The number of distinct prime factors of the largest 4-digit number is
2
3
5
11
Step 1: The largest 4-digit number is \(9999\).
Step 2: Check if \(9999\) is divisible by \(3\).
Sum of digits: \(9+9+9+9 = 36\).
Since \(36\) is divisible by \(3\), \(9999\) is divisible by \(3\).
Step 3: Factor out \(9\) (because \(36\) is also divisible by \(9\)).
\(9999 = 9 \times 1111\).
And \(9 = 3 \times 3 = 3^2\).
Step 4: Now factor \(1111\).
Try \(11\): \(1111 \div 11 = 101\).
So, \(1111 = 11 \times 101\).
Step 5: Check that \(11\) and \(101\) are primes.
\(11\) is a prime number.
\(101\) has no divisors other than \(1\) and \(101\), so it is prime.
Step 6: Write the full prime factorization.
\(9999 = 3^2 \times 11 \times 101\).
Step 7: Count the distinct prime factors.
The distinct primes are \(3\), \(11\), and \(101\).
Number of distinct prime factors = \(3\).
Correct option: B