All Raised to the Power of UFind the values of U, V, C based on the given relationship…useful for upcoming...
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All Raised to the Power of U
Find the values of U, V, C based on the given relationship…useful for upcoming puzzlesFind this Unique UVC Palindrome ( ignoring signs and decimal) from Given Fractional RelationshipThese Two Cubes are The Only Ones That Are All Pure Prime..name themLady Luck Powers Up Every Member to Sum Upto Non-Prime Number. Who am I?UVC wants to give you a Helping Hand in Solving these Unique Set of Pan digital Fractional-Decimal RelationsWe are Unique Friends. Our Special Squares Share a Common Bond…Figure us outDeduce the Component Digits from these three Sets of Symmetric Power RelationsFrom the given Square - Factorial Relationship, deduce the unknownsPlease figure out this Pan digital PrincePan Digital Lucky Seven wants you to figure out all the digits
$begingroup$
$Given$:
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.
mathematics no-computers
asked 8 hours ago
UvcUvc
1,683325
$endgroup$
add a comment |
$begingroup$
$Given$:
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.
mathematics no-computers
asked 8 hours ago
UvcUvc
1,683325
$endgroup$
add a comment |
$begingroup$
$Given$:
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.
mathematics no-computers
asked 8 hours ago
UvcUvc
1,683325
$endgroup$
$Given$:
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.
mathematics no-computers
mathematics no-computers
asked 8 hours ago
UvcUvc
1,683325
asked 8 hours ago
UvcUvc
1,683325
asked 8 hours ago
UvcUvc
1,683325
asked 8 hours ago
UvcUvc
1,683325
asked 8 hours ago
UvcUvc
1,683325
1,683325
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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:
$3^3 + 7^3 + 1^3 = 27 + 343 + 1 = 371$
$4^3 + 0^3 + 7^3 = 64 + 0 + 343 = 407$
answered 7 hours ago
scatterscatter
1913
New contributor
scatter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Explain the logic
$endgroup$
– Uvc
7 hours ago
$begingroup$
@Uvc What logic? Does the math not speak for itself?
$endgroup$
– scatter
7 hours ago
1
$begingroup$
Usually, for this kind of puzzles, deductive reasoning has to be given how you arrived at the final solution
$endgroup$
– Uvc
7 hours ago
$begingroup$
That’s much better
$endgroup$
– Uvc
7 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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:
$3^3 + 7^3 + 1^3 = 27 + 343 + 1 = 371$
$4^3 + 0^3 + 7^3 = 64 + 0 + 343 = 407$
answered 7 hours ago
scatterscatter
1913
New contributor
scatter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Explain the logic
$endgroup$
– Uvc
7 hours ago
$begingroup$
@Uvc What logic? Does the math not speak for itself?
$endgroup$
– scatter
7 hours ago
1
$begingroup$
Usually, for this kind of puzzles, deductive reasoning has to be given how you arrived at the final solution
$endgroup$
– Uvc
7 hours ago
$begingroup$
That’s much better
$endgroup$
– Uvc
7 hours ago
add a comment |
$begingroup$
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:
$3^3 + 7^3 + 1^3 = 27 + 343 + 1 = 371$
$4^3 + 0^3 + 7^3 = 64 + 0 + 343 = 407$
answered 7 hours ago
scatterscatter
1913
New contributor
scatter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Explain the logic
$endgroup$
– Uvc
7 hours ago
$begingroup$
@Uvc What logic? Does the math not speak for itself?
$endgroup$
– scatter
7 hours ago
1
$begingroup$
Usually, for this kind of puzzles, deductive reasoning has to be given how you arrived at the final solution
$endgroup$
– Uvc
7 hours ago
$begingroup$
That’s much better
$endgroup$
– Uvc
7 hours ago
add a comment |
$begingroup$
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:
$3^3 + 7^3 + 1^3 = 27 + 343 + 1 = 371$
$4^3 + 0^3 + 7^3 = 64 + 0 + 343 = 407$
answered 7 hours ago
scatterscatter
1913
New contributor
scatter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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:
$3^3 + 7^3 + 1^3 = 27 + 343 + 1 = 371$
$4^3 + 0^3 + 7^3 = 64 + 0 + 343 = 407$
answered 7 hours ago
scatterscatter
1913
New contributor
scatter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 hours ago
answered 7 hours ago
scatterscatter
1913
New contributor
scatter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 7 hours ago
scatterscatter
1913
answered 7 hours ago
scatterscatter
1913
1913
New contributor
scatter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
scatter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Explain the logic
$endgroup$
– Uvc
7 hours ago
$begingroup$
@Uvc What logic? Does the math not speak for itself?
$endgroup$
– scatter
7 hours ago
1
$begingroup$
Usually, for this kind of puzzles, deductive reasoning has to be given how you arrived at the final solution
$endgroup$
– Uvc
7 hours ago
$begingroup$
That’s much better
$endgroup$
– Uvc
7 hours ago
add a comment |
$begingroup$
Explain the logic
$endgroup$
– Uvc
7 hours ago
$begingroup$
@Uvc What logic? Does the math not speak for itself?
$endgroup$
– scatter
7 hours ago
1
$begingroup$
Usually, for this kind of puzzles, deductive reasoning has to be given how you arrived at the final solution
$endgroup$
– Uvc
7 hours ago
$begingroup$
That’s much better
$endgroup$
– Uvc
7 hours ago
$begingroup$
Explain the logic
$endgroup$
– Uvc
7 hours ago
$begingroup$
Explain the logic
$endgroup$
– Uvc
7 hours ago
$begingroup$
@Uvc What logic? Does the math not speak for itself?
$endgroup$
– scatter
7 hours ago
$begingroup$
@Uvc What logic? Does the math not speak for itself?
$endgroup$
– scatter
7 hours ago
1
1
$begingroup$
Usually, for this kind of puzzles, deductive reasoning has to be given how you arrived at the final solution
$endgroup$
– Uvc
7 hours ago
$begingroup$
Usually, for this kind of puzzles, deductive reasoning has to be given how you arrived at the final solution
$endgroup$
– Uvc
7 hours ago
$begingroup$
That’s much better
$endgroup$
– Uvc
7 hours ago
$begingroup$
That’s much better
$endgroup$
– Uvc
7 hours ago
add a comment |
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