The answer is right under your nose
There is no solution to the equations
5 7 x + 3 1 y + 1 9 z = 3 2 0 4 8 x + 2 5 y + 1 5 z = 1 0 7 a x + b y + c z = a b c
for x , y and z , where x , y and z are real numbers and a , b and c are positive integers less than 10.
Find the value of a + b + c .
The answer is 19.
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1 solution
how do you come to know that it is the only solution for 0 < a , b , c < 1 0 .
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That is a great question to ask. We need a dependent system of equations, and so there is an entire lattice of points of the form α ( 5 7 , 3 1 , 1 9 ) + β ( 4 8 , 2 5 , 1 5 ) that could work. The claim then is that in the cuboid [ 0 , 1 0 ] 3 , we have at most 1 lattice point in it.
This follows by looking at the planar cutout of α ( 5 7 , 3 1 , 1 9 ) + β ( 4 8 , 2 5 , 1 5 ) ⋒ [ 0 , 1 0 ] 3 , and showing that the area is too small to contain 2 lattice points.
Multiplying either equation by up to 5 cannot produce positive integers a,b,c<10. Probably there is another solution if the equations are multiplied by more than 5.
Conditions are not well satisfied according to question,hence many solution can exist for this problem.
I did the same thing, only thought of it as a plane in 3-space. Either way great solution!
Subtract the second equation from the first. Then we get 9x+6y+4z=213. If we substitute a=9, b=6, c=4, 9x6x4=216, a contradiction. Thus a+b+c=9+6+4=19