SOLUTION: There are two differnet triples of positive integers (a,b,c) and (d,e,f) such that a^2 + b^2 + c^2 = 90 and d^2 + e^2 + f^2 = 90. Find the sum a+b+c+d+e+f

Question 1209465: There are two differnet triples of positive integers (a,b,c) and (d,e,f) such that a^2 + b^2 + c^2 = 90 and d^2 + e^2 + f^2 = 90. Find the sum a+b+c+d+e+f
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52780) (Show Source): You can put this solution on YOUR website!
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There are two different triples of positive integers (a,b,c) and (d,e,f) such
that a^2 + b^2 + c^2 = 90 and d^2 + e^2 + f^2 = 90. Find the sum a+b+c+d+e+f.
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There are two basic triples (a,b,c) and (d,e,f) of positive integer numbers such that a^2 + b^2 + c^2 = 90, d^2 + e^2 + f^2 = 90. These triples are (a,b,c) = (8,5,1) and (d,e,f) = (7,5,4). All other triples of this kind are permutations of these triples. Permutations do not change the sum a+b+c+d+e+f, therefore, this sum has a uniques value a + b + c+ d + e + f = 8 + 5 + 1 + 7 + 5 + 4 = 30. ANSWER. The sum a + b + c+ d + e + f is 30.

Solved.

Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

The triples we are looking for are not ordered, so we are looking for two groups of three positive integers for which the sum of the squares of the three integers is 90.

To find the two groups, use a "greedy" algorithm -- an algorithm in which you try the largest numbers first ("take the biggest bites").

The largest positive integer whose square is less than 90 is 9. 9^2=81, so we would need two other integers for which the sum of the squares is 90-81=9. But there are no such integers.

8^2=64; we need two positive integers for which the sum of the squares is 90-64=26. That's easy -- 26 = 25+1, so 5 and 1. So one of the two groups we are looking for is 8, 5, and 1.

Next, 7^2=49; we need two positive integers for which the sum of the squares is 90-49=41. This one take a tiny bit longer to find -- 41 = 25+16, so 5 and 4. And so another of the groups we are looking for is 7, 5, and 4.

We could continue looking for other groups; but the problem tells us there are only two such groups, so our investigation is complete.

The question asks for the sum of the six integers in the two groups.

ANSWER: 8+5+1 + 7+5+4 = 14+16 = 30

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