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Solution

Solution: Work modulo $64$. $49\equiv -15\pmod{64}$ and $49^{1}\equiv49,\ 49^{2}\equiv33,\ 49^{3}\equiv17,\ 49^{4}\equiv1$; hence $49^{n}$ is periodic with period $4$. Also $16n\equiv 0,16,32,48\ (\bmod\ 64)$ for $n\equiv 0,1,2,3$. For each residue class: $n\equiv0$: $1+0+\lambda\equiv0\Rightarrow \lambda\equiv -1$ $n\equiv1$: $49+16+\lambda\equiv1+\lambda\equiv0$ $n\equiv2$: $33+32+\lambda\equiv1+\lambda\equiv0$ $n\equiv3$: $17+48+\lambda\equiv1+\lambda\equiv0$ All give $\lambda\equiv -1\pmod{64}$. The least negative representative is $\boxed{-1}$.