1. **Problem Statement:** We want to find how many possible values there are for the sum $a + b + c$ where $a$, $b$, and $c$ are positive integers such that $abc = 72$. 2. **Understanding the problem:** Since $a$, $b$, and $c$ are positive integers and their product is fixed at 72, we need to find all triples $(a,b,c)$ with $abc=72$ and then find the distinct sums $a+b+c$. 3. **Step 1: Factorize 72.** $$72 = 2^3 \times 3^2$$ 4. **Step 2: Find all positive integer triples $(a,b,c)$ such that $abc=72$.** We consider all factorizations of 72 into three positive integers. To avoid counting duplicates, we can assume $a \leq b \leq c$. 5. **Step 3: Enumerate triples $(a,b,c)$ with $a \leq b \leq c$ and $abc=72$.** - $a=1$: $bc=72$ with $b \leq c$. Possible pairs $(b,c)$: $(1,72), (2,36), (3,24), (4,18), (6,12), (8,9)$ Sums: $1+1+72=74$, $1+2+36=39$, $1+3+24=28$, $1+4+18=23$, $1+6+12=19$, $1+8+9=18$ - $a=2$: $bc=36$ with $b \geq 2$ and $b \leq c$. Possible pairs: $(2,18), (3,12), (4,9), (6,6)$ Sums: $2+2+18=22$, $2+3+12=17$, $2+4+9=15$, $2+6+6=14$ - $a=3$: $bc=24$ with $b \geq 3$ and $b \leq c$. Possible pairs: $(3,8), (4,6)$ Sums: $3+3+8=14$, $3+4+6=13$ - $a=4$: $bc=18$ with $b \geq 4$ and $b \leq c$. Possible pairs: $(4,4.5)$ not integer, so no pairs. - $a=6$: $bc=12$ with $b \geq 6$ and $b \leq c$. Possible pairs: $(6,2)$ but $b \leq c$ and $6 \leq 2$ false, so no pairs. - $a=8$: $bc=9$ with $b \geq 8$ and $b \leq c$. Possible pairs: $(8,1.125)$ no integer pairs. 6. **Step 4: Collect all distinct sums:** $\{74, 39, 28, 23, 22, 19, 18, 17, 15, 14, 13\}$ 7. **Step 5: Count distinct sums:** There are 11 distinct sums. **Final answer:** There are **11** possible values for the sum $a + b + c$ given $abc=72$ and $a,b,c$ positive integers.