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If $(M+A+T+H)^4=MATH,$ then what are M, A, T, and H?
See if you can also find it for $(P+O+W+E+R)^5=POWER$ (I don't think that it has a solution) and $(T+H+E)^3=THE$

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  • $\begingroup$ Just to clarify, in $MATH$, those are digits in a base-10 number? Or multiplied together? $\endgroup$
    – Owen
    Apr 29, 2023 at 12:20

2 Answers 2

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For MATH:

Assuming that M, A, T and H are between 0 and 9, we just need to find the 4th powers with 4 digits :

$6^4 = 1296$ $\Rightarrow$ 1+2+9+6 = 18
$7^4 = 2401$ $\Rightarrow$ 2+4+0+1 = 7
$8^4 = 4096$ $\Rightarrow$ 4+0+9+6 = 19
$9^4 = 6561$ $\Rightarrow$ 6+5+6+1 = 18

So M = 2, A = 4, T = 0 and H = 1

For POWER:

We just need to find the 5th powers with 5 digits:

$7^5 = 16807$ $\Rightarrow$ 1+6+8+0+7 = 22
$8^5 = 32768$ $\Rightarrow$ 3+2+7+6+8 = 26
$9^5 = 59049$ $\Rightarrow$ 5+9+0+4+9 = 27

So it's not possible.

For THE:

We just need to find the cubes with 3 digits:

$5^3 = 125$ $\Rightarrow$ 1+2+5 = 8
$6^3 = 216$ $\Rightarrow$ 2+1+6 = 9
$7^3 = 343$ $\Rightarrow$ 3+4+3 = 10
$8^3 = 512$ $\Rightarrow$ 5+1+2 = 8
$9^3 = 729$ $\Rightarrow$ 7+2+9 = 18

So T = 5, H = 1 and E = 2

Bonus:

$\forall n \in \mathbb{N}^*, (0 + 0 + 0 + ... + 0 + 0 + 0)^n = 000...000$

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2
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Bonus for OF, as in the power OF math:

We just need to find squares with 2 digits:

$4^2 = 16$ $\Rightarrow$ 1+6 = 7
$5^2 = 25$ $\Rightarrow$ 2+5 = 7
$6^2 = 36$ $\Rightarrow$ 3+6 = 9
$7^2 = 49$ $\Rightarrow$ 4+9 = 13
$8^2 = 64$ $\Rightarrow$ 6+4 = 10
$9^2 = 81$ $\Rightarrow$ 8+1 = 9

So O = 8 and F = 1.

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  • 3
    $\begingroup$ Of Alley used to be a London street in an area once owned by "George Villiers, Duke Of Buckingham" and each street took part of his name $\endgroup$
    – Henry
    Apr 28, 2023 at 12:37

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