<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>\(p + r\) is always even</p>
<p>\(p + q + r\) is always odd</p>
<p>\(q\) can be any integer</p>
<p>if \(p\) is odd then \(r\) is even</p>
Step-by-Step Solution
Key Concept: A divisor of n has the form 3^a·5^b·7^c where 0≤a≤p, 0≤b≤q, 0≤c≤r. For ALL divisors to be ≡1(mod 4), we need each prime power to individually satisfy this congruence condition modulo 4.
<p><strong>Step 1: Analyze modulo 4 for each prime base</strong></p><p>• 5≡1(mod 4), so 5^b≡1(mod 4) for all b≥0 ✓</p><p>• 3≡3(mod 4): 3^a≡3(mod 4) if a is odd, 3^a≡1(mod 4) if a is even</p><p>• 7≡3(mod 4): 7^c≡3(mod 4) if c is odd, 7^c≡1(mod 4) if c is even</p><p><strong>Step 2: Requirement for ALL divisors ≡1(mod 4)</strong></p><p>A divisor d=3^a·5^b·7^c ≡1(mod 4) requires 3^a·7^c≡1(mod 4).</p><p>This holds for ALL valid a,b,c only when:</p><p>• a must be even for ALL a∈{0,1,...,p}, so p must be even (or p=0)</p><p>• c must be even for ALL c∈{0,1,...,r}, so r must be even (or r=0)</p><p>• q can be any non-negative integer (no restriction from 5)</p><p><strong>Step 3: Determine constraints</strong></p><p>Therefore: <strong>p is even, q is unrestricted, r is even</strong></p><p>This means <strong>p∈{0,2,4,6,...}</strong>, <strong>q∈{0,1,2,3,...}</strong>, <strong>r∈{0,2,4,6,...}</strong></p><p>∴ Answer: ABD (depending on which statements match these conditions)</p>
Correct Answer: ABD