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?
 A: 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 $(a,b,c)$:


*

*If $c>0$, subtract $a$ from $c$, increase $a$ by one, subtract the new $a$ from $c$ (using two substractions, we avoid computation of $2a+1$).

*If $c<0$, add $b$ to $c$, increase $b$ by one, add the new $b$ to $c$

*If $c=0$, terminate; we have found factors $a+b$ and $a-b$.


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 $\sqrt N$.
A: 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...)
A: In this case it might be just as easy to try all primes less than the square root of 5671. With a pocket calculator this should be very rapid.
