We know the following about the numbers $a$, $b$ and $c$:
If $a + b + c ≥ 1$, determine the number of possible values for $a + b + c$
Puzzling Stack Exchange is a question and answer site for those who create, solve, and study puzzles. It only takes a minute to sign up.Sign up to join this community
We know that:
It then follows that
by adding them up we get $$ 2(a+b+c) = \pm3\pm5\pm9 $$ Since $a+b+c\ge1$, we need this to be greater or equal to $2$.
So the question boils down to finding
how many choices of the signs make $\pm3\pm5\pm9 \ge2$ true. Clearly $9$ cannot have a minus sign, so we have $9\pm3\pm5 \ge2$. This only fails when both remaining signs are minus signs, so there are three possibilities left.
The solutions are therefore:
$$a+b+c = (9+3+5)/2 = 17/2\\a+b+c = (9+3-5)/2 = 7/2\\a+b+c = (9-3+5)/2 = 11/2$$