In the following sum, '$x$' stands for which digit? $$x + 1x + 2x + x3 + x1 = 21x$$

### Concept & Logic This puzzle uses the same unknown digit ($x$) in multiple place values (both as units and tens). Converting each two-digit block into its standard algebraic expansion ($10 \times \text{Tens} + \text{Units}$) transforms the problem into a basic linear equation. ### Step-by-Step Solution * **Given:** $x + 1x + 2x + x3 + x1 = 21x$ * **Place Value Expansion:** Rewrite every term explicitly. * $1x = 10 + x$ * $2x = 20 + x$ * $x3 = 10x + 3$ * $x1 = 10x + 1$ * $21x = 200 + 10 + x = 210 + x$ * **Form the Equation:** Substitute these back into the sum. $$x + (10 + x) + (20 + x) + (10x + 3) + (10x + 1) = 210 + x$$ * **Combine Like Terms:** Group all the constants and all the $x$ terms on the left side. $$x + x + x + 10x + 10x = 23x$$ $$10 + 20 + 3 + 1 = 34$$ So, the left side simplifies to $23x + 34$. * **Solve for x:** $$23x + 34 = 210 + x$$ $$22x = 210 - 34$$ $$22x = 176$$ $$x = \frac{176}{22} = 8$$ ### Exam Strategy & Shortcut **Unit Digit Trick:** You can bypass the full algebra by just looking at the unit digits of the sum. The unit digits being added are: $x + x + x + 3 + 1 = 3x + 4$. The result ($21x$) ends in the digit $x$. This means $(3x + 4)$ must end in $x$. Subtract $x$ from both concepts: $2x + 4$ must end in $0$. For $2x + 4$ to end in $0$, $2x$ must end in $6$. The only single digits where $2x$ ends in $6$ are $x = 3$ and $x = 8$. Test $x = 8$: $8 + 18 + 28 + 83 + 81 = 218$. It matches perfectly! ### Common Pitfall A very common mistake is treating "$1x$" as multiplication ($1 \times x$) instead of place value ($10 + x$). Always remember that in cryptarithms, adjacent symbols represent digits in a multi-digit number, not algebraic multiplication. ### Final Answer Therefore, the correct answer is **8**.