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In the magical circus, today a magician juggles with balls, cones, rings and umbrellas. While performing his art, he sometimes applies a powerful magic word:

  1. Ebrecedebre transforms one ball into one cone and one ring.
  2. Ibricidibri transforms one cone into two rings and one umbrella.
  3. Obrocodobro transforms one ring into one umbrella and three balls.
  4. Ubrucudubru transforms one umbrella into one ball and four cones.

The reversed magic words act in the opposite way:

  1. erbedecerbE transforms one cone and one ring into one ball.
  2. irbidicirbI transforms two rings and one umbrella into one cone.
  3. orbodocorbO transforms one umbrella and three balls into one ring.
  4. urbuducurbU transforms one ball and four cones into one umbrella.

At the current moment, the magician is juggling with a single cone (and no balls, no rings, no umbrellas). Which of the following four situations can be reached by applying these magic words?

  A.  At least five balls, and no cones, rings, umbrellas.  
  B.  At least five cones, and no balls, rings, umbrellas.  
  C.  At least five rings, and no balls, cones, umbrellas.  
  D.  At least five umbrellas, and no balls, cones, rings.
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    $\begingroup$ He just says "Abracadabra!" and the problem magically solves itself. :P $\endgroup$ – Joe Z. Oct 8 '15 at 17:28
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First lets define some variables.
B = Ball, C = Cone, R = Ring and J = Umbrella.

Equations:

  • B = C R
  • C = J 2R
  • R = J 3B
  • J = B 4C

1) 4R = 4B | Proof: 4R = J 3B 3R = C R 3B = 4B
2) R = 4B | Proof: R = J 3B = 4B 4C > (Use Proof 1) > 4C 4R = 4B
3) C = 9B | Proof: C = J 2R > (Use Proof 2) > J 8B = C4 9B > (Use Proof 1) > 5B C4 R4 = 9B
4) J = 37B | Proof: J = B 4C > (Use Proof 3) > J = 37B

Here is solution:

We have single Cone in start.
A. At least five balls, and no cones, rings, umbrellas. C = 9B
B. At least five cones, and no balls, rings, umbrellas. Let x divisble by 4. C = 9B = (9-x)B xR = 9-xB+4xB = 9+3xB = 45B = 5C(where x=12)
C. At least five rings, and no balls, cones, umbrellas.
Let x divisble by 4. C = 9B = (9-x)B xR = 9-xB+4xB = 9+3xB which never be a multiply of 4. Thus not possible.
D. At least five umbrellas, and no balls, cones, rings.
Let x divisble by 4. C = 9B = (9-x)B xR = 9-xB 4xB = 9+3xB = 333B = 9J (where x=108)

Since last part is math problem, i assume previous parts can be explained with mathematical formulas.

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    $\begingroup$ In proof 3, after using proof 2, you should have J 8B. Obviously a typo since your next step is correct. Nice work. $\endgroup$ – Callidus Jan 14 '15 at 13:27
  • $\begingroup$ In part (B), you have (9-x)B xR at one stage, and you later say x = 12. That would mean the juggler had (-3)B 12R at one point, which doesn't make sense. However, there is an easy fix: if you let x = 8, then you get 33B = (6+27)B = 6B 3C. Setting aside 6B 2C, we can take 1C and run the same steps again, this time with x = 4, to get a further 21B = (3+18)B = 3B 2C. We now have a total of (6B 2C) + (3B 2C) = 9B 4C = 5C. (Maybe this is obvious, but I thought I'd mention it.) $\endgroup$ – mathmandan Jan 16 '15 at 16:40
  • $\begingroup$ Also, in part (C): it's not quite obvious to me that this proves the impossibility, because it's not obvious that you have to follow this chain of equalities. I don't see why you couldn't perhaps do some other transformations instead (possibly involving a J or something?) and get a multiple of 4. However, I do have an alternative proof that (C) is impossible. My proof also uses divisibility ideas, so the idea is similar to yours, but I'd like to share it anyway. I'll post it as an answer, though, since there won't be room here. Very nice work! $\endgroup$ – mathmandan Jan 16 '15 at 16:51
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This answer focuses on part (C) of the question. See also the excellent answer by @Shyos.

Let's write E for Ebrecedebre, I for Ibricidibri, and similarly for O and U. Then we'll write e, i, o, u for the "reversed" operations. Following @Shyos, we'll also write B for Ball, C for Cone, R for Ring and J for Umbrella, so that we don't have U meaning two different things. (J is an excellent letter for an Umbrella!)

Each time you do E, it decreases the number of B's by 1, while increasing the number of C's by 1 and the number of R's by 1. On the other hand, if you do an "e", it increases the number of B's by 1 while decreasing C's and R's by 1. Notice that E and e just cancel each other out.

Also, if at any point we did, say, 8 E's followed by 5 e's, the net result would be the same as doing (8-5) = 3 instances of E. So we can count the E's as positive, and then each e is just a negative E.

In part (C), we need a total net increase in the R's by at least 5, and we need a total net decrease in the C's by exactly 1, and no net change in the B's or J's.

Let's write $x,y, z, w$ to mean the net number of applications of E, I, O, U respectively. (Note: since e, i, o, u count as negative, it's possible that $x,y,z,w$ could be negative; however, $x,y,z,w$ must all be integers.) Now the total net change in B's is: $$ \Delta B = -x + 3z + w $$ (Why? Well, remember that each application of E is going to decrease the number of B's by 1. Meanwhile, each application of O will increase the number of B's by 3, and each application of U will increase the number of B's by 1. So 3 times the number of O's, plus 1 times the number of U's, minus 1 times the number of E's, gives the total net change in the number of B's.)

We want $\Delta B = 0$, and $\Delta C = -1$ (since we start with 1 cone and want to end with 0), and similarly $\Delta J =0$. Finally, $\Delta R$ can be any number $k$ as long as $k\geq 5$. We now have:

$$ \begin{eqnarray*} 0 & = & -x + 0y + 3z + w\\ -1 & = & x - y + 0z + 4w\\ k & = & x + 2y - z + 0w\\ 0 & = & 0x + y + z - w\\ \end{eqnarray*} $$

Or in matrix form: $$ \begin{pmatrix} 0\\ -1\\ k \\ 0 \\ \end{pmatrix} = \begin{pmatrix} -1 & 0 & 3 & 1\\ 1 & -1 & 0 & 4\\ 1 & 2 & -1 & 0\\ 0 & 1 & 1 & -1\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ z \\ w \\ \end{pmatrix} $$ We can solve this equation for $x,y,z,w$ in a variety of ways; for example, multiplying both sides by the inverse of the $4\times 4$ matrix. No matter how you do it, the result is: $$ \begin{pmatrix} x\\ y\\ z \\ w \\ \end{pmatrix} = \begin{pmatrix} -{3}/{4} - 2k/3\\ {1}/{4} + {2k}/{3}\\ -1/4 - k/3 \\ k/3 \\ \end{pmatrix} $$ Since $x,y,z,w$ have to be integers, we see that $k$ must be divisible by 3 (from the last entry). Also, from the first entry, the following must be an integer: $$ -\frac{3}{4} - \frac{2k}{3} = \frac{-9 - 8k}{12}, $$ so $(8k+9)$ must be divisible by $12$. This is impossible if $k$ is divisible by $3$, so there is no solution to part (C).

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Abbreviating B = Ball, C = Cone, R = Ring, U = Umbrella: C 2R + U R + 2U + 3B R + 5B + 8C 2B + 4R + 11C 4R + 2U + 3C 5C

So that's B: At least five cones and nothing else. If more cones are wanted, you can transform one cone into 5 more, so (at the very least) you can have 1 + 4n cones for positive integer n.

That feels very manual; maybe it will help someone else see a pattern for the other three. I will add more as I get time.

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