Determine all possible value(s) if a positive integer N such that the two distinct
perfect powers of N will together contain each of the decimal digits 0 to 9 exactly once. None of the perfect powers of N can contain any leading zero.
2 ^ 2 = 4 2 ^ 29 = 536870912
18 ^ 3 = 5832 18 ^ 4 = 104976
69 ^ 2 = 4761 69 ^ 3 = 328509
An exhausive search was done from n = 2 to n = 99380, the largest number whose square would be under the largest pandigital, 9876543210. Each was tested to a power up to that resulting in that largest pandigital number.
The two numbers were presented lower to higher, otherwise each sum would appear twice.
DEFDBL A-Z
PRINT
ll = LOG(1023456789)
lh = LOG(9876543210#)
FOR n = 2 TO 99380
lb = LOG(n)
low = 2
high = -INT(-lh / lb)
FOR pwr1 = low TO high
v1 = INT(n ^ pwr1 + .5)
vs1$ = LTRIM$(STR$(v1))
IF LEN(vs1$) < 10 THEN
good = 1
FOR i = 1 TO LEN(vs1$) - 1
IF INSTR(MID$(vs1$, i + 1), MID$(vs1$, i, 1)) > 0 THEN good = 0: EXIT FOR
NEXT
IF good THEN
togo$ = "1023456789"
FOR i = 1 TO LEN(vs1$)
ix = INSTR(togo$, MID$(vs1$, i, 1))
IF ix > 0 THEN
togo$ = LEFT$(togo$, ix - 1) + MID$(togo$, ix + 1)
END IF
NEXT
IF LEN(togo$) > 1 THEN
IF LEFT$(togo$, 1) = "0" THEN
MID$(togo$, 1, 1) = MID$(togo$, 2, 1)
MID$(togo$, 2, 1) = "0"
END IF
END IF
IF togo$ = "0" THEN GOTO notThis
low2 = INT(LOG(VAL(togo$)) / lb + .5)
togoh$ = "9876543210"
FOR i = 1 TO LEN(vs1$)
ix = INSTR(togoh$, MID$(vs1$, i, 1))
IF ix > 0 THEN
togoh$ = LEFT$(togoh$, ix - 1) + MID$(togoh$, ix + 1)
END IF
NEXT
high2 = -INT(-LOG(VAL(togoh$)) / lb + .5)
FOR pwr2 = low2 TO high2
v2 = INT(n ^ pwr2 + .5)
vs2$ = LTRIM$(STR$(v2))
vAll$ = vs1$ + vs2$
IF LEN(vAll$) = 10 THEN
good2 = 1
FOR i = 1 TO 9
IF INSTR(MID$(vAll$, i + 1), MID$(vAll$, i, 1)) > 0 THEN
good2 = 0: EXIT FOR
END IF
NEXT
IF good2 THEN
IF pwr2 >= pwr1 THEN
PRINT n; "^"; pwr1; "="; v1; TAB(20); n; "^"; pwr2; "="; v2: ct = ct + 1
END IF
END IF
END IF
NEXT
notThis:
END IF
END IF
NEXT
NEXT
PRINT ct
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Posted by Charlie
on 2009-03-23 11:54:31 |
Basic computer solution
