Problem
NT-B2-M09-P007 Two Linear Conditions
#7
★★★☆☆ Level 3 of 5
Find all \(n\) such that \(4n+1\) is divisible by \(9\), and \(5n-2\) is divisible by \(11\).
First solve the two linear congruences for \(n\).
From \(4n+1\equiv 0\pmod 9\), \(4n\equiv 8\), so \(n\equiv 2\pmod 9\). From \(5n-2\equiv 0\pmod {11}\), \(5n\equiv 2\), and \(5^{-1}\equiv 9\pmod {11}\), so \(n\equiv 18\equiv 7\pmod {11}\). Solve \(n=9t+2\), \(9t+2\equiv 7\pmod {11}\), so \(9t\equiv 5\). Since \(9^{-1}\equiv 5\), \(t\equiv 25\equiv 3\pmod {11}\). The answer is \(n\equiv 29\pmod {99}\).
Good for checking modular inverses.