Deduce the Component Digits from these three Sets of Symmetric Power Relations
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Deduce the Component Digits from these three Sets of Symmetric Power Relations
$begingroup$
Use calculator to the minimum.
Under Logical Deduction, these Daunting Relations will quickly melt away to reveal the constituent digits.
All three sets show beautiful symmetric Relations with first and second part of strings with powers on other side.
A, B, C, D, E, F, G, H are single digits..can vary from 0 to 9.
All other letter combinations are Concatenated Numbers. Leading zeroes allowed in the numbers for symmetry sake.
1st Set:
$ABCC$ = $AB^B$ + $CC^B$
$DDCC$ = $DD^B$ + $CC^B$
2nd Set:
$EAEAEE$ = $EAE^B$ + $AEE^B$
$FFEAEE$ = $FFE^B$ + $AEE^B$
3rd Set:
$EGDDBCGC$ = $EGDD^B$ + $BCGC^B$
$FHABBCGC$ = $FHAB^B$ + $BCGC^B$
Further another relationship exists
$EGDD$ + $FHAB$ = $AE*(EAE+FAE)$ = $(AB+DD)*AE^B$
mathematics no-computers
edited Jun 3 at 5:43
JonMark Perry
22.3k643103
asked Jun 3 at 2:58
UvcUvc
1,374221
$endgroup$
add a comment |
$begingroup$
Use calculator to the minimum.
Under Logical Deduction, these Daunting Relations will quickly melt away to reveal the constituent digits.
All three sets show beautiful symmetric Relations with first and second part of strings with powers on other side.
A, B, C, D, E, F, G, H are single digits..can vary from 0 to 9.
All other letter combinations are Concatenated Numbers. Leading zeroes allowed in the numbers for symmetry sake.
1st Set:
$ABCC$ = $AB^B$ + $CC^B$
$DDCC$ = $DD^B$ + $CC^B$
2nd Set:
$EAEAEE$ = $EAE^B$ + $AEE^B$
$FFEAEE$ = $FFE^B$ + $AEE^B$
3rd Set:
$EGDDBCGC$ = $EGDD^B$ + $BCGC^B$
$FHABBCGC$ = $FHAB^B$ + $BCGC^B$
Further another relationship exists
$EGDD$ + $FHAB$ = $AE*(EAE+FAE)$ = $(AB+DD)*AE^B$
mathematics no-computers
edited Jun 3 at 5:43
JonMark Perry
22.3k643103
asked Jun 3 at 2:58
UvcUvc
1,374221
$endgroup$
$begingroup$
Is $AB^B$ = $(AB)^B$ or $A(B^B)$? E.g. if $A=1$ and $B=2$, then is $AB^B$ equal to $(12)^2$ or to $10 + 2^2$?
$endgroup$
– shoover
Jun 3 at 5:07
$begingroup$
All letter combinations other than single letter used are concatenated Numbers as stated.
$endgroup$
– Uvc
Jun 3 at 7:01
add a comment |
$begingroup$
Use calculator to the minimum.
Under Logical Deduction, these Daunting Relations will quickly melt away to reveal the constituent digits.
All three sets show beautiful symmetric Relations with first and second part of strings with powers on other side.
A, B, C, D, E, F, G, H are single digits..can vary from 0 to 9.
All other letter combinations are Concatenated Numbers. Leading zeroes allowed in the numbers for symmetry sake.
1st Set:
$ABCC$ = $AB^B$ + $CC^B$
$DDCC$ = $DD^B$ + $CC^B$
2nd Set:
$EAEAEE$ = $EAE^B$ + $AEE^B$
$FFEAEE$ = $FFE^B$ + $AEE^B$
3rd Set:
$EGDDBCGC$ = $EGDD^B$ + $BCGC^B$
$FHABBCGC$ = $FHAB^B$ + $BCGC^B$
Further another relationship exists
$EGDD$ + $FHAB$ = $AE*(EAE+FAE)$ = $(AB+DD)*AE^B$
mathematics no-computers
edited Jun 3 at 5:43
JonMark Perry
22.3k643103
asked Jun 3 at 2:58
UvcUvc
1,374221
$endgroup$
Use calculator to the minimum.
Under Logical Deduction, these Daunting Relations will quickly melt away to reveal the constituent digits.
All three sets show beautiful symmetric Relations with first and second part of strings with powers on other side.
A, B, C, D, E, F, G, H are single digits..can vary from 0 to 9.
All other letter combinations are Concatenated Numbers. Leading zeroes allowed in the numbers for symmetry sake.
1st Set:
$ABCC$ = $AB^B$ + $CC^B$
$DDCC$ = $DD^B$ + $CC^B$
2nd Set:
$EAEAEE$ = $EAE^B$ + $AEE^B$
$FFEAEE$ = $FFE^B$ + $AEE^B$
3rd Set:
$EGDDBCGC$ = $EGDD^B$ + $BCGC^B$
$FHABBCGC$ = $FHAB^B$ + $BCGC^B$
Further another relationship exists
$EGDD$ + $FHAB$ = $AE*(EAE+FAE)$ = $(AB+DD)*AE^B$
mathematics no-computers
mathematics no-computers
edited Jun 3 at 5:43
JonMark Perry
22.3k643103
asked Jun 3 at 2:58
UvcUvc
1,374221
edited Jun 3 at 5:43
JonMark Perry
22.3k643103
asked Jun 3 at 2:58
UvcUvc
1,374221
edited Jun 3 at 5:43
JonMark Perry
22.3k643103
edited Jun 3 at 5:43
JonMark Perry
22.3k643103
edited Jun 3 at 5:43
JonMark Perry
22.3k643103
22.3k643103
asked Jun 3 at 2:58
UvcUvc
1,374221
asked Jun 3 at 2:58
UvcUvc
1,374221
asked Jun 3 at 2:58
UvcUvc
1,374221
1,374221
$begingroup$
Is $AB^B$ = $(AB)^B$ or $A(B^B)$? E.g. if $A=1$ and $B=2$, then is $AB^B$ equal to $(12)^2$ or to $10 + 2^2$?
$endgroup$
– shoover
Jun 3 at 5:07
$begingroup$
All letter combinations other than single letter used are concatenated Numbers as stated.
$endgroup$
– Uvc
Jun 3 at 7:01
add a comment |
$begingroup$
Is $AB^B$ = $(AB)^B$ or $A(B^B)$? E.g. if $A=1$ and $B=2$, then is $AB^B$ equal to $(12)^2$ or to $10 + 2^2$?
$endgroup$
– shoover
Jun 3 at 5:07
$begingroup$
All letter combinations other than single letter used are concatenated Numbers as stated.
$endgroup$
– Uvc
Jun 3 at 7:01
$begingroup$
Is $AB^B$ = $(AB)^B$ or $A(B^B)$? E.g. if $A=1$ and $B=2$, then is $AB^B$ equal to $(12)^2$ or to $10 + 2^2$?
$endgroup$
– shoover
Jun 3 at 5:07
$begingroup$
Is $AB^B$ = $(AB)^B$ or $A(B^B)$? E.g. if $A=1$ and $B=2$, then is $AB^B$ equal to $(12)^2$ or to $10 + 2^2$?
$endgroup$
– shoover
Jun 3 at 5:07
$begingroup$
All letter combinations other than single letter used are concatenated Numbers as stated.
$endgroup$
– Uvc
Jun 3 at 7:01
$begingroup$
All letter combinations other than single letter used are concatenated Numbers as stated.
$endgroup$
– Uvc
Jun 3 at 7:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Set 1:
$1233 = 12^2 + 33^2$
$8833 = 88^2 + 33^2$
Because:
We can deduce $B=2$. It must be $2$ or $3$ as $11^4gt9999$. If $B=3$, then either $A$ or $C$ is at least $2$, and $22^3>9999$.
I like this bit!
$DDCC-ABCC = 100(DD-AB)$.
But also $DDCC-ABCC=DD^2-AB^2=(DD-AB)(DD+AB)$.
Therefore $DD+AB=100$, and as $B=2$, $D=8$ and $A=1$.
Finally:
Both $1200=12^2$ and $8800-88^2$ equal $1056$. By inspection, we notice $1089$ is the next square, and so $C=3$.
The rest:
$E=0$ ($E=5$ is the only other option, but there is no $F$ that works) AND $F=9$. $G+H=9$, and only $4,5,6,7$ remain, and by inspection $G=5, H=4$.
answered Jun 3 at 5:42
JonMark PerryJonMark Perry
22.3k643103
$endgroup$
$begingroup$
Great!!........
$endgroup$
– Uvc
Jun 3 at 7:08
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
oldest
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active
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votes
$begingroup$
Set 1:
$1233 = 12^2 + 33^2$
$8833 = 88^2 + 33^2$
Because:
We can deduce $B=2$. It must be $2$ or $3$ as $11^4gt9999$. If $B=3$, then either $A$ or $C$ is at least $2$, and $22^3>9999$.
I like this bit!
$DDCC-ABCC = 100(DD-AB)$.
But also $DDCC-ABCC=DD^2-AB^2=(DD-AB)(DD+AB)$.
Therefore $DD+AB=100$, and as $B=2$, $D=8$ and $A=1$.
Finally:
Both $1200=12^2$ and $8800-88^2$ equal $1056$. By inspection, we notice $1089$ is the next square, and so $C=3$.
The rest:
$E=0$ ($E=5$ is the only other option, but there is no $F$ that works) AND $F=9$. $G+H=9$, and only $4,5,6,7$ remain, and by inspection $G=5, H=4$.
answered Jun 3 at 5:42
JonMark PerryJonMark Perry
22.3k643103
$endgroup$
$begingroup$
Great!!........
$endgroup$
– Uvc
Jun 3 at 7:08
add a comment |
$begingroup$
Set 1:
$1233 = 12^2 + 33^2$
$8833 = 88^2 + 33^2$
Because:
We can deduce $B=2$. It must be $2$ or $3$ as $11^4gt9999$. If $B=3$, then either $A$ or $C$ is at least $2$, and $22^3>9999$.
I like this bit!
$DDCC-ABCC = 100(DD-AB)$.
But also $DDCC-ABCC=DD^2-AB^2=(DD-AB)(DD+AB)$.
Therefore $DD+AB=100$, and as $B=2$, $D=8$ and $A=1$.
Finally:
Both $1200=12^2$ and $8800-88^2$ equal $1056$. By inspection, we notice $1089$ is the next square, and so $C=3$.
The rest:
$E=0$ ($E=5$ is the only other option, but there is no $F$ that works) AND $F=9$. $G+H=9$, and only $4,5,6,7$ remain, and by inspection $G=5, H=4$.
answered Jun 3 at 5:42
JonMark PerryJonMark Perry
22.3k643103
$endgroup$
$begingroup$
Great!!........
$endgroup$
– Uvc
Jun 3 at 7:08
add a comment |
$begingroup$
Set 1:
$1233 = 12^2 + 33^2$
$8833 = 88^2 + 33^2$
Because:
We can deduce $B=2$. It must be $2$ or $3$ as $11^4gt9999$. If $B=3$, then either $A$ or $C$ is at least $2$, and $22^3>9999$.
I like this bit!
$DDCC-ABCC = 100(DD-AB)$.
But also $DDCC-ABCC=DD^2-AB^2=(DD-AB)(DD+AB)$.
Therefore $DD+AB=100$, and as $B=2$, $D=8$ and $A=1$.
Finally:
Both $1200=12^2$ and $8800-88^2$ equal $1056$. By inspection, we notice $1089$ is the next square, and so $C=3$.
The rest:
$E=0$ ($E=5$ is the only other option, but there is no $F$ that works) AND $F=9$. $G+H=9$, and only $4,5,6,7$ remain, and by inspection $G=5, H=4$.
answered Jun 3 at 5:42
JonMark PerryJonMark Perry
22.3k643103
$endgroup$
Set 1:
$1233 = 12^2 + 33^2$
$8833 = 88^2 + 33^2$
Because:
We can deduce $B=2$. It must be $2$ or $3$ as $11^4gt9999$. If $B=3$, then either $A$ or $C$ is at least $2$, and $22^3>9999$.
I like this bit!
$DDCC-ABCC = 100(DD-AB)$.
But also $DDCC-ABCC=DD^2-AB^2=(DD-AB)(DD+AB)$.
Therefore $DD+AB=100$, and as $B=2$, $D=8$ and $A=1$.
Finally:
Both $1200=12^2$ and $8800-88^2$ equal $1056$. By inspection, we notice $1089$ is the next square, and so $C=3$.
The rest:
$E=0$ ($E=5$ is the only other option, but there is no $F$ that works) AND $F=9$. $G+H=9$, and only $4,5,6,7$ remain, and by inspection $G=5, H=4$.
answered Jun 3 at 5:42
JonMark PerryJonMark Perry
22.3k643103
edited Jun 3 at 5:54
answered Jun 3 at 5:42
JonMark PerryJonMark Perry
22.3k643103
answered Jun 3 at 5:42
JonMark PerryJonMark Perry
22.3k643103
answered Jun 3 at 5:42
JonMark PerryJonMark Perry
22.3k643103
22.3k643103
$begingroup$
Great!!........
$endgroup$
– Uvc
Jun 3 at 7:08
add a comment |
$begingroup$
Great!!........
$endgroup$
– Uvc
Jun 3 at 7:08
$begingroup$
Great!!........
$endgroup$
– Uvc
Jun 3 at 7:08
$begingroup$
Great!!........
$endgroup$
– Uvc
Jun 3 at 7:08
add a comment |
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$begingroup$
Is $AB^B$ = $(AB)^B$ or $A(B^B)$? E.g. if $A=1$ and $B=2$, then is $AB^B$ equal to $(12)^2$ or to $10 + 2^2$?
$endgroup$
– shoover
Jun 3 at 5:07
$begingroup$
All letter combinations other than single letter used are concatenated Numbers as stated.
$endgroup$
– Uvc
Jun 3 at 7:01