Problem
NT-B2-M09-P006 Three Consecutive Divisibilities
#6
★★★☆☆ Level 3 of 5
Find the least positive \(N\) such that \(N+1\) is divisible by \(4\), \(N+2\) by \(5\), and \(N+3\) by \(9\).
Rewrite the conditions as residues for \(N\).
We need \(N\equiv 3\pmod 4\), \(N\equiv 3\pmod 5\), \(N\equiv 6\pmod 9\). The first two conditions give \(N\equiv 3\pmod {20}\). Let \(N=20t+3\). Then \(20t+3\equiv 6\pmod 9\), so \(2t\equiv 3\pmod 9\). The inverse of \(2\) modulo \(9\) is \(5\), hence \(t\equiv 15\equiv 6\pmod 9\). The least \(N\) is \(20\cdot 6+3=123\).
A standard construction problem before stronger blocks.