Game of Spies

Algebra Level 4

Once upon a time, there were 3 generals ruling their respective kingdoms A, B, C. Each general had an equal number of soldiers under command, and all 3 of them sent some of their men out to become secret soldiers in the other two kingdoms working as their spies.

With their knowledge, general A retained his 17 soldiers in his own kingdom, general B 19 soldiers, and general C 20 soldiers. After the infiltration settled, there were 32 soldiers in kingdom A, 36 soldiers in kingdom B, and 31 soldiers in kingdom C. The numbers of the spies were all distinct in those kingdoms with the maximum of 10 spies per kingdom.

How many spies were there in those kingdoms? Enter U V W X Y Z \overline{UVWXYZ} as your answer, where

\hspace{5mm} U = number of spies from B in A
\hspace{5mm} V = number of spies from C in A
\hspace{5mm} W = number of spies from A in B
\hspace{5mm} X = number of spies from C in B
\hspace{5mm} Y = number of spies from A in C
\hspace{5mm} Z = number of spies from B in C.


The answer is 1059874.

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1 solution

  • Let Ab= spies from A to B and like-wise Ac,Ba,Bc,Ca,Cb.
  • Also it says all the kingdoms has same number soldiers, so (31+36+32)/3=33
  • Now we can form some equations, that are as follows..
  • 17+Ba+Ca=32, So Ba+Ca=15
  • 19+Ab+Cb=36, So Ab+Cb=17
  • 20+Ac+Bc=31, So Ac+Bc=11
  • Now as we have 33 soldiers in each kingdom, we can have 3 more equations as follows..
  • Kingdom A retained 17 soldiers so Ab+Ac=33-17=16
  • Kingdom B retained 19 soldiers so Ba+Bc=33-19=14
  • Kingdom C retained 20 soldiers so Ca+Cb=33-20=13
  • Now we know that the number of spies is unique, which implies all the variables in concern has unique values.
  • Now the logic is to find a number from 15,17,11,16,14,13, which when divided in to 2 groups have least possibilities, considering that maximum number of spies deployed from one kingdom to another is 10.So in our case we have 17 as such number 9+8=10+7=17. also we can consider 16 as number of spies is unique so no 8+8=16.Either way is fine!
  • When the possibilities for Ab and Cb, that is,10,7,9,8 are put up we find the solution with Ab=9, Cb=8.
  • Now we can easily find the rest of the values Ab=9,Ac=7,Ba=10,Bc=4,Ca=5,Cb=8.
  • So the solution to the problem on hand is 1059874.

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