Let N = 65 × 359 × 113. How many distinct prime factors does N have?

Introduction / Context:
This question is about prime factorisation and counting distinct prime factors of a composite number. We are given N as a product of three numbers, 65, 359, and 113, and asked to determine how many different prime numbers appear in the prime factorisation of N. Understanding how to break composite numbers into primes is a fundamental skill in number theory.

Given Data / Assumptions:

• N = 65 × 359 × 113.
• We need to find the distinct prime factors of N.
• Standard definitions: a prime number has exactly two positive divisors, 1 and itself.
• We assume 359 and 113 are primes (they have no smaller prime divisors).

Concept / Approach:
We first factor each component of N into primes. Then we combine all prime factors and remove duplicates to get the set of distinct primes. The number of elements in this set is the answer. The key step is to factor 65, since 359 and 113 are already prime numbers with no non trivial divisors.

Step-by-Step Solution:
Step 1: Factor 65. Note that 65 is divisible by 5 because it ends in 5.Step 2: Compute 65 ÷ 5 = 13, so 65 = 5 × 13.Step 3: Check 13; it is a prime number as it has no divisors other than 1 and 13.Step 4: Check 359. It is not divisible by small primes such as 2, 3, 5, 7, 11, or 13, and is known to be prime.Step 5: Check 113. It is not divisible by 2, 3, 5, 7, or 11, and is also prime.Step 6: Therefore the prime factorisation of N is N = 5 × 13 × 359 × 113.Step 7: The distinct prime factors are 5, 13, 359, and 113.Step 8: Count these primes: there are 4 distinct prime factors.

Verification / Alternative check:
You can verify by multiplying the prime factors back together. First compute 5 × 13 = 65. Then N = 65 × 359 × 113 = (5 × 13) × 359 × 113. Since multiplication is associative and commutative, the order does not matter. The product matches the given definition of N, confirming that we have neither missed nor added any prime factors. No further factorisation is possible because each of 5, 13, 359, and 113 is prime.

Why Other Options Are Wrong:
Option 2: This would imply only two primes, which is impossible since 65 alone contributes two primes, 5 and 13.Option 3: Would require one of 359 or 113 to be composite, which is not the case.Option 5 and Option 6: These overcount the number of distinct primes; there are no additional prime factors beyond 5, 13, 359, and 113.

Common Pitfalls:
Students may forget to factor 65 completely and assume it is prime, which would lead to undercounting. Another mistake is confusing distinct prime factors with the total number of prime factors including multiplicity; here each prime appears once. Always factor each composite completely into primes and then count unique primes only when asked for distinct prime factors.

Final Answer:
The number N has 4 distinct prime factors.