Gardner riddle on mathemagicians A cute riddle (but maybe not so easy!) from Gardner:

At a gathering of mathemagicians, the Grand Master and his 8 disciples are seated at a round table. The Grand Master will judge each of his displices on their newest trick. After he has seen all the performances, the Grand Master hands to each of his disciples a card with their score on it (the score is some integer number of points greater than $0$). In return, The Grand Master then performs the trick himself and allows his disciples to judge his own perforamce. The disciples agree on a score and give the Grand Master a card with his score on it. It turns out that each mathemagician at the table has a different score.

The Grand Master then remarks: "I can think of a number that divides the product of my own score and the score of anyone seated at this table, other than the two people sitting beside me.". Each of the disciples look at their own scorecard and then look around the table, and each discplie remarks: "I can also think of such a number!". All at once, everyone seated at the table announces the number they have in mind. Incredibly, they all say same the same number! "Now that is some trick!" the Grand Master laughs. 

What is the smallest possible number the mathemagicians could have announced?
The Grand Master top-scores with a mere $3^2 \times 5^2 \times 7^2 \times 11 = 121275$ and the other scores are:
$$\begin{array}{|c|c|} 
\hline \text{Disciple} &amp; \text{Score} &amp; \text{Composition} \\
\hline 1 &amp; 80850 &amp; 3 \times 5 \times 5 \times 7 \times 7 \times 11 \times 2  \\
\hline 2 &amp; 53900 &amp; 5 \times 5 \times 7 \times 7 \times 11 \times 2 \times 2  \\
\hline 3 &amp; 32340 &amp; 5 \times 7 \times 7 \times 11 \times 2 \times 2 \times 3  \\
\hline 4 &amp; 19404 &amp; 7 \times 7 \times 11 \times 2 \times 2 \times 3 \times 3  \\
\hline 5 &amp; 13860 &amp; 7 \times 11 \times 2 \times 2 \times 3 \times 3 \times 5  \\
\hline 6 &amp; 9900 &amp; 11 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5  \\
\hline 7 &amp; 6300 &amp; 2 \times 2 \times 3 \times 3 \times 5 \times 5 \times 7  \\
\hline 8 &amp; 22050 &amp; 2 \times 3 \times 3 \times 5 \times 5 \times 7 \times 7  \\
\hline \end{array}
$$
and the new marvellous number is $485100$
(thanks to Prajanan Pate for the improvement)

original answer
So the Grand Master scores highly, of course, getting $5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 = 37182145$
The remaining disciples, in order round the table, get:
$$\begin{array}{|c|c|} 
\hline \text{Disciple} &amp; \text{Score} &amp; \text{Composition} \\
\hline 1 &amp; 14872858 &amp; 2 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 \\
\hline 2 &amp; 6374082 &amp; 2 \times 3 \times 11 \times 13 \times 17 \times 19 \times 23 \\
\hline 3 &amp; 2897310 &amp; 2 \times 3 \times 5 \times 13 \times 17 \times 19 \times 23 \\
\hline 4 &amp; 1560090 &amp; 2 \times 3 \times 5 \times 7 \times 17 \times 19 \times 23 \\
\hline 5 &amp; 1009470 &amp; 2 \times 3 \times 5 \times 7 \times 11 \times 19 \times 23 \\
\hline 6 &amp; 690690 &amp; 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 23 \\
\hline 7 &amp; 510510 &amp; 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \\
\hline 8 &amp; 4849845 &amp; 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \\
\hline \end{array}
$$
and the &quot;incredible&quot; number is $23\#=223092870$.
... and the field is open for a lower number!
What about 2, 2, 3, 3, 5, 5, 2, 7 11 = 138600