6
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Solve this magical equation:

$$ (M+A+G+I+C) \times (M+A+G+I+C) \times (M+A+G+I+C) = \overline{MAGIC} $$

Each letter represents a separate digit.

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12
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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 ($27$)

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  • 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 '19 at 17:51
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(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:
$$\begin{array} \\ Number & Cube & Sum & Solution \\ 22 & 10648 & 19 & No \\ 23 & 12167 & 17 & No \\ 24 & 13824 & 18 & No \\ 25 & 15625 & 19 & No \\ 26 & 17576 & 26 & Yes! \\ 27 & 19683 & 27 & Yes! \\ 28 & 21952 & 19 & No \\ 29 & 24389 & 26 & No \\ 30 & 27000 & 9 & No \\ 31 & 29791 & 28 & No \\ 32 & 32768 & 26 & No \\ 33 & 35937 & 27 & No \\ 34 & 39304 & 19 & No \\ 35 & 42875 & 26 & No \\ \end{array} $$

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

$$27^3=19683$$

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  • $\begingroup$ 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 May 4 '19 at 4:18
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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.

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