Find magical solution to magical equationLagrange's Four Square CryptarithmBaseball games alphameticWheat field alphameticLetters = numbersAlphametic with moduloSolve XAB + YZCD = ZEXYOne year celebrationDecipher this puzzleCITEMAHPLA Reverse AlphameticNo spanish required
Magical Modulo Squares
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Find magical solution to magical equation
Lagrange's Four Square CryptarithmBaseball games alphameticWheat field alphameticLetters = numbersAlphametic with moduloSolve XAB + YZCD = ZEXYOne year celebrationDecipher this puzzleCITEMAHPLA Reverse AlphameticNo spanish required
$begingroup$
Solve this magical equation...
(M+A+G+I+C) x (M+A+G+I+C) x (M+A+G+I+C) = MAGIC.
Each letter represents a separate digit.
alphametic
edited May 3 at 16:09
GentlePurpleRain♦
17.3k569138
asked May 3 at 15:39
UvcUvc
1804
New contributor
Uvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Solve this magical equation...
(M+A+G+I+C) x (M+A+G+I+C) x (M+A+G+I+C) = MAGIC.
Each letter represents a separate digit.
alphametic
edited May 3 at 16:09
GentlePurpleRain♦
17.3k569138
asked May 3 at 15:39
UvcUvc
1804
New contributor
Uvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Solve this magical equation...
(M+A+G+I+C) x (M+A+G+I+C) x (M+A+G+I+C) = MAGIC.
Each letter represents a separate digit.
alphametic
edited May 3 at 16:09
GentlePurpleRain♦
17.3k569138
asked May 3 at 15:39
UvcUvc
1804
New contributor
Uvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Solve this magical equation...
(M+A+G+I+C) x (M+A+G+I+C) x (M+A+G+I+C) = MAGIC.
Each letter represents a separate digit.
alphametic
alphametic
edited May 3 at 16:09
GentlePurpleRain♦
17.3k569138
asked May 3 at 15:39
UvcUvc
1804
New contributor
Uvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited May 3 at 16:09
GentlePurpleRain♦
17.3k569138
asked May 3 at 15:39
UvcUvc
1804
New contributor
Uvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited May 3 at 16:09
GentlePurpleRain♦
17.3k569138
edited May 3 at 16:09
GentlePurpleRain♦
17.3k569138
edited May 3 at 16:09
GentlePurpleRain♦
17.3k569138
17.3k569138
asked May 3 at 15:39
UvcUvc
1804
New contributor
Uvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked May 3 at 15:39
UvcUvc
1804
asked May 3 at 15:39
UvcUvc
1804
1804
New contributor
Uvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Uvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Answer
$M=1, A=9, G=6, I=8, C=3$
Method
The equation simplifies to $(M+A+G+I+C)^3 = MAGIC$. The term in brackets is at most $45$ and must be at least $22$ for the cube to have five digits. It also makes sense to restrict to the case where all the digits are distinct. This happens for the cubes of $22, 24, 27, 29, 32, 35, 38, 41$. Among these only the digits in the cube of $27$ add up to the number itself ($27$)
answered May 3 at 15:44
hexominohexomino
48.8k4144230
$endgroup$
1
$begingroup$
For the interested: The answer could also be 26 (26^3 = 17576, 1+7+5+7+6=26) if it weren't for the restriction that each digit be different.
$endgroup$
– Engineer Toast
May 3 at 17:51
add a comment |
$begingroup$
(M+A+G+I+C) x (M+A+G+I+C) x (M+A+G+I+C) = MAGIC
Assumptions:
- $M ne 0$ because that would make a 5 digit number starting with $0$.
- All the digits of $MAGIC$ are unique
Let $S=M+A+G+I+C$. The cube of $S$ is a 5 digit number. Since $21 lt sqrt[3]10000 lt 22$ and $46 lt sqrt[3]100000 lt 47$, we know that $22 le S le 46$. But the maximum sum for 5 different digits is $9+8+7+6+5=35$. Thus, we can further restrict the range to $22 le S le 35$.
There are now
12 numbers that we need to check:
$$beginarray \ Number & Cube & Sum & Solution \ 22 & 10648 & 19 & No \ 23 & 12167 & 17 & No \ 24 & 13824 & 18 & No \ 25 & 15625 & 19 & No \ 26 & 17576 & 26 & Yes! \ 27 & 19683 & 27 & Yes! \ 28 & 21952 & 19 & No \ 29 & 24389 & 26 & No \ 30 & 27000 & 9 & No \ 31 & 29791 & 28 & No \ 32 & 32768 & 26 & No \ 33 & 35937 & 27 & No \ 34 & 39304 & 19 & No \ 35 & 42875 & 26 & No \ endarray $$
So there are ...
2 solutions! But if you look at the $S=26, MAGIC=17576$, we see that $A=I=7$ has a repeated digit.
Thus, the only valid solution is:
$$MAGIC=19683$$
The sum is then
$$M+A+G+I+C=1+9+6+8+3=27$$
And the cube is
$$27^3=19683$$
edited 2 days ago
Rubio♦
31.1k668189
answered May 3 at 19:01
TreninTrenin
7,5541644
$endgroup$
$begingroup$
Please, use spoilers. Also, while more verbose, this follows more or less exactly the same path of @hexonimo's answer, covering the same steps just in more words. As a general suggestion: if you're posting an answer that's largely the same as an existing one, acknowledge the prior answer and indicate how yours differs, improves upon, or adds relevant detail to the answer already provided; if you can't really explain why your answer is not essentially a duplicate, that's probably a good sign it isn't adding anything to what's already been said, and shouldn't be posted.
$endgroup$
– Rubio♦
2 days ago
add a comment |
$begingroup$
Answer is
MAGIC = 19683
and
M+A+G+I+C = 27
SOLUTION
(M+A+G+I+C)^3 = MAGIC
22 is the first one that give 5 digit cube.
So checked for each number above 22 and 27 satisfied the equation MAGIC=(M+A+G+I+C)^3.
answered May 3 at 15:48
AeJeyAeJey
11.4k248111
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Answer
$M=1, A=9, G=6, I=8, C=3$
Method
The equation simplifies to $(M+A+G+I+C)^3 = MAGIC$. The term in brackets is at most $45$ and must be at least $22$ for the cube to have five digits. It also makes sense to restrict to the case where all the digits are distinct. This happens for the cubes of $22, 24, 27, 29, 32, 35, 38, 41$. Among these only the digits in the cube of $27$ add up to the number itself ($27$)
answered May 3 at 15:44
hexominohexomino
48.8k4144230
$endgroup$
1
$begingroup$
For the interested: The answer could also be 26 (26^3 = 17576, 1+7+5+7+6=26) if it weren't for the restriction that each digit be different.
$endgroup$
– Engineer Toast
May 3 at 17:51
add a comment |
$begingroup$
Answer
$M=1, A=9, G=6, I=8, C=3$
Method
The equation simplifies to $(M+A+G+I+C)^3 = MAGIC$. The term in brackets is at most $45$ and must be at least $22$ for the cube to have five digits. It also makes sense to restrict to the case where all the digits are distinct. This happens for the cubes of $22, 24, 27, 29, 32, 35, 38, 41$. Among these only the digits in the cube of $27$ add up to the number itself ($27$)
answered May 3 at 15:44
hexominohexomino
48.8k4144230
$endgroup$
1
$begingroup$
For the interested: The answer could also be 26 (26^3 = 17576, 1+7+5+7+6=26) if it weren't for the restriction that each digit be different.
$endgroup$
– Engineer Toast
May 3 at 17:51
add a comment |
$begingroup$
Answer
$M=1, A=9, G=6, I=8, C=3$
Method
The equation simplifies to $(M+A+G+I+C)^3 = MAGIC$. The term in brackets is at most $45$ and must be at least $22$ for the cube to have five digits. It also makes sense to restrict to the case where all the digits are distinct. This happens for the cubes of $22, 24, 27, 29, 32, 35, 38, 41$. Among these only the digits in the cube of $27$ add up to the number itself ($27$)
answered May 3 at 15:44
hexominohexomino
48.8k4144230
$endgroup$
Answer
$M=1, A=9, G=6, I=8, C=3$
Method
The equation simplifies to $(M+A+G+I+C)^3 = MAGIC$. The term in brackets is at most $45$ and must be at least $22$ for the cube to have five digits. It also makes sense to restrict to the case where all the digits are distinct. This happens for the cubes of $22, 24, 27, 29, 32, 35, 38, 41$. Among these only the digits in the cube of $27$ add up to the number itself ($27$)
answered May 3 at 15:44
hexominohexomino
48.8k4144230
edited May 3 at 15:50
answered May 3 at 15:44
hexominohexomino
48.8k4144230
answered May 3 at 15:44
hexominohexomino
48.8k4144230
answered May 3 at 15:44
hexominohexomino
48.8k4144230
48.8k4144230
1
$begingroup$
For the interested: The answer could also be 26 (26^3 = 17576, 1+7+5+7+6=26) if it weren't for the restriction that each digit be different.
$endgroup$
– Engineer Toast
May 3 at 17:51
add a comment |
1
$begingroup$
For the interested: The answer could also be 26 (26^3 = 17576, 1+7+5+7+6=26) if it weren't for the restriction that each digit be different.
$endgroup$
– Engineer Toast
May 3 at 17:51
1
1
$begingroup$
For the interested: The answer could also be 26 (26^3 = 17576, 1+7+5+7+6=26) if it weren't for the restriction that each digit be different.
$endgroup$
– Engineer Toast
May 3 at 17:51
$begingroup$
For the interested: The answer could also be 26 (26^3 = 17576, 1+7+5+7+6=26) if it weren't for the restriction that each digit be different.
$endgroup$
– Engineer Toast
May 3 at 17:51
add a comment |
$begingroup$
(M+A+G+I+C) x (M+A+G+I+C) x (M+A+G+I+C) = MAGIC
Assumptions:
- $M ne 0$ because that would make a 5 digit number starting with $0$.
- All the digits of $MAGIC$ are unique
Let $S=M+A+G+I+C$. The cube of $S$ is a 5 digit number. Since $21 lt sqrt[3]10000 lt 22$ and $46 lt sqrt[3]100000 lt 47$, we know that $22 le S le 46$. But the maximum sum for 5 different digits is $9+8+7+6+5=35$. Thus, we can further restrict the range to $22 le S le 35$.
There are now
12 numbers that we need to check:
$$beginarray \ Number & Cube & Sum & Solution \ 22 & 10648 & 19 & No \ 23 & 12167 & 17 & No \ 24 & 13824 & 18 & No \ 25 & 15625 & 19 & No \ 26 & 17576 & 26 & Yes! \ 27 & 19683 & 27 & Yes! \ 28 & 21952 & 19 & No \ 29 & 24389 & 26 & No \ 30 & 27000 & 9 & No \ 31 & 29791 & 28 & No \ 32 & 32768 & 26 & No \ 33 & 35937 & 27 & No \ 34 & 39304 & 19 & No \ 35 & 42875 & 26 & No \ endarray $$
So there are ...
2 solutions! But if you look at the $S=26, MAGIC=17576$, we see that $A=I=7$ has a repeated digit.
Thus, the only valid solution is:
$$MAGIC=19683$$
The sum is then
$$M+A+G+I+C=1+9+6+8+3=27$$
And the cube is
$$27^3=19683$$
edited 2 days ago
Rubio♦
31.1k668189
answered May 3 at 19:01
TreninTrenin
7,5541644
$endgroup$
$begingroup$
Please, use spoilers. Also, while more verbose, this follows more or less exactly the same path of @hexonimo's answer, covering the same steps just in more words. As a general suggestion: if you're posting an answer that's largely the same as an existing one, acknowledge the prior answer and indicate how yours differs, improves upon, or adds relevant detail to the answer already provided; if you can't really explain why your answer is not essentially a duplicate, that's probably a good sign it isn't adding anything to what's already been said, and shouldn't be posted.
$endgroup$
– Rubio♦
2 days ago
add a comment |
$begingroup$
(M+A+G+I+C) x (M+A+G+I+C) x (M+A+G+I+C) = MAGIC
Assumptions:
- $M ne 0$ because that would make a 5 digit number starting with $0$.
- All the digits of $MAGIC$ are unique
Let $S=M+A+G+I+C$. The cube of $S$ is a 5 digit number. Since $21 lt sqrt[3]10000 lt 22$ and $46 lt sqrt[3]100000 lt 47$, we know that $22 le S le 46$. But the maximum sum for 5 different digits is $9+8+7+6+5=35$. Thus, we can further restrict the range to $22 le S le 35$.
There are now
12 numbers that we need to check:
$$beginarray \ Number & Cube & Sum & Solution \ 22 & 10648 & 19 & No \ 23 & 12167 & 17 & No \ 24 & 13824 & 18 & No \ 25 & 15625 & 19 & No \ 26 & 17576 & 26 & Yes! \ 27 & 19683 & 27 & Yes! \ 28 & 21952 & 19 & No \ 29 & 24389 & 26 & No \ 30 & 27000 & 9 & No \ 31 & 29791 & 28 & No \ 32 & 32768 & 26 & No \ 33 & 35937 & 27 & No \ 34 & 39304 & 19 & No \ 35 & 42875 & 26 & No \ endarray $$
So there are ...
2 solutions! But if you look at the $S=26, MAGIC=17576$, we see that $A=I=7$ has a repeated digit.
Thus, the only valid solution is:
$$MAGIC=19683$$
The sum is then
$$M+A+G+I+C=1+9+6+8+3=27$$
And the cube is
$$27^3=19683$$
edited 2 days ago
Rubio♦
31.1k668189
answered May 3 at 19:01
TreninTrenin
7,5541644
$endgroup$
$begingroup$
Please, use spoilers. Also, while more verbose, this follows more or less exactly the same path of @hexonimo's answer, covering the same steps just in more words. As a general suggestion: if you're posting an answer that's largely the same as an existing one, acknowledge the prior answer and indicate how yours differs, improves upon, or adds relevant detail to the answer already provided; if you can't really explain why your answer is not essentially a duplicate, that's probably a good sign it isn't adding anything to what's already been said, and shouldn't be posted.
$endgroup$
– Rubio♦
2 days ago
add a comment |
$begingroup$
(M+A+G+I+C) x (M+A+G+I+C) x (M+A+G+I+C) = MAGIC
Assumptions:
- $M ne 0$ because that would make a 5 digit number starting with $0$.
- All the digits of $MAGIC$ are unique
Let $S=M+A+G+I+C$. The cube of $S$ is a 5 digit number. Since $21 lt sqrt[3]10000 lt 22$ and $46 lt sqrt[3]100000 lt 47$, we know that $22 le S le 46$. But the maximum sum for 5 different digits is $9+8+7+6+5=35$. Thus, we can further restrict the range to $22 le S le 35$.
There are now
12 numbers that we need to check:
$$beginarray \ Number & Cube & Sum & Solution \ 22 & 10648 & 19 & No \ 23 & 12167 & 17 & No \ 24 & 13824 & 18 & No \ 25 & 15625 & 19 & No \ 26 & 17576 & 26 & Yes! \ 27 & 19683 & 27 & Yes! \ 28 & 21952 & 19 & No \ 29 & 24389 & 26 & No \ 30 & 27000 & 9 & No \ 31 & 29791 & 28 & No \ 32 & 32768 & 26 & No \ 33 & 35937 & 27 & No \ 34 & 39304 & 19 & No \ 35 & 42875 & 26 & No \ endarray $$
So there are ...
2 solutions! But if you look at the $S=26, MAGIC=17576$, we see that $A=I=7$ has a repeated digit.
Thus, the only valid solution is:
$$MAGIC=19683$$
The sum is then
$$M+A+G+I+C=1+9+6+8+3=27$$
And the cube is
$$27^3=19683$$
edited 2 days ago
Rubio♦
31.1k668189
answered May 3 at 19:01
TreninTrenin
7,5541644
$endgroup$
(M+A+G+I+C) x (M+A+G+I+C) x (M+A+G+I+C) = MAGIC
Assumptions:
- $M ne 0$ because that would make a 5 digit number starting with $0$.
- All the digits of $MAGIC$ are unique
Let $S=M+A+G+I+C$. The cube of $S$ is a 5 digit number. Since $21 lt sqrt[3]10000 lt 22$ and $46 lt sqrt[3]100000 lt 47$, we know that $22 le S le 46$. But the maximum sum for 5 different digits is $9+8+7+6+5=35$. Thus, we can further restrict the range to $22 le S le 35$.
There are now
12 numbers that we need to check:
$$beginarray \ Number & Cube & Sum & Solution \ 22 & 10648 & 19 & No \ 23 & 12167 & 17 & No \ 24 & 13824 & 18 & No \ 25 & 15625 & 19 & No \ 26 & 17576 & 26 & Yes! \ 27 & 19683 & 27 & Yes! \ 28 & 21952 & 19 & No \ 29 & 24389 & 26 & No \ 30 & 27000 & 9 & No \ 31 & 29791 & 28 & No \ 32 & 32768 & 26 & No \ 33 & 35937 & 27 & No \ 34 & 39304 & 19 & No \ 35 & 42875 & 26 & No \ endarray $$
So there are ...
2 solutions! But if you look at the $S=26, MAGIC=17576$, we see that $A=I=7$ has a repeated digit.
Thus, the only valid solution is:
$$MAGIC=19683$$
The sum is then
$$M+A+G+I+C=1+9+6+8+3=27$$
And the cube is
$$27^3=19683$$
edited 2 days ago
Rubio♦
31.1k668189
answered May 3 at 19:01
TreninTrenin
7,5541644
edited 2 days ago
Rubio♦
31.1k668189
edited 2 days ago
Rubio♦
31.1k668189
edited 2 days ago
Rubio♦
31.1k668189
31.1k668189
answered May 3 at 19:01
TreninTrenin
7,5541644
answered May 3 at 19:01
TreninTrenin
7,5541644
answered May 3 at 19:01
TreninTrenin
7,5541644
7,5541644
$begingroup$
Please, use spoilers. Also, while more verbose, this follows more or less exactly the same path of @hexonimo's answer, covering the same steps just in more words. As a general suggestion: if you're posting an answer that's largely the same as an existing one, acknowledge the prior answer and indicate how yours differs, improves upon, or adds relevant detail to the answer already provided; if you can't really explain why your answer is not essentially a duplicate, that's probably a good sign it isn't adding anything to what's already been said, and shouldn't be posted.
$endgroup$
– Rubio♦
2 days ago
add a comment |
$begingroup$
Please, use spoilers. Also, while more verbose, this follows more or less exactly the same path of @hexonimo's answer, covering the same steps just in more words. As a general suggestion: if you're posting an answer that's largely the same as an existing one, acknowledge the prior answer and indicate how yours differs, improves upon, or adds relevant detail to the answer already provided; if you can't really explain why your answer is not essentially a duplicate, that's probably a good sign it isn't adding anything to what's already been said, and shouldn't be posted.
$endgroup$
– Rubio♦
2 days ago
$begingroup$
Please, use spoilers. Also, while more verbose, this follows more or less exactly the same path of @hexonimo's answer, covering the same steps just in more words. As a general suggestion: if you're posting an answer that's largely the same as an existing one, acknowledge the prior answer and indicate how yours differs, improves upon, or adds relevant detail to the answer already provided; if you can't really explain why your answer is not essentially a duplicate, that's probably a good sign it isn't adding anything to what's already been said, and shouldn't be posted.
$endgroup$
– Rubio♦
2 days ago
$begingroup$
Please, use spoilers. Also, while more verbose, this follows more or less exactly the same path of @hexonimo's answer, covering the same steps just in more words. As a general suggestion: if you're posting an answer that's largely the same as an existing one, acknowledge the prior answer and indicate how yours differs, improves upon, or adds relevant detail to the answer already provided; if you can't really explain why your answer is not essentially a duplicate, that's probably a good sign it isn't adding anything to what's already been said, and shouldn't be posted.
$endgroup$
– Rubio♦
2 days ago
add a comment |
$begingroup$
Answer is
MAGIC = 19683
and
M+A+G+I+C = 27
SOLUTION
(M+A+G+I+C)^3 = MAGIC
22 is the first one that give 5 digit cube.
So checked for each number above 22 and 27 satisfied the equation MAGIC=(M+A+G+I+C)^3.
answered May 3 at 15:48
AeJeyAeJey
11.4k248111
$endgroup$
add a comment |
$begingroup$
Answer is
MAGIC = 19683
and
M+A+G+I+C = 27
SOLUTION
(M+A+G+I+C)^3 = MAGIC
22 is the first one that give 5 digit cube.
So checked for each number above 22 and 27 satisfied the equation MAGIC=(M+A+G+I+C)^3.
answered May 3 at 15:48
AeJeyAeJey
11.4k248111
$endgroup$
add a comment |
$begingroup$
Answer is
MAGIC = 19683
and
M+A+G+I+C = 27
SOLUTION
(M+A+G+I+C)^3 = MAGIC
22 is the first one that give 5 digit cube.
So checked for each number above 22 and 27 satisfied the equation MAGIC=(M+A+G+I+C)^3.
answered May 3 at 15:48
AeJeyAeJey
11.4k248111
$endgroup$
Answer is
MAGIC = 19683
and
M+A+G+I+C = 27
SOLUTION
(M+A+G+I+C)^3 = MAGIC
22 is the first one that give 5 digit cube.
So checked for each number above 22 and 27 satisfied the equation MAGIC=(M+A+G+I+C)^3.
answered May 3 at 15:48
AeJeyAeJey
11.4k248111
answered May 3 at 15:48
AeJeyAeJey
11.4k248111
answered May 3 at 15:48
AeJeyAeJey
11.4k248111
answered May 3 at 15:48
AeJeyAeJey
11.4k248111
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Uvc is a new contributor. Be nice, and check out our Code of Conduct.
Uvc is a new contributor. Be nice, and check out our Code of Conduct.
Uvc is a new contributor. Be nice, and check out our Code of Conduct.
Uvc is a new contributor. Be nice, and check out our Code of Conduct.
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