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Very difficult math question?
A row vector has 5 elements all real numbers. The sum of all the numbers is 36. The odd position numbers are in the ratio 1:3:5. The sum of the odd position numbers equals the difference of the even position numbers. The largest positive number is 3 times the smallest positive number. What are the numbers?
1 Answer
- Julius NLv 79 years agoFavorite Answer
a+b+c+d+e=36
c=3a
e=5a
a+c+e = d-b
d=3a
(if a is the smallest positive number?)
Standard form: a+b+c+d+e=36; -3a+c=0; -5a+e=0; a+b+c-d+e=0; -3a+d=0;
substitute/eliminate a = -1b-1c-1d-1e+36
substitute/eliminate b = -4/3c-1d-1e+36
substitute/eliminate c = +3/5e-0
substitute/eliminate d = +18
substitute/eliminate e = +30
Solution: a = 6; b = -36; c = 18; d = 18; e = 30;
(a, b, c, d, e) = (6, -36, 18, 18, 30)
but what if the smallest positive number were b, then we would have:
d = 3b
Standard form: a+b+c+d+e=36; -3a+c=0; -5a+e=0; a+b+c-d+e=0; -3b+d=0;
substitute/eliminate a = -1b-1c-1d-1e+36
substitute/eliminate b = -4/3c-1d-1e+36
substitute/eliminate c = +3/5e-0
substitute/eliminate d = +18
substitute/eliminate e = +20/3
Solution: a = 4/3; b = 6; c = 4; d = 18; e = 20/3;
(a, b, c, d, e) = (1.3333, 6, 4, 18, 6.6667)
but a is smaller than b, so this is not a valid solution.