Question

Let N be the number of ordered triples of (A, B, C) such that $A \cup B \cup C = \{1, 2, 3,....,2003\}$ and $A \cap B \cap C = \phi$. Then number of divisors of N is equal to

Answer 1

2004^2

Answer 2

Let $S = \{1, 2, 3, \ldots, 2003\}$. The size of the set $S$ is $n = 2003$.

We are looking for the number of ordered triples $(A, B, C)$ such that:

1. $A \cup B \cup C = S$
2. $A \cap B \cap C = \phi$

Let's consider an arbitrary element $x \in S$. For each element $x$, we need to determine which of the sets A, B, or C it belongs to. There are $2^3 = 8$ possible ways an element $x$ can be distributed among the three sets A, B, C:

1. $x \in A, x \notin B, x \notin C$ (belongs to A only)
2. $x \notin A, x \in B, x \notin C$ (belongs to B only)
3. $x \notin A, x \notin B, x \in C$ (belongs to C only)
4. $x \in A, x \in B, x \notin C$ (belongs to A and B, but not C)
5. $x \in A, x \notin B, x \in C$ (belongs to A and C, but not B)
6. $x \notin A, x \in B, x \in C$ (belongs to B and C, but not A)
7. $x \in A, x \in B, x \in C$ (belongs to A, B, and C)
8. $x \notin A, x \notin B, x \notin C$ (belongs to none of A, B, C)

Now, let's apply the given conditions:

• Condition 1: $A \cup B \cup C = S$. This means every element $x \in S$ must belong to at least one of A, B, or C. Therefore, state 8 ($x \notin A \cup B \cup C$) is not allowed.

• Condition 2: $A \cap B \cap C = \phi$. This means no element $x \in S$ can belong to all three sets A, B, and C simultaneously. Therefore, state 7 ($x \in A \cap B \cap C$) is not allowed.

So, for each element $x \in S$, there are $8 - 2 = 6$ allowed ways for it to be distributed among the sets A, B, C.

Since there are $n = 2003$ elements in $S$, and the choice for each element is independent, the total number of ordered triples $(A, B, C)$ is $6 \times 6 \times \ldots \times 6$ (2003 times).

Thus, $N = 6^{2003}$.

Next, we need to find the number of divisors of $N$.

First, express $N$ in its prime factorization form: $N = 6^{2003} = (2 \times 3)^{2003} = 2^{2003} \times 3^{2003}$.

For an integer $k = p_1^{e_1} p_2^{e_2} \cdots p_m^{e_m}$, where $p_i$ are distinct prime numbers and $e_i$ are their exponents, the number of divisors is given by the product of (exponent + 1) for each prime factor: $(e_1+1)(e_2+1)\cdots(e_m+1)$.

In our case, $p_1 = 2$, $e_1 = 2003$, and $p_2 = 3$, $e_2 = 2003$.

Number of divisors of $N = (2003+1)(2003+1)$

Number of divisors of $N = (2004)(2004)$

Number of divisors of $N = 2004^2$.

The final answer is $2004^2$.