It completely depends the datatype that you have assigned for the variables 'a' , 'b' , and 'c'. Check the compiler that you are using for the size of the datatype in bytes. Add them and thus you will get the answer.
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∙ 13y agob+b+b+c+c+c+c =3b+4c
2b + 2c or 2(b + c)
And how does this relate to coins?
A+c= 2a+b
Because there is no way to define the divisors, the equations cannot be evaluated.
b+b+b+c+c+c+c =3b+4c
b + b + b + c + c + c + c = 3b + 4c
It is impossible to give any decimal/numeric value if we are not given the values of at least one variable, so the answer is B + B + B + C + C + C.
2b + 2c or 2(b + c)
sizeof (int) will tell you (in bytes). It's often 2, 4 or 8 bytes.
And how does this relate to coins?
a= (+a) or a= (-) b= 2a b= 2a c= (-a) c= (+a)
A+c= 2a+b
Try 'sizeof', it will return the size in bytes.
If a + b + c + d + 80 + 90 = 100, then a + b + c + d = -70.
If: a = b+c+d Then: c = a-b-d
Because there is no way to define the divisors, the equations cannot be evaluated.