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I have not dealt with problems of this type:

Three integers $a$, $b$, and $c$ are written on a blackboard. Then one of them is erased and replaced by the sum of the other two diminished by 1. This operation is repeated a finite number of times until we have the final result, which is $17$, $\,1967$, and $1983$.

Could $a,b,c$ have been $(2,2,2)$ i.e. starting from $2$, $2$, $2$, can we reach $17$, $\,1967$, and $1983$?

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A general strategy for these sort of problems is to look for some invariant i.e. making use of the three numbers, find a quantity such that the quantity remains unchanged under the specified operations. – user17762 Oct 4 '12 at 5:08
The operation is somewhat more transparent if you systematically replace any number $a$ by $a-1$: then a (transformed) number is replaced by the sum of the two others. – Marc van Leeuwen Oct 4 '12 at 6:46
up vote 6 down vote accepted

HINT: The first step from $(2,2,2)$ must give you $(2,2,3)$. Now prove that every triple that you can reach from here has two even numbers and one odd number.

In attacking this problem I looked at the triples that could be reached from $(2,2,2)$ in three or four steps and looked for something that they had in common. As Marvis suggested in the comments, you want to find something that’s preserved by the operation, and in this case it turns out to be something very simple.

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So, I was barking up the wrong tree. What I tried to do was take this $(17,k_1,k_2)$ and notice that the sum increases by 32.But then, it was not particularly helpful.It turns out it was something much simpler. – user37450 Oct 4 '12 at 5:15

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