# All Raised to the Power of U

$$Given$$:

C, I, L, U, V are all distinct digits and can vary from 0 to 9.

$$LIV$$, $$UVC$$ are concatenated Numbers.

$$U^U$$ + $$V^U$$ + $$C^U$$ = $$UVC$$

$$L^U$$ + $$I^U$$ + $$V^U$$ = $$LIV$$

Solve for all the digits.

Immediately,

The first line includes a digit raised to itself, summed with two other numbers that resulted in a three digit number (beginning with that digit). This immediately ruled out anything above $$4^4$$ as $$5^5 > 599$$. $$U = 0,1,2$$ was impossible as even the highest possible sum, $$2^2 + 8^2 + 9^2$$ is less than $$200$$. So U was either 3 or 4. $$U = 4$$ was impossible, as $$5^4 > 499$$ and $$4^4 + 3^4 + 2^4 < 400$$, leaving no possible way to reach a number with the required first digit. So $$U = 3$$. The rest fell into place via trial and error.

$$C, I, L, U, V =$$

$$1, 0, 4, 3, 7$$

Equations:

$$3^3 + 7^3 + 1^3 = 27 + 343 + 1 = 371$$
$$4^3 + 0^3 + 7^3 = 64 + 0 + 343 = 407$$

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• Explain the logic – Uvc Jun 12 at 19:17
• @Uvc What logic? Does the math not speak for itself? – scatter Jun 12 at 19:18
• Usually, for this kind of puzzles, deductive reasoning has to be given how you arrived at the final solution – Uvc Jun 12 at 19:19
• That’s much better – Uvc Jun 12 at 19:30