Number puzzle : "You can't determine my sum."

Question

Albert said to Bob, "I have two unequal positive integers; the smaller is at least 2; the larger is at most 25. I will only tell you their product." So he did. Later, Albert has forgotten the numbers but remembers the other conditions, and remembers their sum. He says to Bob."I see no way you can determine my sum." WHAT WAS THE SUM?............This was told to me by Michael Korenberg (Ph.D.--matrix theory), who may have composed it. My first thought was that it's paradoxical, but it's not.

Answer 1

When Albert tells Bob that, "I see no way you can determine the sum." what he is actually saying is the following, where $S$ is the sum of the two numbers.

• For every pair of numbers $2 \leq a < b \leq 25$ such that $a + b = S$, there is a different pair of numbers $2 \leq c < d \leq 25$ such that $a \cdot b = c \cdot d$. (Note that necessarily $a + b \neq c + d$.)

(The existence of this competing pair $c,d$ is what prevents Bob from knowing the sum if the original numbers happened to be $a,b$.)

So we are looking for the sum $S$ which has this property.

For example, $S \neq 15$ because $2 + 13 = 15$ and $2 \cdot 13 = 26$ and this is the only pair of numbers in the range which has this product. (So if the original numbers happened to be $2,13$ — which is possible as far as Albert is concerned at that point in the story — then Bob would be able to determine the sum because there is only one possible pair with this product which is known to Bob.)

You can brute force this without too much difficulty to show that $S = 11$. There are only four pairs of numbers in the range which have this sum.

• $2 + 9 = 11$, and $2 \cdot 9 = 18 = 3 \cdot 6$.
• $3 + 8 = 11$, and $3 \cdot 8 = 24 = 4 \cdot 6$.
• $4 + 7 = 11$, and $4 \cdot 7 = 28 = 2 \cdot 14$.
• $5 + 6 = 11$, and $5 \cdot 6 = 30 = 2 \cdot 15$.