Four Integers from Products
Four positive integers $A$, $B$, $C$, and $D$ satisfy:
$AB = 16, \quad BC = 14, \quad CD = 63$
Find $A + B + C + D$.
Hints
- The variable $B$ appears in two constraints: $AB = 16$ and $BC = 14$. What does that say about $B$?
- $B$ must be a factor of both 16 and 14. List the factors of each and find the common ones.
- With $B$ pinned down, solve for $C$ from $BC = 14$, then $D$ from $CD = 63$, then $A$ from $AB = 16$. Check that all values are positive integers.
Worked Solution
How to Think About It: You have three product equations and four unknowns -- at first glance under-determined. The trick is that the constraints are chained: $B$ appears in both $AB = 16$ and $BC = 14$, and $C$ appears in both $BC = 14$ and $CD = 63$. So the middle variable $B$ must simultaneously be a factor of 16 and a factor of 14. That is a strong constraint: factors of 16 are $\{1, 2, 4, 8, 16\}$ and factors of 14 are $\{1, 2, 7, 14\}$. The only overlap is $B \in \{1, 2\}$.
Quick Estimate: Since $BC = 14$ and $C$ must be a positive integer, if $B = 1$ then $C = 14$, but then $CD = 63$ gives $D = 63/14$, which is not an integer. So $B = 2$, giving $C = 7$ and $D = 9$. Then $A = 16/2 = 8$. Sum: $8 + 2 + 7 + 9 = 26$. Done.
Formal Solution:
Step 1: Constrain $B$. Since $AB = 16$, $B$ divides 16: $B \in \{1, 2, 4, 8, 16\}$. Since $BC = 14$, $B$ divides 14: $B \in \{1, 2, 7, 14\}$. The intersection is $B \in \{1, 2\}$.
Step 2: Eliminate $B = 1$. If $B = 1$, then $C = 14/1 = 14$, and $D = 63/14 = 4.5$. Not a positive integer. Ruled out.
Step 3: Confirm $B = 2$. Then $C = 14/2 = 7$, and $D = 63/7 = 9$. Both are positive integers. Also $A = 16/2 = 8$. All four values are positive integers. Consistent.
Step 4: Sum. $A + B + C + D = 8 + 2 + 7 + 9 = 26$.
Answer: $A + B + C + D = 26$
Intuition
The key move here is identifying the choke point: $B$ is over-constrained by two product equations, which dramatically limits its possible values. Rather than trying all combinations of $(A, B)$ pairs from $AB = 16$, you immediately intersect the factor sets and reduce to two candidates. This "find the common factor" approach is a general technique for constrained integer problems -- whenever a variable appears in multiple equations, its feasible set is the intersection of all individual feasibility sets.
This type of problem tests methodical case elimination rather than any deep math. In a real interview, the danger is getting bogged down in algebra. The right approach is to identify the most constrained variable first ($B$, not $A$ or $D$, because $B$ sits in the middle), pin it down quickly, and propagate. That discipline -- work from most-constrained to least-constrained -- applies broadly in constraint satisfaction problems, combinatorics, and even in optimization (identify binding constraints first).