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# OSL4#36

+1
77
4
+893

IDK know to do this.

Jul 27, 2019

#1
0

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

Jul 27, 2019
#2
+893
+3

Shouldn't there be two values for x, the smallest and the largest, to find the answer?

dgfgrafgdfge111  Jul 28, 2019
#3
+103689
+3

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

Jul 28, 2019
#4
+893
+1

Thanks Melody!!!!

dgfgrafgdfge111  Jul 28, 2019