Conjurer's trick

**le_su14** · September 18th 2007, 06:23 AM

Hi, everyone! I have a exciting puzzle here:

"A conjurer invites a audience on stage. Then the conjurer goes to the tormentor and his partner then asks the audience to write a series of numbers on a board. He(the partner) erases 2 adjacent numbers in that series. Then, the conjurer come back to the stage and try to guess which numbers are erased.

Find the least length of the series of numbers to the conjurer always guesses correctly?

"

Let's discuss about this puzzle. I haven't solved it yet.

**topsquark** · September 18th 2007, 06:37 AM

Originally Posted by le_su14

I have a exciting puzzle but I can't solve it. Help me, please!

A conjurer invites a audience on stage. Then the conjurer goes to the tormentor and his partner then asks the audience to write a series of numbers on a board. He(the partner) erases 2 adjacent numbers in that series. Then, the conjurer come back to the stage and try to guess which numbers are erased.

Find the least length of the series of numbers to the conjurer always guesses correctly?

Can't be done, in general.

However I suspect you are learning about arithmatic and geometric series, which have the following formulas (respectively)
$a_n = a_0 + nd$
$a_n = a_0r^n$

Each of these equations has two unknowns in it ( $a_0, d$ and $a_0,r$ respectively) so we need two numbers in the series in order to solve for them. BUT we can always write both an arithmatic and geometric series between any two numbers, so to distinguish between the cases we need an extra number. So your answer is likely that you need three numbers from the series.

However, not all series are so nice. For example, the Fibbonacci series
$a_n = a_{n - 1} + a_{n - 2};~a_0 = 1, ~ a_1 = 1$
fits neither of these types of series, and of course the series of numbers 3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, ... representing the digits of $\pi$ follow no known pattern. So your problem is insolvable in general.

-Dan

**le_su14** · September 18th 2007, 11:08 AM

Originally Posted by topsquark

Can't be done, in general.

However I suspect you are learning about arithmatic and geometric series, which have the following formulas (respectively)
$a_n = a_0 + nd$
$a_n = a_0r^n$

Each of these equations has two unknowns in it ( $a_0, d$ and $a_0,r$ respectively) so we need two numbers in the series in order to solve for them. BUT we can always write both an arithmatic and geometric series between any two numbers, so to distinguish between the cases we need an extra number. So your answer is likely that you need three numbers from the series.

However, not all series are so nice. For example, the Fibbonacci series
$a_n = a_{n - 1} + a_{n - 2};~a_0 = 1, ~ a_1 = 1$
fits neither of these types of series, and of course the series of numbers 3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, ... representing the digits of $\pi$ follow no known pattern. So your problem is insolvable in general.

-Dan

Thank for your reply, Dan!

This puzzle is given by my teacher. I'm learning about advance discrete.
In this puzzle, the series of numbers is writen randomly by the audience. So, maybe there is no rule in this series. A key of this problem is that which 2 adjacent numbers are chosen in this random series by the partner.
How can he choose it?

**le_su14** · September 20th 2007, 09:41 AM

Hi topsquark,
The answer is very interesting.
The partner needs at least 101 numbers. Let $a = (a_1 + a_3 +...+ a_{101})$ mod 10; $b = (a_2 + a_4 + a_{100})$ mod 10.
The first erased position: $p = 10*a + b$ . (a = 0, b = 0 means p = 100)

The conjurer uses the value of p and a' ( where $a' = a_1 + a_3 +...+ a_{p-1} + a_{p+1} +...+ a_{101}$ ) to perceive a, b and $a_p$ .

Similarly, he uses p and b' (where $b' = a_0 + a_3 +...+ a_{p-1} + a_{p+1} +...+ a_{101}$ ) to perceive $a_{p+1}$ .

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