Problem
NT-B1-M04-P022 Divisibility by \(3n+2\)
#22
★★★★☆ Level 4 of 5
Find all positive \(n\) such that \(3n+2\mid n^2+5n+9\).
Force \(3n+2\) to divide a constant.
If \(3n+2\mid n^2+5n+9\), then it divides \(9(n^2+5n+9)=9n^2+45n+81\). Subtract \((3n+2)^2=9n^2+12n+4\): we get \(3n+2\mid33n+77\). Now \(3(33n+77)-33(3n+2)=165\), so \(3n+2\mid165\). The positive divisors of \(165\) congruent to \(2\) modulo \(3\) are \(5\) and \(11\). Thus \(3n+2=5\) or \(11\), giving \(n=1\) or \(3\). Both values check.
A strong problem: the linear combination must be constructed by the student.