Permutations & Combinations
Permutation and Combination
star_batch_jee_advanced_2025
Grade 11
Given that the divisors of $n = 3^p \cdot 5^q \cdot 7^r$ are of the form $4\lambda + 1, \lambda \geq 0$. Then
Step-by-Step Solution
Key Concept: A divisor $d = 3^a \cdot 5^b \cdot 7^c$ satisfies $d \equiv 1 \pmod{4}$ if and only if the exponent sum $a + c$ of the factors congruent to $-1 \pmod{4}$ is even.
For all divisors of $n = 3^p \cdot 5^q \cdot 7^r$ to be of the form $4\lambda + 1$, every divisor $3^a \cdot 5^b \cdot 7^c$ (where $0 \leq a \leq p, 0 \leq b \leq q, 0 \leq c \leq r$) must satisfy $3^a \cdot 5^b \cdot 7^c \equiv 1 \pmod{4}$. Since $3 \equiv -1 \pmod{4}$, $5 \equiv 1 \pmod{4}$, and $7 \equiv -1 \pmod{4}$, we need $(-1)^a \cdot 1^b \cdot (-1)^c \equiv 1 \pmod{4}$, which means $(-1)^{a+c} \equiv 1 \pmod{4}$. This is only possible if $a + c$ is always even, requiring $p = 0$ or $p$ even, AND $r = 0$ or $r$ even, meaning $p + r$ must be even. The parameter $q$ can be any non-negative integer since $5 \equiv 1 \pmod{4}$.
Correct Answer: 1,2,3