3
$\begingroup$

I am unable to figure out how to solve this question @Joffan has given an answer but he hasn't mentioned much regarding his approach. I requested him but he still didn't give a clear answer. This is the question :

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?

I would appreciate if someone can either write an original answer explaining how to solve the above question or explain to me, the thinking process and the steps needed, to reach @Joffan 's answer .

A big bonus would be an an intuitive answer. An intuitive answer, according to me, is one that is not just easy to understand but also makes one go, "Oh yes.. that was so obvious. Why didn't I think of solving it this way. "

This is what I have understood so far : Let's call the persons around the table, $A_1 , A_2 .....A_9$ . Let's call the announced number, X. Also,

let's assume that $A_1$ * 7= X. Now, $A_1 * A_9$ is not divisible by X. This means that $A_9$ does not have 7 as a factor, otherwise $A_1 * A_9$ would have been divisible by X.

Similarly, $A_2$ cannot have 7 as a factor, otherwise $A_1 * A_2$ would be divisible by X. However, $A_2 * A_9$ need to be divisible by X. But, since both $A_2$ and $A_9$ don't have 7 as a factor, therefore, $A_2 * A_9$ cannot be divisible by X. This means that our assumption that $A_1$ has only one missing factor is wrong. Thus, $A_1$ needs to have, at least 2 missing factors.
I just cannot figure out how to progress further from here.

$\endgroup$
0
2
$\begingroup$

I can give some insight into Joffan's answer if that helps, which follows Milo Brandt's comment underneath the question.

First of all let's look at the original answer given by Joffan

Let's say I take any nine distinct primes $p_1, p_2, \ldots, p_9$ and I assign to each mathemagician, as their score, a product of seven of the primes which are consecutive when considered cyclically.

Explicitly, considering the mathemagicians in a clockwise order around the table:

Grand Master has the score $p_3 p_4p_5p_6p_7p_8p_9$
Mathemagician 1 has the score $p_4p_5p_6p_7p_8p_9p_1$
Mathemagician 2 has the score $p_5p_6p_7p_8p_9p_1p_2$
Mathemagician 3 has the score $p_6p_7p_8p_9p_1p_2p_3$
Mathemagician 4 has the score $p_7p_8p_9p_1p_2p_3p_4$
Mathemagician 5 has the score $p_8p_9p_1p_2p_3p_4p_5$
Mathemagician 6 has the score $p_9p_1p_2p_3p_4p_5p_6$
Mathemagician 7 has the score $p_1p_2p_3p_4p_5p_6p_7$
Mathemagician 8 has the score $p_2p_3p_4p_5p_6p_7p_8$

Notice that, under this arrangement, the product of the scores of any two mathemagicians not seated next to each other is divisible by $p_1p_2p_3p_4p_5p_6p_7p_8p_9$. However, if two mathemagicians are seated next to each other, there is some $p_i$ which does not divide their product.

So this will work for any set of nine primes which are distinct and, in Joffan's origin answer, they pick them to be the first nine primes.

However, this begs the question, can we improve on this by making some of the $p_i$s equal?

Apparently, after discussion with "Prajanan Pate", Joffan realises that some of these can be set to be equal to keep this strategy working but no more than two can be equal to each other.

In particular, we could put $p_1=p_2$, $p_3=p_4$, $p_5=p_6$, $p_7=p_8$, and the arrangement above will still maintain the property that the product of two non-adjacent mathemagician's scores will be divisible by $p_1p_2p_3p_4p_5p_6p_7p_8p_9$ while the product of two adjacent mathemagician's scores will not be.

To get the smallest announced number, Joffan assigns the smallest primes appropriately. In particular, $$p_1=p_2=2\,\,\,,\,\,\,p_3=p_4=3\,\,\,,\,\,\,p_5=p_6=5\,\,\,,\,\,\,p_7=p_8=7\,\,\,,\,\,\,p_9=11$$ and the number announced is $$2^2 \times 3^2 \times 5^2 \times 7^2 \times 11 = 485100$$

$\endgroup$
8
  • $\begingroup$ Why can't more than 2 $p_i$s be equal ? $\endgroup$ – Hemant Agarwal Apr 1 at 11:17
  • $\begingroup$ @HemantAgarwal Try to make 3 $p_i$s equal and see what happens for yourself. $\endgroup$ – hexomino Apr 1 at 11:19
  • $\begingroup$ Let's assume that $$p_1=p_2=p_3=2\,\,\,,\,\,\,p_4=p_5=3\,\,\,,\,\,\,p_6=p_7=5\,\,\,,\,\,\,p_8=p_9=7$$ Then, I can't see any conditions getting violated. What am I missing ? Secondly, I can intuitively understand that $p_i$s will be all prime numbers . But, if I were to explain this to someone, then I don't know how to do it . What is a simple explanation behind why all the 9 numbers need to be prime ? $\endgroup$ – Hemant Agarwal Apr 1 at 13:24
  • 1
    $\begingroup$ @HemantAgarwal In the solution you propose, the product of the Grand Master and mathemagician 8's scores is divisible by $p_1p_2p_3p_4p_5p_6p_7p_8p_9$ and they are sitting next to each other. $\endgroup$ – hexomino Apr 1 at 13:27
  • 2
    $\begingroup$ @HemantAgarwal The answer above is certainly not a proof but more of an explanation of Joffan's answer which is clearly working off intuition. If I think of a nice proof I'll add it in but no guarantees. $\endgroup$ – hexomino Apr 1 at 13:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.