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This puzzle highlights some cyclic power Relations.

Please provide detailed reasoning of derivation of digits from various power relations given below:

$Given$:

$A$, $C$, $L$, $N$, $O$, $S$, $U$, $V$ are all distinct digits varying from 0 to 9.

$SOLV$, $UVC$, $LSU$, $VAN$ are concatenated Numbers.

1) $S^L$ + $O^L$ + $L^L$ + $V^L$ = $UVC$

2) $U^L$ + $V^L$ + $C^L$ = $LSU$

3) $L^L$ + $S^L$ + $U^L$ = $VAN$

4) $V^L$ + $A^L$ + $N^L$ = $UVC$

Back to Step 2..Repeat and Rinse Cycle of nice Power Relations.

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  • $\begingroup$ Don’t be scared of number of equations. Deductive logic and reasonable assumptions will get you to the right result. This is no more difficult than some of my previous ones. $\endgroup$ – Uvc Jun 14 at 10:43
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I believe the following values are consistent with the equations:

A = 6, C = 7, L = 3, N = 0, O = 4, S = 5, U = 2, V = 1

My first discovery is that

If L is 4 or greater, one of the remaining digits must be 6 or greater since there are eight of them total, and 6^4 exceeds 1000. Meanwhile, if L is 2 or less, this also does not work because of the second equation, where U^2+V^2+C^2 = 200 + SU. Assuming maximum values for U, V, and C, you would get 81+64+49 = 194, which is insufficient. Therefore, L must be equal to 3.

Working from Equations 1 and 4:

S^3+O^3+3^3 = A^3+N^3, as both equations equal to UVC and the V^L's cancel out. I feel like there may have been a more elegant way to figure this out, but I recall that 3^3 + 4^3 + 5^3 = 6^3, so I believe S and O are 4 and 5 while A and N are 0 and 6.

This gives us:

216 + V^3 = UVC The remaining possibilities for V are 2, 1, 7, 8, and 9, however we can rule out the latter two from equation 2 as 8^3 exceeds 400 and therefore cannot sum to LSU. 7 does not work as you result with 559, therefore V would have to be both 5 and 7. 2 does not work as your result would be 224, and both C and S/O cannot be 4. Lastly V=1 gives you a result of 217, making U = 2, V = 1, and C = 7. So far we are consistent.

Now back to equation 2:

We have 2^3 + 1^3 + 7^3 = LSU, resulting in LSU = 352, giving us L = 3 and U = 2, which is consistent, and S = 5, leaving O to equal 4. We are now left with A and N.

Finally with equation 3:

3^3 + 5^3 + 2^3 = VAN, resulting in VAN = 160. Once again confirming V = 1, as well as A = 6 and N = 0, consistent with our original hypothesis from Equations 1 and 4.

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  • $\begingroup$ Excellent deductive reasoning leading to the right result..this is one of my favorite..most of the people are put off when they see bunch of equations $\endgroup$ – Uvc Jun 19 at 22:21

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