Find the number of positive divisors of 2002.
$ 2002 = 2^1 \cdot 7^1 \cdot 11^1 \cdot 13^1 \qquad \Rightarrow \qquad t(2002) = (1 + 1)(1 + 1)(1 + 1)(1 + 1) = \boxed{16}. $
16
Level 3
Number Theory
Problem: Find the number of positive divisors of 2002.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the number of positive divisors of 2002, first, we need to factor 2002 into its prime components. We start by dividing 2002 by the smallest prime number, which is 2. Since 2002 is even, we have:

2002 ÷ 2 = 1001.

Next, we factor 1001. We check divisibility by the next smallest prime, which is 3. The sum of the digits of 1001 is 2, which is not divisible by 3, so we move to the next prime, 5. Since 1001 does not end in 0 or 5, it is not divisible by 5. We then check 7:

1001 ÷ 7 = 143.

Now we factor 143. We check divisibility by 11:

143 ÷ 11 = 13.

At this point, we have fully factored 2002 into prime factors: 

2002 = 2^1 × 7^1 × 11^1 × 13^1.

To find the number of positive divisors, we use the formula for the number of divisors based on the prime factorization. If a number n is expressed as p1^e1 × p2^e2 × ... × pk^ek, the number of positive divisors d(n) is given by:

d(n) = (e1 + 1)(e2 + 1)...(ek + 1).

In our case, the exponents of the prime factors are all 1:

e1 = 1 (for 2),
e2 = 1 (for 7),
e3 = 1 (for 11),
e4 = 1 (for 13).

Applying the formula:

d(2002) = (1 + 1)(1 + 1)(1 + 1)(1 + 1) = 2 × 2 × 2 × 2 = 16.

Thus, the final answer is 16.<|END_OF_LATENT|> The final answer is $16$.