# Pan Digital Split among Two Powers

$$Given$$:

$$AB$$, $$DBCE$$,$$AGFPQR$$ 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.

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 $$18^4 = 104976$$

• That’s right... – Uvc Jul 7 at 10:17
• Accept after a while, so as to get other puzzlers' attention : ) – Ak19 Jul 7 at 10:17
• Sure............ – Uvc Jul 7 at 10:19