Pattern recognition problem: if $2 \star 8 \star 8 = 161642\dots$ then M=?

Question

The following sequence was given to me by one of my friend:

$$\color{red}{2 \star 8 \star 8 = 161642}$$ $$\color{blue}{4 \star 9 \star 7 = 362843}$$ $$\color{red}{7 \star 5 \star 9 = 356344}$$ $$\color{blue}{9 \star 6 \star 8 = 547245}$$ $$\color{red}{5 \star 7 \star 9 = 354546}$$ $$\color{blue}{3 \star 9 \star 9 = 272748}$$ $$\color{red}{4 \star 8 \star 9 = \text{M}}$$ $\color{red}{\text{M}}$ = ?

My Thoughts:

The first two digits of R.H.S can be found by $a.b$ , where L.H.S = $a \star b \star c$ ,
i.e, For 1st relation: $2.8=16 = 1^{st}$ two digits of $\color{green}{16}1642$

Similarly,the second two digits of R.H.S can be found by $a.c$ , where L.H.S = $a \star b \star c$ ,
i.e, For 2nd relation: $4.7=28 = 2^{nd} $ two digits of $36 \color{green}{28}43$

$$\implies M= 3236 \, \_ \, \_$$

How do I find the $3^{rd}$ two digits of R.H.S?

Answer 1

The last pair of digits is given by the formula:

$a+2b+3c$

so for $4*8*9$ we get

$4+16+27=47$

Answer 2

M=

$323647$

because

the first four digits are as you give. The fifth digit is $4$. The final digit is $1$+the lowest digit in the opening local sequence that hasn't already been used by the previous entries in the global sequence. In this case $2+1=3,3+1=4$ and $4+1=5$ have already been used previously, leaving $6+1=7$ as the last digit.

For example, the lowest digit in $16164$ is $1$, so the last digit is $1+1=2$. The lowest digit in $36284$ is $2$, so last digit is $3$. The lowest digit in $54724$ is $2$, but $3$ has already been used, so we try with the next lowest digit, $4$, to give the last digit as $5$.