Question 1173182: There are three triples of positive integers (a,b,c, d,e,f, and g,h,i) such that a^2+b^2+c^2 = d^2+e^2+f^2 = g^2+h^2+i^2 = 101. Evaluate the expression abc+def+ghi.
I've found three
1. 101 = 1^2 + 6^2 + 8^2
2. 101 = 2^2 + 4^2 + 9^2
3. 101 = 4^2 + 6^2 + 7^2
They all share terms so I don't think these are options.
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
Three triples are
(a,b,c) = (1,6,8) : 1^2 + 6^2 + 8^2 = 101
(d,e,f) = (2,4,9) : 2^2 + 4^2 + 9^2 = 101
(g,h,i) = (4,6,7) : 4^2 + 6^2 + 7^2 = 101
The sum abc + def + ghi = 1*6*8 + 2*4*9 + 4*6*7 = 288. ANSWER
Solved.
Answer by Edwin McCravy(20055)  (Show Source):
You can put this solution on YOUR website!
Triples that share 1 or 2 numbers are still DIFFERENT triples. Triples would have
to share ALL THREE numbers to be the same triple.
Edwin
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