Given positive integers a, b, and c with a +b+c=20
Determine the number of possible integer values for a+b/c
Answers
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Given : positive integers a, b, and c such that a + b + c = 20
To Find : number of possible integer values for a+b/c
Solution:
a + b + c = 20
a , b & c are positive integers
0 < a , b ,c < 19
a + b/c is integer
=> b/c is integer
case 1 - c is 1
b/c can have value from 1 to 18 where c = 1 b from 1 to 18 as from 18 to 1
a + b/c = 19
case 1 : b = c
=> b/c = 1
b = c = 1 , a = 18 , a + b/c = 19
b = c = 2, a = 16 , a + b/c = 17
b = c = 3, a = 14 , a + b/c = 15
and so on
19 , 17 , 15 , 13 , 11 , 9 , 7 , 5 , 3 , 1
case 2 - c is 1
b/c can have value from 1 to 18 where c = 1 , b from 1 to 18 then a from 18 to 1
a + b/c = 19
case 2 - c is 2
b/c can have value from 1 ,2 , ....8 for b = 2, 4 , 6 , ......16 where a = 16 , 14 , 12 ,........2 .
a + b/c =17 , 16 , 15 , 14 , 13 , 12 , 11 , 10
case 3 - c is 3
b/c can have value from 1 ,2 , ....5 for b =3, 6 , 9, 12 15 where a = 14 , 11 ,8 , 5 , 2.
a + b/c = 15 13 , 11 , 9 , 7
case 4 - c is 4
b/c can have value from 1 ,2 ,3 for b =4, 8 ,12 where a = 12 , 8 , 4.
a + b/c = 13 , 10 , 7
case 5 - c is 5
b/c can have value from 1 ,2 for b =5, 10 where a = 10, 5
a + b/c = 11 , 7
case 6 - c is 6
b/c can have value from 1 ,2 for b =6, 12 where a = 8 , 2
a + b/c = 9 , 4
further b = 7 , c = 7 , a = 6 => a + b/c = 7
b = 8 , c = 8 , a =4 => a + b/c =5
b = 9 , c = 9 , a = 2 => a + b/c = 3
Combining all
a + b/c = 1 , 3 , 4 , 5 , 7 . 9 ,10 ,11 , 12 , 13 , 14 , 15 , 16 , 17 , 19
number of possible integer values for a+b/c = 15
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