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 Pandigital and Pretty Powerful III (Posted on 2009-05-12)
Determine all possible triplet(s) (P, Q, N) of positive integers, with P < Q and N ≥ 3, such that the decimal representations of PN and QN will together contain each of the digits 0 to 9 exactly once. Neither PN nor QN can contain any leading zero.

 See The Solution Submitted by K Sengupta No Rating

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 Version in Basic Comment 4 of 4 |
Since 2^36 exceeds any 10-digit number we need check no higher n.

Once p^n exceeds a 5-digit number for a given n, we need not check that nor succeeding p, as q is larger and the two powers together will have more than 10 digits.

The following program checks all the possibilities within those constraints:

DEFDBL A-Z
CLS
FOR n = 3 TO 36
p = 0
DO
p = p + 1
ppwr = INT(p ^ n + .5)
part1\$ = LTRIM\$(STR\$(ppwr))
IF LEN(part1\$) > 5 THEN EXIT DO
good = 1
FOR i = 1 TO LEN(part1\$) - 1
IF INSTR(i + 1, part1\$, MID\$(part1\$, i, 1)) > 0 THEN good = 0: EXIT FOR
NEXT
IF good THEN
q = p
DO
q = q + 1
qpwr = INT(q ^ n + .5)
part2\$ = LTRIM\$(STR\$(qpwr))
w\$ = part1\$ + part2\$
IF LEN(w\$) > 10 THEN EXIT DO
good = 1
FOR i = 1 TO LEN(w\$) - 1
IF INSTR(i + 1, w\$, MID\$(w\$, i, 1)) > 0 THEN good = 0: EXIT FOR
NEXT
IF good AND LEN(w\$) = 10 THEN PRINT p, q, n, part1\$; " "; part2\$
LOOP
END IF
LOOP

NEXT

finding only

21            93            3            9261 804357

meaning (P, Q, N) = (21, 93, 3)
and 21^3 = 9261 and 93^3 = 804357.

 Posted by Charlie on 2009-05-12 15:11:21

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