If $\lfloor a-b\rfloor =2$, $\lfloor b-c\rfloor =3$, $\lfloor c-d\rfloor =4$, what is the sum of all possible values of $\lfloor a - d\rfloor?$

Question

Let $a,b,c,d$ be real numbers with $\lfloor a - b\rfloor = 2 , \lfloor b - c\rfloor = 3 , \lfloor c - d\rfloor = 4$. What is the sum of all possible values of $\lfloor a - d\rfloor ?$ ($\lfloor \cdot \rfloor$ denote the greatest integer function)

What I Tried:

I got that:- $$\rightarrow 3 > a - b \geq 2$$ $$\rightarrow 4 > b - c \geq 3$$ $$\rightarrow 5 > c - d \geq 4$$ But I seem to find no way to use these. One think I thought is to first find the maximum $(a - d)$ , then find the minimum $(a - d)$, and then consider all the integer values between the maximum and the minimum. The question is, how do you find that?

Can anyone help me?

Answer 1

$a-d = (a-b) + (b-c) + (c-d)$

Now $[a-b]\le (a-b) < [a-b]+1$ etc. so

$[a-b] +[b-c]+[c-d] \le (a-b) +(b-c)+(c-d) = (a-d) < [a-b] + [b-c]+[c-d] + 3$ so

$2 + 3 + 4 \le (a-d) < 2+3+4 + 3$

$9 \le a-d < 12$ so $a-d$ may be $9,10$ or $11$.

So the sum of all possible values (why do they ask these questions???) is $9+10+11 = 30$.