solve a + b + c + d + e = 320
a + b + c = 283
a + e = 119 for a, b, c, d, e
c = -a - b + 283 and d = a - 82 and e = 119 - a
a =101, b=97, c=85, d=19, e=18
+937 Shouldn't there be two values for x, the smallest and the largest, to find the answer?
+118677 Let the numbers be
a, b,c,d, and X
But it is better to replace c+d with Y
So our positive integer numbers are a, b, Y, X where a < b < Y/2 < X
A+b+Y/2+X = 320
Y/2+X=283
so
a+b = 37
283/3 = 94.3...
93, 94, 96 are the closest integers that add to 283
93+94 = 187
so
a+b = 37
Let
\(a=18-\alpha,\quad b=19+\alpha,\quad Y=187-\delta, \quad X=96+\delta\)
Now
\(a+X=119\\ 18-\alpha+96+\delta=119\\ \delta=5+\alpha\)
\(a=18-\alpha,\quad b=19+\alpha,\quad Y=187-5-\alpha, \quad X=96+5+\alpha\\ a=18-\alpha,\quad b=19+\alpha,\quad Y=182-\alpha, \quad X=101+\alpha\\\)
the smallest value of alpha is 0 and the largest is 17
So the difference between the greatest and least values of X is 17
Answer: 17
