How to factor 5671? The other day I wanted to factor 5671 in my head. (It turns out to be $53\cdot107$, but I did not know this at the time.)  I quickly ruled out the easy divisors, 2, 3, 5, 7, 11, and 13. At this point I saw no obvious way to proceed short of a very tedious trial division.
But I did notice that 5671 has a residue of 1 modulo 2, 3, 5, and 7. Is there any way to use this coincidence to reduce or simplify the problem of finding a factorization of 5671, perhaps by ruling out certain types of divisors?

Just to elaborate on the suggestion of using the information about $n = 5671 \equiv 1 \mod {2,3,5,7}$ to find a factoration of $5671$: 
Suppose $n = p \cdot q$ for some factors $p$ and $q$, and $n \equiv 1 \mod{2,3,5,7}$. What does this tell us? Well, it gives us four equations in $p$ and $q$:
\begin{align}
p \cdot q &\equiv 1 \mod 2 \\
p \cdot q &\equiv 1 \mod 3 \\
p \cdot q &\equiv 1 \mod 5 \\
p \cdot q &\equiv 1 \mod 7
\end{align}
But since $2,3,5,7$ are primes, any element (but $0$) has an inverse, and so this tells us that:
\begin{align}
q &\equiv p^{-1} \mod 2 \qquad \qquad p,q \not\equiv 0 \mod 2 \\
q &\equiv p^{-1} \mod 3 \qquad \qquad p,q \not\equiv 0 \mod 3 \\
q &\equiv p^{-1} \mod 5 \qquad \qquad p,q \not\equiv 0 \mod 5 \\
q &\equiv p^{-1} \mod 7 \qquad \qquad p,q \not\equiv 0 \mod 7 
\end{align}
In fact, any choice of $p \mod {2,3,5,7}$ is possible, as long as none of those numbers are $0$, and it leads to a unique set of values for $q \mod{2,3,5,7}$. But we already knew that $p, q \not\equiv 0 \mod {2,3,5,7}$, as otherwise one of those numbers $2,3,5,7$ would be a divisor of $n$. 
So, a long story short (TL;DR): Knowing that $5671 \equiv 1 \mod {2,3,5,7}$ does not really help you find any of its divisors. It only tells you that $2,3,5,7$ are not divisors of $n$.
(Once you find $p = 53 \equiv (1,2,3,4) \mod {(2,3,5,7)}$, this does tell you that $q \equiv {(1,2,2,2)} \mod {(2,3,5,7)}$. But I suppose that once you know $p$, finding $q$ should not be too hard anyway...)
