$$a$$, $$b$$, $$c$$, $$d$$ are digits in 2014 such that $$\{a, b, c, d\} = \{2, 0, 1, 4\}$$. Then the number of different values $$\left(\left(a^b\right)^c\right)^d$$ takes is

Solution

$$\left(\left(a^b\right)^c\right)^d = a^{bcd}$$. If $$a = 0$$, the exponent is $$2 \times 1 \times 4 = 8$$ and the value is 0. If $$a$$ is 1, 2 or 4, then 0 is one of $$b$$, $$c$$, $$d$$, so the exponent is 0 and the value is 1. Only the values 0 and 1 occur, so there are 2 different values.

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