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 |
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re(3): results so far, but no proof in sight
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 | Comment 4 of 7 |  |
(In reply to re(2): results so far, but no proof in sight by Ady TZIDON)
I've managed to establish an upper bound for C as follows:
let C have n digits. Then A,B can have at most n digits. Thus P(A) and P(B) can be at most 9^n each. Thus
P(A)+P(B)<=2*9^n. However, for C to have n digits, we require
C>=10^(n-1)
thus
2*9^n>=P(A)+P(B)=C>=10^(n-1)
2*9^n>=10^(n-1)
20*9^n>=10^n
(10/9)^n<=20
n*log(10/9)<=log(20)
n<=log(20)/log(10/9)
n<=28
Unfortunately this upper limit does not lend itself to a timely exhaustive search. However, I have a feeling that there is more clever methods which would allow for a much smaller upper limit and thus permitting an exhaustive search.
| Posted by Daniel on 2013-08-04 15:42:58 |
