Mentally finding factors of an integer of up to three-digits. I’m looking for the easiest methods to find factors, not necessarily prime, of positive integers $<1000$, preferably mentally, but using pen and blank paper when required.
Although these techniques might be useful for a certain televised Numbers Game, I’m just looking for methods faster than trial division.
I take for granted full knowledge of the twelve by twelve multiplication tables, and recognition of primes up to $101$
This, with the addition of $7\cdot13=91$ should cover much of the factorisation up to two digits.
Notation
Using the idea of hundreds, tens and units, so $n=100H+10T+U$ is the number, I’ll use $H,T,U$ for the individual digits,  $HT$ for the first two digits and $TU$ for the last two.
For example, $n=456$ gives $H=4$, $T=5$, $U=6$, $HT=45$ and $TU=56$
Some easy published divisibility tests
If $U=0$ then $10|n$
If $(TU)\equiv 0{\pmod{25}}$ then ${25}|n$
If $U=5$ then $5|n$
If $U\equiv 0{\pmod 2}$ then $2|n$
If $(H+T+U)\equiv 0{\pmod 9}$ then $9|n$
If $(H+T+U)\equiv 0{\pmod 3}$ then $3|n$
Merged published divisibility tests for primes $p=7$ to $47$
With $z$ taken from the following, if $(HT+zU)\equiv 0{\pmod p}$ then $p|n$ 
$$(p,z)$$
$$(7,-2)$$
$$(11,-1)$$
$$(13,4)$$
$$(17,-5)$$
$$(19,2)$$
$$(23,7)$$
$$(29,3)$$
$$(31,-3)$$
$$(37,-11)$$
$$(41,-4)$$
$$(43,13)$$
$$(43,-30)$$
$$(47,-14)$$
For some examples in the range, the test may need to be repeated on the result.
As $31^2=961$, the last few test are of limited use.
Alternative divisibility test for $11$
If $(H+U-T)=0$ or $11$ then $11|n$
Difference of two squares
Using $a^2-b^2\equiv(a+b)(a-b)$ helps in a few cases. For example,
$$899=900-1=(30+1)(30-1)=31\cdot29$$
$$91=100-9=(10+3)(10-3)=13\cdot7$$
$$391=400-9=(20+3)(20-3)=23\cdot17$$
My question
I was tempted to ask what other divisibility or factorisation techniques exist that are potential easier than trial division, but “easier” is subjective; so I questions is:.
What other mental divisibility or factorisation techniques exist for an integer up to three digits? 
Clarification update 11 May 2018
Sorry, I don’t feel I’ve sufficiently emphasised the importance of time, just assuming that easier was faster. 
Full factorisation is ideal; partial factorisation is much better than nothing, so $663=3\cdot13\cdot17$ is best, but $663=3\cdot221$ is better than nothing.
Obtaining more than one small prime factor with one test is desirable, where the prime factors are obvious, as per the tests for $10,25$ and $9$, but a full list of all possible factors of $n$ is not needed. 
 A: As you have stated:
Checking a number is even will automatically give you the factor of $2$.
Summing the digits, if the digit sum is a multiple of $3$ or $9$, the number is also a multiple of $3$ or $9$. 
Multiples of $5$ end in $5$, multiples of $10$ end in $0$. 

For my extra method, let's consider $473$. It has no obvious factors based on those rules, so here I would use prime multiple subtraction. Start with $7$, the multiple of $7$ that ends in $3$ is $7\cdot9=63$, so subtract $63$ to get $410$. $41$ is not a multiple of $7$ or $9$, so neither are factors. Then consider $11$. $3\cdot 11=33$ so subtract $33$ to get $440$. $440=40\cdot 11$, hence we can see that $473=43\cdot 11$.

Note for $11$ that if you have $ab\cdot11$ , where $a,b$ represent the digits, the result is $a(a+b)b$, again to mean digits: e.g. $35\cdot11=385$. When $a+b>10$, it is harder to spot but still possible. What you should look for then is: $a(a+b-11)b$, and this is $(a-1)b \cdot 11$. E.g. for $528$, $5+8-11=2, \therefore 528=11*48$

If you can identify numbers as concatenations of multiples  of the same number, they will be a multiple of that number. E.g. $654$ contains $6$ and $54$, both of which are multiples of $6$, and so we can do $6\div6=1$, and $54\div6=09$, therefore $654\div6=109$
