Lucky Guess

Since the only integers that have 3 positive factors are squares of primes $p$ , (as $p^{2}$ has factors $1, p$ and $p^{2}$ ), we are looking for integers of the form $n = p^{2} - 1 \lt 200$ which have 16 positive factors. As $p = 13$ is the largest primes such that $p^{2} - 1 \lt 200$ , our options for $n$ are:

• $2^{2} - 1 = 3 \Longrightarrow 2$ factors,

• $3^{2} - 1 = 2^{3} \Longrightarrow 4$ factors,

• $5^{2} - 1 = 24 = 2^{3} \times 3 \Longrightarrow (3 + 1)(1 + 1) = 8$ factors,

• $7^{2} - 1 = 48 = 2^{4} \times 3 \Longrightarrow (4 + 1)(1 + 1) = 10$ factors,

• $11^{2} - 1 = 120 = 2^{3} \times 3 \times 5 \Longrightarrow (3 + 1)(1 + 1)(1 + 1) = 16$ factors,

• $13^{2} - 1 = 168 = 2^{3} \times 3 \times 7 \Longrightarrow (3 + 1)(1 + 1)(1 + 1) = 16$ factors.

So the only two suitable values for $n$ are $120$ and $168$ , wiich sum to $\boxed{288}$ .

See this wiki for an explanation of the number-of-factor calculations made above.

---

Further Discussion:

Looking at the different ways to find $n = p_{1}^{a_{1}}p_{2}^{a_{2}}....p_{k}^{a_{k}}$ so that $(a_{1} + 1)(a_{2} + 1)...(a_{k} + 1) = 16$ is another approach. With $16 = 16 \times 1$ the least integer we would be looking at is $2^{15} \gt 200$ . With $16 = 8 \times 2$ the least would be $2^{7} \times 3 \gt 200$ . With $16 = 4 \times 4$ we would have $2^{3} \times 3^{3} = 216 \gt 200$ . With $16 = 4 \times 2 \times 2$ we end up with only the two numbers found in the above solution being less than $200$ , with the next smallest being $3^{3} \times 2 \times 5 = 270$ . Finally, with $16 = 2 \times 2 \times 2 \times 2$ the least would be $2 \times 3 \times 5 \times 7 = 210$ . So the only two suitable numbers that Alicia and Benjamin can choose from are $120$ and $168$ , both of which happen to be of the form $p^{2} - 1$ for some prime $p$ . Thus the information about being one less than an integer with 3 positive factors was in this case redundant, but starting from this condition did nevertheless make for a quicker solution.

Further, since all primes $p \gt 3$ are of the form $6k \pm 1$ we can show that $p^{2} - 1 = (p - 1)(p + 1)$ is divisible by $24 = 2^{3} \times 3$ . Also, at the very least one of $p - 1$ or $p + 1$ will introduce one more prime into the factorization other than $2$ and $3$ , which means that to keep the number of factors of $p^{2} - 1$ down to $16$ we must have one of $p - 1$ or $p + 1$ equal to $12$ . So in fact $120 = 10 \times 12$ and $168 = 12 \times 14$ are the $only$ possible positive integers for Alicia and Benjamin to choose from, so the upper bound of $200$ is ultimately superfluous.

Let the numbers be $A$ and $B$ . Since $A+1$ and $B+1$ have an odd number of factors, they must be perfect squares, and since they have three factors, the square of a prime number. The squares of primes under 200 are 4, 9, 25, 49, 121, and 169. $A$ and $B$ must be fairly large to have 16 factors; so we check 120 and 168. $120 = 2^3 \cdot 3^1 \cdot 5^1$ , which has $(3+1)(1+1)(1+1) = 16$ factors. $168 = 2^3 \cdot 3^1 \cdot 7^1$ , which also has $(3+1)(1+1)(1+1) = 16$ factors. Our answer is $120 + 168 = 288$ .