$Given$:
$AB$, $DBCE$,$AGFPQR$ are three concatenated numbers with all distinct digits varying from zero to nine.
$AB^C$ = $DBCE$
$AB^F$ = $AGFPQR$
Deduce all the digits through logical reasoning only.
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The letters are
A = 1, B = 8 ,C = 3, D = 5, E = 2, F = 4, G = 2, P = 9, Q = 7, R = 6
Explanation
A at the ten's place can't be zero. C and F being exponents can't be zero. B, E, R at the one's place can't be zero.
So, out of G, P and Q one of them is zero.
Working from, $$AB^C = DBCE$$ I started with the power $3$.
$21^3 = 9261$ is the maximum value with 4 digits. So from 21 in back, $20^3 = 8000$ is not possible so is 19. $18^3 = 5832$ seems as a possible solution.
Now $$18^4 = 104976$$
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$\begingroup$ Accept after a while, so as to get other puzzlers' attention : ) $\endgroup$ – Ak. Jul 7 '19 at 10:17
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