$, check $496 \div 31 = 16$ exactly. 5. **Conclusion:** The digit $x$ that makes $34x1_5$ divisible by 31 is $\boxed{4}$.