If we multiply the 5-digit number $RAFFI$ by a single-digit $I$, then the resulting 6-digit number $MAAAAM$ lies below $100000*I$.
Hence its starting digit $M$ is smaller than $I$:
(1). The digits $M$ and $I$ satisfy $M<I$.
The last digit $M$ of $MAAAAM$ coincides with the last digit of $I^2$.
Testing $I=0,1,2,3,4,5,6,7,8,9$ yields $M=0,1,4,9,6,5,6,9,4,1$.
Combining this with $M<I$ from (1), we conclude:
(2). Either (2a) $I=8$ and $M=4$, or (2b) $I=9$ and $M=1$.
Let us get rid of case (2b): In this case $R\ge2$, as digit $1$ is already blocked by $M=1$.
Hence $MAAAAM=I*RAFFI>9*20000=180000$, which together with $M=1$ implies $A=8$. But then $MAAAAM=188881$ is not divisible by $I=9$; a contradiction.
So only case (2a) with $I=8$ and $M=4$ remains.
Since $MAAAAM$ is a multiple of $I=8$, also its last three digits $AAM$ are a multiple of $8$.
Out of the numbers $004,114,224,334,554,664,774,994$, only $224$ and $664$ work.
These two options $A=2$ and $A=6$ yield for $RAFFI=MAAAAM/8$ the respective values $52778$ (good) and $58333$ (bad). We summarize:
$8*52778=422224$ is the unique solution to Raffi's cryptarithm.