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Solution to the Math Riddle Fall Newsletter 08-09
Here is an algebraic solution.
Let x1 = the 1st number in our list, x2 = the 2nd number, … x5 = the 5th number
x5 + x3 = 14
x4 = x2 + 1
x1 = 2x2 – 1
x2 + x3 = 10
x1 + x2 + x3 + x4 + x5 = 30
Rewrite our last equation in terms of a single variable.
(x1) + (x2 + x3) + (x4) + (x5) = 30
(2x2 – 1) + (10) + (x2 + 1) + (14 – x3) = 30
(2x2 – 1) + (10) + (x2 + 1) + (14 – (10 - x2)) = 30
4x2 + 14 = 30
x2 = 4
Substitution gives us our five numbers in the following order: 7,4,6,5,8