# The Sums and Product Puzzle | Solution

The answer: 2 and 9.

Is that what you came up with? Here’s how we get there. Get ready—this is a real tour de force of logic!

First of all, the numbers are greater than 1.

The second student has already been told the sum and knows it’s less than 14. Here are the possible sums he will consider right away (and what he will consider as options for the two numbers):

4 (2 + 2)

5 (2 + 3)

6 (2 + 4, or 3 + 3)

7 (2 + 5, or 3 + 4)

8 (2 + 6, or 3 + 5, or 4 + 4)

9 (2 + 7, or 3 + 6, or 4 + 5)

10 (2 + 8, or 3 + 7, or 4 + 6, or 5 + 5)

11 (2 + 9, or 3 + 8, or 4 + 7, or 5 + 6)

12 (2 + 10, or 3 + 9, or 4 + 8, or 5 + 7, or 6 + 6)

13 (2 + 11, or 3 + 10, or 4 + 9, or 5 + 8, or 6 + 7)

If a product is made up of only two prime factors (primes being 2, 3, 5, 7, 11, 13, etc.—i.e., numbers that are divisible by only 1 and themselves), then the sum of its factors is unique.

For example, in our list above, 4 can only be made by 2 + 2. The product 2×2 is also 4. So, the second student would assume that if the first student was told that the product is “4,” then he would know the sum.

The same happens with 5: 5 can only be made by 2 + 3. The product 2×3 is 6. If the first student was told that the product is 6, he would have broken it down into a product of two prime numbers, 2 and 3, and concluded the numbers as 2 and 3.

The second student would go through the list of all remaining possibilities and eliminate any pairs of prime numbers:

Sum: 6 (2 + 4~~, or 3 + 3~~)

Sum: 7 (~~2+ 5, or ~~3 + 4)

Sum: 8 (2 + 6~~, or 3 + 5~~, or 4 + 4)

Sum: 9 (~~2 + 7, or ~~3 + 6, or 4 + 5)

Sum: 10 (2 + 8~~, or 3 + 7~~, or 4 + 6~~, or 5 + 5~~)

Sum: 11 (2 + 9, or 3 + 8, or 4 + 7, or 5 + 6)

Sum: 12 (2 + 10, or 3 + 9, or 4 + 8~~, or 5 + 7~~, or 6 + 6)

Sum: 13 (~~2 + 11, or ~~3 + 10, or 4 + 9, or 5 + 8, or 6 + 7)

But the second student is also very smart. He would also think about what possible products the remaining numbers could make, so he can guess what the first student might be thinking:

Sum: 6 (2 + 4 **[product: 8]**)

Sum: 7 (3 + 4 **[product: 12]**)

Sum: 8 (2 + 6 **[product: 12]**, or 4 + 4 **[product: 16]**)

Sum: 9 (3 + 6 **[product: 18]**, or 4 + 5 **[product: 20]**)

Sum: 10 (2 + 8 **[product: 16]**, or 4 + 6 **[product: 24]**)

Sum: 11 (2 + 9 **[product: 18]**, or 3 + 8 **[product: 24]**, or 4 + 7 **[product: 28]**, or 5 + 6 **[product: 30]**)

Sum: 12 (2 + 10 **[product: 20]**, or 3 + 9 **[product: 27]**, or 4 + 8 **[product: 32]**, or 6 + 6 **[product: 36]**)

Sum: 13 (3 + 10 **[product: 30]**, or 4 + 9 **[product: 36]**, or 5 + 8 **[product: 40]**, or 6 + 7 **[product: 42]**)

Now, he will also eliminate any products that only show up once, because otherwise, the first student would have been able to guess the numbers (so, cut 8, 27, 28, 40, and 42—and as a result, that eliminates sum 6 as a possibility):

Sum: 7 (3 + 4 **[product: 12]**)

Sum: 8 (2 + 6 **[product: 12]**, or 4 + 4 **[product: 16]**)

Sum: 9 (3 + 6 **[product: 18]**, or 4 + 5 **[product: 20]**)

Sum: 10 (2 + 8 **[product: 16]**, or 4 + 6 **[product: 24]**)

Sum: 11 (2 + 9 **[product: 18]**, or 3 + 8 **[product: 24]**, or 5 + 6 **[product: 30]**)

Sum: 12 (2 + 10 **[product: 20]**, or 4 + 8 **[product: 32]**, or 6 + 6 **[product: 36]**)

Sum: 13 (3 + 10 **[product: 30]**, or 4 + 9 **[product: 36]**)

But also, knowing that the first student knows the sum is less than 14, he’d eliminate any products that *could possibly* have factors adding to greater than 14:

2×12 = **24** = 3×8 = 4×6

2×14 = **28** = 4×7

2×15 = **30** = 3×10 = 5×6

2×16 = **32** = 4×8

2×18 = **36** = 3×12 = 4×9 = 6×6

2×20 = **40** = 4×10 = 5×8

2×21 = **42** = 3×14 = 6×7

His updated list now looks like this:

Sum: 7 (3 + 4 **[product: 12]**)

Sum: 8 (2 + 6 **[product: 12]**, or 4 + 4 **[product: 16]**)

Sum: 9 (3 + 6 **[product: 18]**, or 4 + 5 **[product: 20]**)

Sum: 10 (2 + 8 **[product: 16]**)

Sum: 11 (2 + 9 **[product: 18]**)

Sum: 12 (2 + 10 **[product: 20]**)

Now, since the first student was able to guess the number, this means his product can’t be 12 either or else he wouldn’t know if the second student’s sum is 7 or 8:

Sum: 8 (4 + 4 **[product: 16]**)

Sum: 9 (3 + 6 **[product: 18]**, or 4 + 5 **[product: 20]**)

Sum: 10 (2 + 8 **[product: 16]**)

Sum: 11 (2 + 9 **[product: 18]**)

Sum: 12 (2 + 10 **[product: 20]**)

The product can’t be 16 either, since otherwise, the first student wouldn’t be able to know if the second student’s sum is 8 or 10:

Sum: 9 (3 + 6 **[product: 18]**, or 4 + 5 **[product: 20]**)

Sum: 11 (2 + 9 **[product: 18]**)

Sum: 12 (2 + 10 **[product: 20]**)

Last, there are two factors at work in how the students know the number:

1. If the product were 20, then the first student would have been stuck (not knowing if the second student had a sum of 18 or 20)

2. If the product is 18, then that eliminates the possibility of a product of 20—which eliminates the possibility of a sum of 9 or 12.

The only possibility, then, is a sum of 11, with the product being 18. Which means the numbers have to be 2 and 9.

And there is the true beauty of deductive logic in the the form of addition and multiplication!

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