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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?

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    $\begingroup$ Hi, @Suresh, and welcome to PSE! When you post someone else's puzzle on the site, please always attribute the puzzle to the original creator. (Mentioning the book's and/or creator's name is usually enough.) Thanks, and happy puzzling! $\endgroup$ – Bass Mar 6 at 8:18
  • $\begingroup$ @Bass actually i don't know it's creator; one of my friend has given this question; so how would i mention its creator name? :( $\endgroup$ – Suresh Mar 6 at 10:13
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    $\begingroup$ I tried googling the numbers, but the only match other than this page is a Korean forum post from last August. $\endgroup$ – Jaap Scherphuis Mar 6 at 12:28
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    $\begingroup$ @Suresh You don't have to exhaustively search to find the original source if it's not readily available, but if it's not your original content then you should note where you found it. ("Source: sent to me by a friend" is acceptable, if that's honestly the best you can do.) $\endgroup$ – Rubio Mar 6 at 18:22
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The last pair of digits is given by the formula:

$a+2b+3c$

so for $4*8*9$ we get

$4+16+27=47$

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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 $5$.

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    $\begingroup$ Yeah, always with the sample size :-/. This seems to be a valid solution, but I think Jaap's must be the intended one, because it can be applied to any three numbers without the full knowledge of the other samples. +1ing anyway. $\endgroup$ – Bass Mar 6 at 8:12
  • $\begingroup$ Sorry I am not able to understand your reason for the $6^{th}$ digit , i.e, "the lowest digit in the opening local sequence that hasn't already been used by the previous entries in the global sequence." Can you please explain this statement in details... $\endgroup$ – Suresh Mar 6 at 8:13
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    $\begingroup$ @Suresh; I've added an explanation into my post. $\endgroup$ – JonMark Perry Mar 6 at 8:34

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