How many integers in the set {1, 2, 3......, 100} have exactly 3 divisors?

Correct — C. A number with exactly 3 divisors must be the square of a prime.

Divisor count: prime exponents a, b, … give (a + 1)(b + 1)… divisors
3 is prime, so the product is a single factor: a + 1 = 3, a = 2
So n = p², with divisors 1, p and p²

List the prime squares up to 100:
2² = 4, 3² = 9, 5² = 25, 7² = 49
Next prime: 11² = 121 > 100, so stop

Count: 4 integers → option (c)

Why the others are wrong

• (a)
  2 — 2 stops at 4 and 9. But 25 = 5² (divisors 1, 5, 25) and 49 = 7² (divisors 1, 7, 49) also have exactly three divisors.
• (b)
  8 — 8 overcounts. A qualifying number is p² ≤ 100, so p ≤ 10, and the primes up to 10 are just 2, 3, 5 and 7.
• (d)
  6 — 6 would need two more prime squares within 100. The next is 11² = 121, and a composite square such as 16 has five divisors (1, 2, 4, 8, 16).

Divisors pair up: each d that divides n pairs with n⁄d. The count is therefore odd only for perfect squares, where √n pairs with itself.

For exactly three, the product (a + 1)(b + 1)… must equal 3, which forces a single prime raised to the power 2.

So the question becomes how many primes p have p² ≤ 100. That means p ≤ 10: 2, 3, 5 and 7.

Key facts

• A number whose prime factorisation has exponents a, b, c, … has (a + 1)(b + 1)(c + 1)… divisors.
• A number has an odd number of divisors exactly when it is a perfect square.
• The numbers with exactly 3 divisors are the prime squares: 4, 9, 25, 49, 121, …

Common traps

• Counting every perfect square: 16, 36, 64, 81 and 100 have 5, 9, 7, 5 and 9 divisors, not 3.
• Counting 1 as a perfect square that qualifies, though it has a single divisor.

No directly related past PYQ was found.