Digit fifth powers
- Task
Task desciption is taken from Project Eulet(https://projecteuler.net/problem=30)
Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
Ring
<lang ring> see "working..." + nl
sumEnd = 0 sumList = [] limitStart = 10000 limitEnd = 199999
for n = limitStart to limitEnd
sum = 0 nStr = string(n) for m = 1 to len(nStr) sum = sum + pow(number(nStr[m]),5) next if sum = n add(sumList,n) sumEnd += n ok
next
see "The sum of all the numbers that can be written as the sum of fifth powers of their digits:" + nl for n = 1 to len(sumList)-1
see "" + sumList[n] + " + "
next see "" + sumList[n] + " = " + sumEnd + nl see "done..." + nl </lang>
- Output:
working... The sum of all the numbers that can be written as the sum of fifth powers of their digits: 54748 + 92727 + 93084 + 194979 = 435538 done...
XPL0
Since 1 is not actually a sum, it should not be included. Thus the answer should be 443839. <lang XPL0>\upper bound: 6*9^5 = 354294 \7*9^5 is still only a 6-digit number, so 6 digits are sufficient
int A, B, C, D, E, F, \digits, A=LSD
A5, B5, C5, D5, E5, F5, \digits to 5th power A0, B0, C0, D0, E0, F0, \digits multiplied by their decimal place N, \number that can be written as the sum of its 5th pwrs S; \sum of all numbers
[S:= 0;
for A:= 0, 9 do \for all digits
[A5:= A*A*A*A*A; A0:= A; for B:= 0, 9 do [B5:= B*B*B*B*B; B0:= B*10; for C:= 0, 9 do [C5:= C*C*C*C*C; C0:= C*100; for D:= 0, 9 do [D5:= D*D*D*D*D; D0:= D*1000; for E:= 0, 9 do [E5:= E*E*E*E*E; E0:= E*10000; for F:= 0, 3 do [F5:= F*F*F*F*F; F0:= F*100000; [N:= F0 + E0 + D0 + C0 + B0 + A0; if N = A5 + B5 + C5 + D5 + E5 + F5 then [S:= S + N; IntOut(0, N); CrLf(0); ]; ]; ]; ]; ]; ]; ]; ];
CrLf(0); IntOut(0, S); CrLf(0); ]</lang>
- Output:
0 4150 1 4151 93084 92727 54748 194979 443840