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Question #40626

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of all prime numbers $p \le 367$ . For each $p \in S$ , let $n_p$ be the exponent of $p$ in the prime factorization of $367!$ . Find the number of primes $p \in S$ such that the binomial expansion of $\left( x^2 + \frac{1}{x^3} \right)^{n_p}$ does \textbf{not} contain a term independent of $x$ .

Options

A.
64
B.
62
C.
65
D.
54

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Term Independent of x|Exponent of Prime in n!|Complementary Counting

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of binomial expansion constraints with number theory (Legendre's formula).

The primary trap is the brute-force calculation of valid primes.

Students who do not employ complementary counting will be bogged down calculating prime distributions across 12 different fractional intervals where $n_p \not\equiv 0 \pmod 5$ .

Furthermore, the '367' trap tests if students correctly identify $\pi(367)=73$ and remember to check higher powers of small primes ( $p^2, p^3$ ) in Legendre's formula, which alters the modulo 5 parity for primes like 2, 3, and 7.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Binomial Theorem - Class 11
Concept	Term Independent of x
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free