There are three sixdigit numbers such that each is a 4th power of its sum of digits .
1. List them.
2. Each one has a certain peculiarity (or more than one). Try to define it.
3. Ignoring the 6digits constraint , how many integers like that exist?
DECLARE FUNCTION sod# (n#)
DEFDBL AZ
FOR n = 0 TO 26873856 ' (8*9)^4
s = sod(n)
s = s * s * s * s
IF s = n THEN PRINT n
NEXT n
FUNCTION sod (n)
st$ = LTRIM$(STR$(n))
t = 0
FOR i = 1 TO LEN(st$)
t = t + VAL(MID$(st$, i, 1))
NEXT
sod = t
END FUNCTION
finds
0
1
2401
234256
390625
614656
1679616
By inspection you can see the three 6digit solutions.
There can't be any more as the largest sod for an 8digit number is 8*9; when raised to the 4th power, it's 26873856. Then (9*9)^4 and (10*9 )^4 are also only 8digit numbers so no 9 or 10digit number will work, nor any larger number as the number of digits in the power will always be smaller than the number of digits in the original number.

Posted by Charlie
on 20120814 15:24:35 