1. **State the problem:** Solve for $c$ in the equation $$\frac{1}{4}(1+3c) + \frac{1}{4}(1-c) + \frac{1}{4}(1+2c) + \frac{1}{4}(1-4c) = 1.$$\n\n2. **Write the equation clearly:**\n$$\frac{1}{4}(1+3c) + \frac{1}{4}(1-c) + \frac{1}{4}(1+2c) + \frac{1}{4}(1-4c) = 1.$$\n\n3. **Distribute $\frac{1}{4}$ to each term inside the parentheses:**\n$$\frac{1}{4} \times 1 + \frac{1}{4} \times 3c + \frac{1}{4} \times 1 - \frac{1}{4} \times c + \frac{1}{4} \times 1 + \frac{1}{4} \times 2c + \frac{1}{4} \times 1 - \frac{1}{4} \times 4c = 1.$$\n\n4. **Simplify each term:**\n$$\frac{1}{4} + \frac{3c}{4} + \frac{1}{4} - \frac{c}{4} + \frac{1}{4} + \frac{2c}{4} + \frac{1}{4} - \frac{4c}{4} = 1.$$\n\n5. **Combine like terms:**\nConstants: $$\frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 1.$$\nCoefficients of $c$: $$\frac{3c}{4} - \frac{c}{4} + \frac{2c}{4} - \frac{4c}{4} = \frac{3c - c + 2c - 4c}{4} = \frac{0}{4} = 0.$$\n\n6. **Rewrite the equation:**\n$$1 + 0 = 1.$$\n\n7. **Interpretation:**\nThe equation simplifies to $$1 = 1,$$ which is always true regardless of $c$.\n\n**Therefore, $c$ can be any real number.**