The Power of Math

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 $$(T+H+E)^3=THE$$

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

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:

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

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.

• 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 Apr 28, 2023 at 12:37