Ponder This Challenge - July 1999 - Labeling bowling pins

Ponder This Challenge:

While bowling, you notice that the ten pins (arranged in a triangle) are labeled with the different integers 0,1,...,9, and that if two pins are side-by-side, the sum of their labels (reduced modulo 10) is equal to the label of the pin in front and between them.

For example:             1    2    4    5                3    6    9                  9    5                     4

Add the labels of the two pins in the second row to obtain the label of the one pin in the first row: (9+5=14 = 4 mod 10). But in the example some of the pins have the same label; in the solution, they don't. While you are busy noticing this, your opponent makes a strike. How are the pins labeled? |

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Solution

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  If the four back pins are labeled A,B,C,D, then we have:

  A                 B                 C                  D  A+B             B+C            C+D   A+2B+C        B+2C+D     A+3B+3C+D (reduced mod 10). Since the labels are (in some order) the integers 0 through 9, their sum is 45 (mod 10). So we get an equation: 4A + 9B + 9C + 4D = 5(mod 10) The pin labeled 0 must be in front, since otherwise it would cause equal labels to be assigned to the pin to its side and the one in front of and between them. So we have a second equation: A + 3B + 3C + D = 0(mod 10) Combining, we find D = 5-A (mod 10) C = 5-B (mod 10). So our setup is

  A              B             5-B               5-A A+B             5           -A-B   A+B+5       5-A-B       0

  with everything being considered mod 10. The opponent just bowled another strike. We'd better hurry up. Now A is neither 0 nor 5, so -A is not equal to A. Where can the pin labeled -A go? Not in the first row (the lone pin there is already labeled 0). Not in the second row, since the two pins there add to 0 (or 10), so if one was -A the other would be A. Not in the third row, because having (-A) next to (5) would cause the pin in front of them to be labeled (5-A), duplicating one in the back row. So the (-A) goes in the back row: either B=-A or 5-B=-A. Now, if B=-A then the left-hand pin of the third row is 0, duplicating the head pin. The only possibility is that 5-B=-A, which means B=5+A, so that the setup is:

  A        5+A       -A      -5-A 5+2A      5      5-2A   2A      -2A     0

  Furthermore this will work with any A except 5 or 0 (which would duplicate an existing label). For example with A=4 we get

  4        9       6      1 3      5      7   8      2     0

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