Understanding the Problem: Pairs and Remainders Modulo 15

The question asks about the number of pairs of distinct elements (a, b) from a set S of fourteen natural numbers such that their product (ab) leaves a remainder of 1 when divided by 15. This condition can be written mathematically as \(ab \equiv 1 \pmod{15}\).

For the product of two integers ab to be congruent to 1 modulo 15, both a and b must be coprime to 15. A number is coprime to 15 if its greatest common divisor (GCD) with 15 is 1. The prime factors of 15 are 3 and 5. So, a number is coprime to 15 if it is not divisible by 3 and not divisible by 5.

The natural numbers coprime to 15 have residues modulo 15 that are also coprime to 15. These residues are:

• 1 (gcd(1, 15) = 1)
• 2 (gcd(2, 15) = 1)
• 4 (gcd(4, 15) = 1)
• 7 (gcd(7, 15) = 1)
• 8 (gcd(8, 15) = 1)
• 11 (gcd(11, 15) = 1)
• 13 (gcd(13, 15) = 1)
• 14 (gcd(14, 15) = 1)

Let's find the pairs of residues (x, y) such that \(xy \equiv 1 \pmod{15}\):

• \(1 \times 1 = 1 \pmod{15}\)
• \(2 \times 8 = 16 \equiv 1 \pmod{15}\)
• \(4 \times 4 = 16 \equiv 1 \pmod{15}\)
• \(7 \times 13 = 91 \equiv 1 \pmod{15}\)
• \(8 \times 2 = 16 \equiv 1 \pmod{15}\)
• \(11 \times 11 = 121 \equiv 1 \pmod{15}\)
• \(13 \times 7 = 91 \equiv 1 \pmod{15}\)
• \(14 \times 14 = 196 = 13 \times 15 + 1 \equiv 1 \pmod{15}\)

So, for a pair (a, b) from S to satisfy \(ab \equiv 1 \pmod{15}\), the residues of a and b modulo 15, say r(a) and r(b), must form one of the pairs from the list above. Since a and b must be distinct elements from the set S (a ≠ b), we are looking for ordered pairs (a, b) from S such that a ≠ b and \(r(a) r(b) \equiv 1 \pmod{15}\).

Counting Pairs Based on Modular Residues

Let \(N_i\) be the number of elements in the set S that are congruent to \(i \pmod{15}\), for \(i \in \{1, 2, 4, 7, 8, 11, 13, 14\}\). Let \(N_{not\_coprime}\) be the number of elements in S that are not coprime to 15.

Since S has fourteen distinct natural numbers, the sum of the counts for all possible residues must be 14:

\(\sum_{i \in \{0, \dots, 14\}} N_i = 14\)

The pairs (a, b) from S with a ≠ b satisfying \(ab \equiv 1 \pmod{15}\) must have a and b coprime to 15. We count these pairs based on their residues:

• If \(a \equiv 1 \pmod{15}\) and \(b \equiv 1 \pmod{15}\). We need two distinct numbers from S congruent to 1 mod 15. If there are \(N_1\) such numbers in S, the number of ordered pairs (a, b) with a ≠ b is \(N_1(N_1-1)\).
• If \(a \equiv 4 \pmod{15}\) and \(b \equiv 4 \pmod{15}\). Similarly, if there are \(N_4\) such numbers, the number of pairs is \(N_4(N_4-1)\).
• If \(a \equiv 11 \pmod{15}\) and \(b \equiv 11 \pmod{15}\). If there are \(N_{11}\) such numbers, the number of pairs is \(N_{11}(N_{11}-1)\).
• If \(a \equiv 14 \pmod{15}\) and \(b \equiv 14 \pmod{15}\). If there are \(N_{14}\) such numbers, the number of pairs is \(N_{14}(N_{14}-1)\).
• If \(a \equiv 2 \pmod{15}\) and \(b \equiv 8 \pmod{15}\), or \(a \equiv 8 \pmod{15}\) and \(b \equiv 2 \pmod{15}\). If there are \(N_2\) numbers congruent to 2 mod 15 and \(N_8\) numbers congruent to 8 mod 15 in S, any number from the first group can be paired with any number from the second group. Since \(2 \not\equiv 8 \pmod{15}\), distinct numbers from these two groups are always distinct values. The number of ordered pairs (a, b) with \(a \equiv 2\) and \(b \equiv 8\) is \(N_2 N_8\). The number of ordered pairs (a, b) with \(a \equiv 8\) and \(b \equiv 2\) is \(N_8 N_2\). The total number of pairs for this case is \(N_2 N_8 + N_8 N_2 = 2N_2N_8\).
• If \(a \equiv 7 \pmod{15}\) and \(b \equiv 13 \pmod{15}\), or \(a \equiv 13 \pmod{15}\) and \(b \equiv 7 \pmod{15}\). Similarly, the number of pairs is \(2N_7N_{13}\).

The total number of ordered pairs (a, b) with a ≠ b such that \(ab \equiv 1 \pmod{15}\) is the sum of pairs from all these cases:

Total Pairs = \(N_1(N_1-1) + 2N_2N_8 + N_4(N_4-1) + 2N_7N_{13} + N_{11}(N_{11}-1) + N_{14}(N_{14}-1)\)

where \(N_i \ge 0\) are integers and \(\sum_{i \in \{1, 2, 4, 7, 8, 11, 13, 14\}} N_i \le 14\).

Analyzing the Possible Number of Pairs

Let's look at the formula for the total number of pairs:

• The term \(N_i(N_i-1)\) represents the number of ordered pairs of distinct elements from a group of size \(N_i\). If \(N_i \ge 2\), this is a product of two consecutive integers, which is always even. If \(N_i < 2\), the term is 0, which is also even.
• The term \(2N_iN_j\) is always even because of the factor of 2.

Since the total number of pairs is a sum of even numbers, the total number of pairs must always be an even number.

Looking at the options provided: 3, 5, 6.

• Option 1 is 3, which is odd. It is impossible to get 3 pairs.
• Option 2 is 5, which is odd. It is impossible to get 5 pairs.
• Option 3 is 6, which is even. This might be possible.

Let's check if 6 is a possible number of pairs. We need the sum \(N_1(N_1-1) + 2N_2N_8 + \dots\) to equal 6, with \(\sum N_i \le 14\) for \(i \in \{1, 2, 4, 7, 8, 11, 13, 14\}\).

Consider the case where only \(N_1\) is non-zero (among the coprime residues). If \(N_1 = 3\), the number of pairs is \(N_1(N_1-1) = 3(3-1) = 3 \times 2 = 6\). We can construct a set S of 14 distinct natural numbers such that exactly 3 of them are congruent to 1 mod 15 and the other 11 are not coprime to 15 (e.g., S = {1, 16, 31, 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24}). In this set, \(N_1=3\), and \(N_i=0\) for \(i \in \{2, 4, 7, 8, 11, 13, 14\}\). The number of pairs is 6. Thus, 6 is a possible number of pairs.

Conclusion Based on Question Phrasing

Our analysis shows that 3 and 5 are impossible, while 6 is a possible number of pairs. If the question is interpreted as "Which of the following is a possible number of pairs?", then 6 would be the correct answer.

However, the phrasing "The possible number of pairs ... is" might imply that there is a unique number of pairs that results from any set of fourteen natural numbers. As we have shown, the number of pairs depends heavily on the specific elements within the set S (specifically, their distribution modulo 15). For instance:

• If S = {1, 16, 3, 5, ..., (10 other numbers not coprime to 15)}, \(N_1=2\), pairs = \(2(1)=2\).
• If S = {1, 16, 31, 3, 5, ..., (9 other numbers not coprime to 15)}, \(N_1=3\), pairs = \(3(2)=6\).
• If S contains no numbers coprime to 15, the number of pairs is 0.

Since the number of pairs varies depending on the set S, there is no single value that represents "the" possible number of pairs for *any* set of fourteen natural numbers. Given this interpretation, none of the options (3, 5, or 6) can be "the" possible number of pairs.

Therefore, based on the phrasing of the question and the provided options, the most fitting answer is that none of the given options represent "the" unique possible number of pairs that holds true for any set of fourteen natural numbers. While 6 is a possible value for a *specific* set S, it is not the *only* possible value across all sets S.

Revision Table: Key Concepts

Additional Information: Properties of Modulo Arithmetic

Understanding modular arithmetic is crucial for solving this problem. The property used here is that if \(a \equiv r_a \pmod{15}\) and \(b \equiv r_b \pmod{15}\), then \(ab \equiv r_a r_b \pmod{15}\). This allows us to analyze the product of numbers by just looking at their residues modulo 15.

Also, the concept of multiplicative inverses modulo n is important. An inverse exists only for numbers coprime to n. If a is not coprime to n, then gcd(a, n) > 1. For any b, gcd(ab, n) will have a common factor of gcd(a, n), so gcd(ab, n) > 1. Thus, ab cannot be congruent to 1 modulo n, because if \(ab \equiv 1 \pmod{n}\), then \(ab - 1 = kn\) for some integer k, which implies \(ab - kn = 1\). By Bezout's identity, this linear Diophantine equation has integer solutions for a and k if and only if gcd(b, n) divides 1, meaning gcd(b, n) = 1. Wait, this logic is slightly off. If \(ab \equiv 1 \pmod{n}\), then \(ab\) and \(n\) are coprime, which implies \(a\) and \(n\) must be coprime, and \(b\) and \(n\) must be coprime. So if \(ab \equiv 1 \pmod{15}\), then gcd(a, 15) = 1 and gcd(b, 15) = 1.