Determine all possible triplets (a,b,c) of positive integers, with a>b>c, such that:
- a+bc < b+ca < c+ab are three expressions in arithmetic sequence, and:
- a+b-c=22
The 3 expressions are in arithmetic progression, permitting an equality.
Say the difference is r, then: a+bc+r = b+ca = c+ab-r
Get rid of c=(a+b-22), then : (a-23)a=(b-23)b+r, b^2+2r+22=23b, a^2+r+22=23a
Note: From a+bc < b+ca < c+ab, a+b-c=22, it follows that 1<a<22
Thus b^2+2r+22=23b, a^2+r+22=23a, 0<b<a<22 actually has only 2 solutions: {a,b,r} = {19,10,54}{19,13,54}.
(The range is helpful because otherwise there are infinitely many.)
19+bc+54 = b+c*19 = c+19b-54 gives {b,c} = {10,7}{13,10}
So the full solution set is {a,b,c,r} = {19,10,7,54}(19,13,10,54}
Nice problem.
Edited on September 22, 2022, 12:21 am
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Posted by broll
on 2022-09-22 00:09:17 |
Possible Solution
