How many integer solutions does inequality x+y+z+t<=15 have that satisfy the following conditions:x>=5, y>=-3,z>=2, and t>=-7

Answer 1

How many integer solutions does inequality x+y+z+t<=15 have that satisfy the following conditions:x>=5, y>=-3,z>=2, and t>=-7

The smallest value of (x+y+z+t) is (5-3+2-7) = (-3).

The largest value of (x+y+z+t) is (15, given).

Let’s reword the problem:

   Let:

        a = x-5        so a >= 0

        b = y+3       so b >= 0

        c = z-2         so c >= 0

        d = t+7         so t >=0

   (x+y+z+t) <= 15

   (a+5 + b-3 + c+2 + d-7) <=15

    (a+b+c+d) -3 <= 15

    (a+b+c+d) <= 18

Combining the above,

    0 <= a <= 18

    0 <= b <= 18

    0 <= c <= 18

    0 <= d <= 18

And

    (a+b+c+d) <= 15

Now, more specifically:

   When a=0,   0 <= (b+c+d) <=18

   When a=1,   0 <= (b+c+d) <=17

    . . .

   When a=18, (b+c+d) = 0

    When (a+b) = 0,   0 <= (c+d) <= 18

  etc.

As Mark M. pointed out, this is nested loops (iteration), which is easy in a computer program:

Count = 0

FOR x = 5 TO 23

  FOR y = -3 TO 11

     FOR z = 2 TO 20

        FOR t = -7 TO 11

           IF ( (x+y+z+t) <= 15 ) Count = Count + 1

        END

     END

  END

END

PRINT (Count)

END      

This prints 7315 for the value of Count.

Note the distribution:

For

x = 13,   6600 cases

x = 14,    221 cases

. . .

x = 22,   4 cases

x = 23,    1 case

Answer 2

By subtracting we get: 5 ≤ x ≤ 23 (23 - 5 + 1 = 19 integers)

-3 ≤ y ≤ 15 ( 15 + 3 + 1 = 19 integers)

2 ≤ z ≤ 20 (20 - 2 + 1 = 19 integers);

-7 ≤ t ≤ 11 (11 + 7 + 1 = 19 intergest)

-3 ≤ x + y + z + t ≤ 15

Integers values for x + y + z + t is 15 + 3 + 1 = 19