SOLUTION: There are two triples of positive integers (a,b,c and d,e,f) such that a^2+b^2+c^2 = 86 and d^2+e^2+f^2 = 86. Numbers should not repeat within each triplet. Evaluate the expression

Question 1151145: There are two triples of positive integers (a,b,c and d,e,f) such that a^2+b^2+c^2 = 86 and d^2+e^2+f^2 = 86. Numbers should not repeat within each triplet. Evaluate the expression | abc-def |
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20060) (Show Source): You can put this solution on YOUR website!
There are two triples of positive integers (a,b,c and d,e,f) such that
a^2+b^2+c^2 = 86 and d^2+e^2+f^2 = 86. Numbers should not repeat within each
triplet. Evaluate the expression | abc-def |

We'll assume a < b < c and d < e < f a,b,c,d,e,f all have to be 9 or less. The squares are 1,4,9,16,25,36,49,64,81 Three of these must have sum 86 We try c = 9 a² + b² + c² = 86 a² + b² + 9² = 86 a² + b² + 81 = 86 a² + b² = 5 We see that the first two, 1 and 4, have sum 5 so (a,b,c) or (d,e,f) can be (1,4,9) We try c = 8 a² + b² + c² = 86 a² + b² + 8² = 86 a² + b² + 64 = 86 a² + b² = 22 No two of these 1,4,9,16,25,36,49,64,81 add to 22, so we try c = 7 a² + b² + c² = 86 a² + b² + 7² = 86 a² + b² + 49 = 86 a² + b² = 37 We see that 1+36=37, so (a,b,c) or (d,e,f) can be (1,6,7) Let's let (a,b,c) = (1,4,9) and (d,e,f) = (1,6,7) You calculate | abc-def | Edwin

Answer by ikleyn(52803) (Show Source): You can put this solution on YOUR website!
.

1) 86 = 1^2 + 2^2 + 9^2 2) 86 = 1^2 + 6^2 + 7^2. So, a= 1, b= 2, c= 9; d= 1, e= 6, f= 7. abc = 18, def = 42, | abc - def | = 42 - 18 = 24. ANSWER

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