Determine all possible triplets (A,B,C) of positive integers with A ≤ B ≤ C that satisfy this system of equations:
A = P(B)+ P(C), B = P(A) + P(C), and C = P(A) + P(B)
Prove that there are no others.
| No Solution Yet | Submitted by K Sengupta |
| No Rating |
results so far, but no proof in sight
|
 | Comment 1 of 7 |
So far the following (a,b,c) work: (5,5,10),(36,36,36) and (498,498,576).
The program has checked up to a+b=167903, last time I pressed a key to get a progress report.
DECLARE FUNCTION p# (x#)
DEFDBL A-Z
FOR tot = 2 TO 999999
FOR a = 1 TO tot / 2
IF INKEY$ > "" THEN PRINT tot, a
pa = p(a)
b = tot - a
IF b >= pa THEN
pb = p(b):
c = pa + pb
IF c >= b THEN
pc = p(c)
IF a = pb + pc AND b = pa + pc THEN
PRINT a, b, c
END IF
END IF
END IF
NEXT a
NEXT tot
FUNCTION p (x)
prod = 1
n = x
WHILE n > 0
prod = prod * (n MOD 10)
n = n \ 10
WEND
p = prod
END FUNCTION
| Posted by Charlie on 2013-08-04 12:53:46 |
