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?
On
As two comments have already pointed towards the Fermat factorization method, let's spell out how one might employ it here in a way suitable for mental operation.
We observe that $49<56<81$, so $\sqrt{5671}\approx 70$ (very roughly). More precisely, we have $5671 = 70^2-0^2+771$ and start with the triple $(a,b,c)=(70,0,771)$. Now repeatedly do the following with the current triple $(a,b,c)$:
In our case the sequence of computation runs like this: $$(70,0,771)\\(71,0,630)\\(72,0,487)\\(73,0,342)\\(74,0,195)\\(75,0,46)\\(76,0,-106)$$ Note that the steps until now were only necessary because we started with an awfully rough estimate for $\sqrt N$. Next we use a littel shortcut as we step immediately to $b=\sqrt{|c|}$ and replace $c$ with $c+b^2$ $$(76,10,-6)\\(77,11,15)\\(78,11,-140)\\(78,12,-117)\\(78,13,-92)\\(78,14,-65)\\(78,15,-36)\\(78,16,-5)\\(78,17,27)\\(79,17,-130)\\\vdots\\(80,27,0)$$ Admittedly, it does take a while until one reaches a factorization (I left out eleven more steps here), but at least the single steps are very trivial (only addition and subtractions). One can devise some speed-ups for cases when $|c|\gg a,b$, but I'll leave that as an exercise. Depending on your memorizing capabilities, all this may not be suitable for actually doing a mental factorization, but it is sure good enough for doing it on a paper napkin. Of course, fastest results are obtained when $b$ is small, i.e. the factors are close to $\sqrt N$.
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...)