Permutations & Combinations
Number theory constraints
Grade 11
Question:
<p>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,</p>
<p>(a) \(p + r\) is always even</p>
<p>(b) \(p + q + r\) is even or odd</p>
<p>(c) \(q\) can be any integer</p>
<p>(d) if \(p\) is even, then \(r\) is odd</p>
Step-by-Step Solution
Key Concept: Use properties of numbers modulo 4 to determine constraints on exponents in prime factorization.
<p>Since $3^p = (4-1)^p = 4\lambda_1 + (-1)^p$, and for divisors to be of form $4\lambda + 1$, we need $(-1)^p \equiv 1 \pmod{4}$, which means $p$ is even.</p><p>Similarly, $7^r = (8-1)^r = 4\lambda_3 + (-1)^r$, so $r$ must be even.</p><p>Thus $p + r$ is always even (sum of two even numbers).</p><p>Since $5 \equiv 1 \pmod{4}$, $q$ can be any non-negative integer.</p><p>Therefore $p + q + r = \text{even} + q + \text{even} = \text{even or odd}$ depending on $q$.</p>
Correct Answer: a, b, c