Multiplication without figures I have taken the 12th problem of this pdf, do you know any way to resolve it without using brute force?
Simply I have to replace '*' of this multiplication below with correct digits, in order to have the factors corrected.
$$\begin{alignedat}{2}*&*4&&\,\times{}\\ &**&&={}\\\hline 44&*4&&  \end{alignedat} $$
Each '*' can be a different digit.
 A: The most straightforward way I can think of is to consider all ten possibilities for the * in 44*4, factor each into primes, and see which ones have prime factors that can recombine into a two-digit number and a three-digit number that ends in a $4$:
$$\begin{align}
4404&=2\cdot2\cdot3\cdot367\\
4414&=2\cdot2207\\
4424&=2\cdot2\cdot2\cdot7\cdot79\\
4434&=2\cdot3\cdot739\\
4444&=2\cdot2\cdot11\cdot101\\
4454&=2\cdot17\cdot131\\
4464&=2\cdot2\cdot2\cdot2\cdot3\cdot3\cdot31\\
4474&=2\cdot2237\\
4484&=2\cdot2\cdot19\cdot59\\
4494&=2\cdot3\cdot7\cdot107
\end{align}$$
There is more than one answer:

 $4444=11\cdot404\qquad 4464=36\cdot124\qquad4494=21\cdot214$

However, this seems a bit brute-forcey. In particular, how do we know $2207$ and $2237$ are primes?  If you're trying to do everything by hand (which I didn't!), you've got your work cut out for you.
A: *

*Hints of bruteforceless method without factorising:


$\ \ \ \ \  \  abc\ \ $
$*\ \ \ \  \ de$
$=xyzt$
1- $t=ce=4e$, returning to rule of congruence $e$ is either $1$ or $6$ whilst $r(t)=2$ the retained
2- Since $x=ad+r(y)=4$ so either ${a,d}=2,r(y)=0$ or $a=1$ and $d={1,2,3}$ or vice versa with $r(y)$ is respectively {3,2,1}
3- Sine $ae+bd=4$ starting with case $e=1$
$a+bd+r(z)=4+r(y)$ if $(a,d)=(2,2),r(y)=0$ then $b=1$ and $r(z)=0$ so $z=dc+be=4*2+1*1=9$ is a valid solution
