I f a 2 + b 3 + c 6 = 2 0 0 9 + 9 1 8 2 + 2 7 2 3 + 7 2 6 a n d a , b , c ∈ Z + f i n d t h e v a l u e o f a + b + c
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Squaring both sides,
2 a 2 + 3 b 2 + 6 c 2 + 2 a b 6 + 2 a c 1 2 + 2 b c 1 8 = 2 0 0 9 + 9 1 8 2 + 2 7 2 3 + 7 2 6
2 a 2 + 3 b 2 + 6 c 2 + 2 a b 6 + 4 a c 3 + 6 b c 2 = 2 0 0 9 + 9 1 8 2 + 2 7 2 3 + 7 2 6
Since a , b and c are positive integers,
2 a 2 + 3 b 2 + 6 c 2 = 2 0 0 9 ( ∗ )
2 a b 6 = 7 2 6
4 a c 3 = 2 7 2 3
6 b c 2 = 9 1 8 2
a b = 3 6 = 4 × 9
a c = 6 8 = 4 × 1 7
b c = 1 5 3 = 9 × 1 7
a = 4 , b = 9 , c = 1 7
Checking our solution against ( ∗ ) ,
2 a 2 + 3 b 2 + 6 c 2 = 2 × 1 6 + 3 × 8 1 + 6 × 2 8 9 = 2 0 0 9
Therefore,
a + b + c = 3 0
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Square both sides will gives 2 a 2 + 3 b 2 + 6 c 2 + 2 a b 6 + 4 a c 3 + 6 b c 2 = 2 0 0 9 + 9 1 8 2 + 2 7 2 3 + 7 2 6 By observing the coefficient a b = 3 6 , a c = 6 8 , b c = 1 5 3 . Multiply them all to ( a b c ) 2 = ( 6 × 2 × 3 × 1 7 ) 2 . As all variables are positive, then a b c = 6 1 2 . Solve for each gives ( a , b , c ) = ( 4 , 9 , 1 7 ) so a + b + c = 4 + 9 + 1 7 = 3 0