Passengers in the Ships!

Logic Level 4

There are twelve ships situated on a 10 × 10 10 \times 10 grid. The ships are denoted by the letters A A through L L , and each ship consists of three cells of the grid in either a horizontal or a vertical line, as shown in the diagram. Each ship contains a certain number of passengers. There are also some numbers in the last row and the last column of the diagram. These numbers represent the total number of passengers on all the ships intersected by that row or column.

For example: The two ships B B and L L intersect the right-most column, so together they contain 9 passengers. The two ships G G and L L intersect the bottom-most row, so together they contain 6 6 passengers.

Given that there are no passengers on two of the ships and the remaining ten ships contain 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 1,2,3,4,5,6,7,8,9 and 10 10 passengers, let the number of passengers each of the twelve ships A A to L L contain be represented by their respective denoted letters. Then find the value of:

A B + C + D E + F G + H I + J − K + L L = ? \large{AB + C + \dfrac{D}{E} + F^G + \dfrac{H}{I} + J - K + L^L = \ ?}


The answer is 127.

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Satyajit Mohanty by Satyajit Mohanty , 24, India

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1 solution

Rayner Chuang
Sep 13, 2015

Firstly, we write out all known equations:

Let letters {A,B,C,D,E,F,G,H,I,J,K,L} denote the number of passengers in their corresponding ships, and xyz to be the total number of passengers in ships x, y and z.

  • BDE=17
  • EIJ=15
  • FIK=19
  • GL=6
  • AF=8
  • ADFG=21
  • CIH=6
  • BL=9

We can then also derive DG=13 (since AF=8, ADFG=21)

From DG=13 and GL=6, we can determine that G can only be 4, 5 or 6.

Case 1: G=4

  • DG=13 => D=9
  • GL=6 => L=2
  • BL=9 => B=7
  • BDE=17 => E=1

A or F cannot be 10, since AF=8. C or I or H cannot be 10, since CIH=6. If J=10, I=4 (from EIJ=15). However, this means that CH=2, which does not have a solution since they cannot both have 1 passenger each, or a ship with 2 passengers as L already has 2 passengers. By contradiction, K=10 .

FI=9 => F or I is 3 or 6. If I=3, then CH=3 which has no solution. Hence F=3 , I=6 . Then, C and H both have zero passengers. AF=8 => A=5 , EIJ=15 => J=8 .

Therefore, A=5, B=7, C=0, D=9, E=1, F=3, G=4, H=0, I=6, J=8, K=10, L=2

5 × 7 + 0 + 9 1 + 3 4 + 0 6 + 8 − 10 + 2 2 = 127 5\times 7+0+\frac { 9 }{ 1 } +{ 3 }^{ 4 }+\frac { 0 }{ 6 } +8-10+{ 2 }^{ 2 }=\boxed{127}

Case 2: G=5

  • GL=6 => L=1
  • BL=9 => B=8 (contradiction)

Hence, G ≠ 5 G\neq 5

Case 3: G=6

  • DG=13 => D=7
  • GL=6 => L=0
  • BL=9 => B=9
  • BDE=17 => E=1
  • EIJ=15 => IJ=14 (no solution)

Hence, G ≠ 6 G\neq 6

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