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Question #40744

Difficulty: 8/10 | 🟢 Free

The Question

Let $E_p(N)$ denote the exponent of the highest power of prime $p$ dividing $N$ . Consider two parallel lines $L_1: Ax + By + C_1 = 0$ and $L_2: Ax + By + C_2 = 0$ , where: $A = E_2\left( \sum_{r=0}^{32767} \binom{32767}{r}^2 \right), \quad B = E_3\left( \sum_{r=0}^{6560} \binom{6560}{r}^2 \right)$ $C_1 = E_5\left( \sum_{r=0}^{125} (-1)^{125-r} \binom{125}{r} r^{125} \right), \quad C_2 = E_5\left( \sum_{r=0}^{265} (-1)^{265-r} \binom{265}{r} r^{265} \right)$ Find the distance between the lines $L_1$ and $L_2$ .

Options

A.
2
B.
$\frac{33}{17}$
C.
$\frac{28}{17}$
D.
$\frac{35}{17}$

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

Distance Between Parallel Lines|Binomial Coefficient Summation Properties|Exponent of Prime in n!

Why This Question Is Tricky

Common Mistakes

This question integrates high-order binomial summations with number theory (Legendre's formula and Kummer's theorem) and coordinate geometry.

The primary trap lies in the evaluation of $C_1$ and $C_2$ .

Students often compute prime exponents by only considering the first one or two powers of the prime (e.g., calculating $E_5(265!)$ as $53 + 10 = 63$ , missing the $125$ divisor).

This precision error directly alters the constant terms of the parallel lines, leading to distractor options like $33/17$ .

Furthermore, the problem requires recognizing the inclusion-exclusion identity for $n!$ and applying base-conversion logic to evaluate $\binom{2n}{n}$ , acting as a rigorous multi-concept filter typical of JEE Advanced.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Straight Lines - Class 11
Concept	Distance Between Parallel Lines
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free