Find the values of U, V, C based on the given relationship…useful for upcoming puzzlesTheoretical Puzzle: Making a binary puzzle with a unique solutionFewest possible weighings to determine which ball is heavier/lighterFind the number given its remaindersMinimum number of tries to find the balance!A String-based Puzzle: Can you get from the string “baa” to “bacaccacacccc”?Proving the existence of a given matrixFind the equality with all digitsForm the biggest squaring numberFive Powers of Fives Produce Unique Pandigital Number…Solve for X..Tell me YBe a Wizard..Find the missing digits of this 41 digit Humongous Pallindromic Square!
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Find the values of U, V, C based on the given relationship...useful for upcoming puzzles
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Find the values of U, V, C based on the given relationship…useful for upcoming puzzles
Theoretical Puzzle: Making a binary puzzle with a unique solutionFewest possible weighings to determine which ball is heavier/lighterFind the number given its remaindersMinimum number of tries to find the balance!A String-based Puzzle: Can you get from the string “baa” to “bacaccacacccc”?Proving the existence of a given matrixFind the equality with all digitsForm the biggest squaring numberFive Powers of Fives Produce Unique Pandigital Number…Solve for X..Tell me YBe a Wizard..Find the missing digits of this 41 digit Humongous Pallindromic Square!
$begingroup$
Given:
U, V, C are 3 distinct digits..values can vary from 1 to 9.
CU is a concatenated number
Solve for U,V, C from the following relationship:
$U^V$ X $V^U $ = $CU $
This will give some basis to upcoming Unique Pan digital
Fraction problems.
mathematics logical-deduction no-computers
asked 2 hours ago
UvcUvc
70812
$endgroup$
add a comment |
$begingroup$
Given:
U, V, C are 3 distinct digits..values can vary from 1 to 9.
CU is a concatenated number
Solve for U,V, C from the following relationship:
$U^V$ X $V^U $ = $CU $
This will give some basis to upcoming Unique Pan digital
Fraction problems.
mathematics logical-deduction no-computers
asked 2 hours ago
UvcUvc
70812
$endgroup$
add a comment |
$begingroup$
Given:
U, V, C are 3 distinct digits..values can vary from 1 to 9.
CU is a concatenated number
Solve for U,V, C from the following relationship:
$U^V$ X $V^U $ = $CU $
This will give some basis to upcoming Unique Pan digital
Fraction problems.
mathematics logical-deduction no-computers
asked 2 hours ago
UvcUvc
70812
$endgroup$
Given:
U, V, C are 3 distinct digits..values can vary from 1 to 9.
CU is a concatenated number
Solve for U,V, C from the following relationship:
$U^V$ X $V^U $ = $CU $
This will give some basis to upcoming Unique Pan digital
Fraction problems.
mathematics logical-deduction no-computers
mathematics logical-deduction no-computers
asked 2 hours ago
UvcUvc
70812
asked 2 hours ago
UvcUvc
70812
asked 2 hours ago
UvcUvc
70812
asked 2 hours ago
UvcUvc
70812
asked 2 hours ago
UvcUvc
70812
70812
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
U = 2, V = 3, C = 7: 23 × 32 = 72.
U and V can't be too big, if U is 2 the product becomes greater than 100 even for V = 4. On the other hand, if U or V = 1, the equation becomes U = CU or V = CU, single digit on the left and two digits on the right, which is also impossible.
If we try 2 and 3, which are not too big and not too small, we get the product 72.
answered 2 hours ago
Mariia MykhailovaMariia Mykhailova
2,138426
$endgroup$
1
$begingroup$
ninjaed by you... :(, nvm, have an upvote!
$endgroup$
– Omega Krypton
2 hours ago
add a comment |
$begingroup$
If $U=1, 1*V=CU$ (not possible)
If $V=1, U*1=CU$ (not possible)
When $U=2$,
$2^V*V^2<=92$
$V<=3$
When $V=1, 2^1*1^2=2$ (rejected)
When $V=2, U=V$ (rejected)
When $V=3, 2^3*3^2=72$ (possible)
When U=3,
$3^V*V^3<=93$
$V<=3$
When $V=1, 3^1*1^3=3$ (rejected)
When $V=2, 3^2*2^3=72$ (rejected)
When $V=3, U=V$ (rejected)
When U=4+,
$4^V*V^4<=94$
$V=1$ (rejected)
Final Answer:
When $U=2, V=3,$ and $C=7, 2^3*3^2=72$
answered 2 hours ago
Omega KryptonOmega Krypton
6,3812953
$endgroup$
$begingroup$
ninja-ed by @Mariia, yet just wanna post this as a more complete explanation
$endgroup$
– Omega Krypton
2 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
U = 2, V = 3, C = 7: 23 × 32 = 72.
U and V can't be too big, if U is 2 the product becomes greater than 100 even for V = 4. On the other hand, if U or V = 1, the equation becomes U = CU or V = CU, single digit on the left and two digits on the right, which is also impossible.
If we try 2 and 3, which are not too big and not too small, we get the product 72.
answered 2 hours ago
Mariia MykhailovaMariia Mykhailova
2,138426
$endgroup$
1
$begingroup$
ninjaed by you... :(, nvm, have an upvote!
$endgroup$
– Omega Krypton
2 hours ago
add a comment |
$begingroup$
U = 2, V = 3, C = 7: 23 × 32 = 72.
U and V can't be too big, if U is 2 the product becomes greater than 100 even for V = 4. On the other hand, if U or V = 1, the equation becomes U = CU or V = CU, single digit on the left and two digits on the right, which is also impossible.
If we try 2 and 3, which are not too big and not too small, we get the product 72.
answered 2 hours ago
Mariia MykhailovaMariia Mykhailova
2,138426
$endgroup$
1
$begingroup$
ninjaed by you... :(, nvm, have an upvote!
$endgroup$
– Omega Krypton
2 hours ago
add a comment |
$begingroup$
U = 2, V = 3, C = 7: 23 × 32 = 72.
U and V can't be too big, if U is 2 the product becomes greater than 100 even for V = 4. On the other hand, if U or V = 1, the equation becomes U = CU or V = CU, single digit on the left and two digits on the right, which is also impossible.
If we try 2 and 3, which are not too big and not too small, we get the product 72.
answered 2 hours ago
Mariia MykhailovaMariia Mykhailova
2,138426
$endgroup$
U = 2, V = 3, C = 7: 23 × 32 = 72.
U and V can't be too big, if U is 2 the product becomes greater than 100 even for V = 4. On the other hand, if U or V = 1, the equation becomes U = CU or V = CU, single digit on the left and two digits on the right, which is also impossible.
If we try 2 and 3, which are not too big and not too small, we get the product 72.
answered 2 hours ago
Mariia MykhailovaMariia Mykhailova
2,138426
answered 2 hours ago
Mariia MykhailovaMariia Mykhailova
2,138426
answered 2 hours ago
Mariia MykhailovaMariia Mykhailova
2,138426
answered 2 hours ago
Mariia MykhailovaMariia Mykhailova
2,138426
2,138426
1
$begingroup$
ninjaed by you... :(, nvm, have an upvote!
$endgroup$
– Omega Krypton
2 hours ago
add a comment |
1
$begingroup$
ninjaed by you... :(, nvm, have an upvote!
$endgroup$
– Omega Krypton
2 hours ago
1
1
$begingroup$
ninjaed by you... :(, nvm, have an upvote!
$endgroup$
– Omega Krypton
2 hours ago
$begingroup$
ninjaed by you... :(, nvm, have an upvote!
$endgroup$
– Omega Krypton
2 hours ago
add a comment |
$begingroup$
If $U=1, 1*V=CU$ (not possible)
If $V=1, U*1=CU$ (not possible)
When $U=2$,
$2^V*V^2<=92$
$V<=3$
When $V=1, 2^1*1^2=2$ (rejected)
When $V=2, U=V$ (rejected)
When $V=3, 2^3*3^2=72$ (possible)
When U=3,
$3^V*V^3<=93$
$V<=3$
When $V=1, 3^1*1^3=3$ (rejected)
When $V=2, 3^2*2^3=72$ (rejected)
When $V=3, U=V$ (rejected)
When U=4+,
$4^V*V^4<=94$
$V=1$ (rejected)
Final Answer:
When $U=2, V=3,$ and $C=7, 2^3*3^2=72$
answered 2 hours ago
Omega KryptonOmega Krypton
6,3812953
$endgroup$
$begingroup$
ninja-ed by @Mariia, yet just wanna post this as a more complete explanation
$endgroup$
– Omega Krypton
2 hours ago
add a comment |
$begingroup$
If $U=1, 1*V=CU$ (not possible)
If $V=1, U*1=CU$ (not possible)
When $U=2$,
$2^V*V^2<=92$
$V<=3$
When $V=1, 2^1*1^2=2$ (rejected)
When $V=2, U=V$ (rejected)
When $V=3, 2^3*3^2=72$ (possible)
When U=3,
$3^V*V^3<=93$
$V<=3$
When $V=1, 3^1*1^3=3$ (rejected)
When $V=2, 3^2*2^3=72$ (rejected)
When $V=3, U=V$ (rejected)
When U=4+,
$4^V*V^4<=94$
$V=1$ (rejected)
Final Answer:
When $U=2, V=3,$ and $C=7, 2^3*3^2=72$
answered 2 hours ago
Omega KryptonOmega Krypton
6,3812953
$endgroup$
$begingroup$
ninja-ed by @Mariia, yet just wanna post this as a more complete explanation
$endgroup$
– Omega Krypton
2 hours ago
add a comment |
$begingroup$
If $U=1, 1*V=CU$ (not possible)
If $V=1, U*1=CU$ (not possible)
When $U=2$,
$2^V*V^2<=92$
$V<=3$
When $V=1, 2^1*1^2=2$ (rejected)
When $V=2, U=V$ (rejected)
When $V=3, 2^3*3^2=72$ (possible)
When U=3,
$3^V*V^3<=93$
$V<=3$
When $V=1, 3^1*1^3=3$ (rejected)
When $V=2, 3^2*2^3=72$ (rejected)
When $V=3, U=V$ (rejected)
When U=4+,
$4^V*V^4<=94$
$V=1$ (rejected)
Final Answer:
When $U=2, V=3,$ and $C=7, 2^3*3^2=72$
answered 2 hours ago
Omega KryptonOmega Krypton
6,3812953
$endgroup$
If $U=1, 1*V=CU$ (not possible)
If $V=1, U*1=CU$ (not possible)
When $U=2$,
$2^V*V^2<=92$
$V<=3$
When $V=1, 2^1*1^2=2$ (rejected)
When $V=2, U=V$ (rejected)
When $V=3, 2^3*3^2=72$ (possible)
When U=3,
$3^V*V^3<=93$
$V<=3$
When $V=1, 3^1*1^3=3$ (rejected)
When $V=2, 3^2*2^3=72$ (rejected)
When $V=3, U=V$ (rejected)
When U=4+,
$4^V*V^4<=94$
$V=1$ (rejected)
Final Answer:
When $U=2, V=3,$ and $C=7, 2^3*3^2=72$
answered 2 hours ago
Omega KryptonOmega Krypton
6,3812953
edited 1 hour ago
answered 2 hours ago
Omega KryptonOmega Krypton
6,3812953
answered 2 hours ago
Omega KryptonOmega Krypton
6,3812953
answered 2 hours ago
Omega KryptonOmega Krypton
6,3812953
6,3812953
$begingroup$
ninja-ed by @Mariia, yet just wanna post this as a more complete explanation
$endgroup$
– Omega Krypton
2 hours ago
add a comment |
$begingroup$
ninja-ed by @Mariia, yet just wanna post this as a more complete explanation
$endgroup$
– Omega Krypton
2 hours ago
$begingroup$
ninja-ed by @Mariia, yet just wanna post this as a more complete explanation
$endgroup$
– Omega Krypton
2 hours ago
$begingroup$
ninja-ed by @Mariia, yet just wanna post this as a more complete explanation
$endgroup$
– Omega Krypton
2 hours ago
add a comment |
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