Q. In mathematica country 1,2,3,4....,8,9 are nine cities. Cities which form a no. that is divisible by 3 are connected by air planes. (e.g. cities 1 & 2 form no. 12 which divisible by 3 then 1 is connected to city 2). Find the total no. of ways you can go to 8 if you are allowed to break the journies

• Acc to me its 7
  Because cities divisible by 3 are- 12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90.
  now numbers that include 8 are-18,48,78,81,84,87
  because 18 and 81 are same connection so finally we have-18,48,78
  now as we have to do break journey so we can reach to 8 by cities related to 1,4,7 thus from 24,54,27,57
  thus there are 7 ways(18-1to8,48,78,24-2to4/4to8,54-5to4/4to8,27-2to7/7to8,57-5to7/7to8)
• 10 years agoHelpfull: Yes(22) No(6)
• total routes with different origins are 41
  1->2->4->5->7->8
  1->2->4->8
  1->5->4->2->7->8
  1->5->4->8
  1->5->7->2->4->8
  1->5->7->8
  1->8
  
  2->1->5->4->8
  2->1->5->7->8
  2->1->8
  2->4->5->1->8
  2->4->5->7->8
  2->4->8
  2->7->5->1->8
  2->7->5->4->8
  2->7->8
  
  4->2->1->5->7->8
  4->2->1->8
  4->2->7->5->1->8
  4->2->7->8
  4->5->1->2->7->8
  4->5->1->8
  4->5->7->2->1->8
  4->8
  
  5->1->2->4->8
  5->1->8
  5->1->2->7->8
  5->4->2->1->8
  5->4->2->7->8
  5->4->8
  5->7->2->1->8
  5->7->8
  
  7->2->1->5->4->8
  7->2->1->8
  7->2->4->5->1->8
  7->2->4->8
  7->5->1->2->4->8
  7->5->1->8
  7->5->4->2->1->8
  7->5->4->8
  7->8
  
• 10 years agoHelpfull: Yes(9) No(3)
• total 9 ways
• 10 years agoHelpfull: Yes(7) No(2)
• it should be 9
  
  guys 12 and 21 are same similarly 51 and 15 and so on
  
  so the nums which is divisible by 3 are
  12
  15
  18
  24
  27
  36
  39
  45
  48
  57
  69
  78
  
  
  and the air planes which connect country 8 are
  
  
  18 48 78
  12 45 57
  15
  24
  27
  
  
• 10 years agoHelpfull: Yes(6) No(0)
• Correct answer is 5
• 10 years agoHelpfull: Yes(5) No(0)
• cause if u stuck at 3 or 6 u can't go to 8 anyway
  
• 10 years agoHelpfull: Yes(3) No(1)
• the possible routes are-12,15,18,21,24,27,36,39,42,45,48,51,54,57,63,69,72,75,78,81,84,87,93,96
  nowroutes from station 1 to 8 cud be
  18,
  12-24-48,
  12-27-78,
  15-54-48,
  15-57-78,
  routes from station 2 to 8 is
  21-18
  24-48
  27-78
  21-15-57-78
  21-15-54-48
  route 3 to 8...
  no route
  route 4 to 8 ..
  48
  42-21-18
  42-27-78
  45-57-78
  ...............
  route from 6 to 8..
  no route
  route from 7 to 8
  78
  75-54-48
  75-51-18
  72-21-18
  72-24-48
  ...so total 20 routes are possible
• 10 years agoHelpfull: Yes(2) No(0)
• Its 15.
  Solution
  cities divisible by 3 are- 12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90.
  1 -> 8
  4 -> 8
  7 -> 8
  2 -> 4 -> 8
  5 -> 4 -> 8
  2 -> 7 -> 8
  5 -> 7 -> 8
  1 -> 2 -> 4 -> 8
  7 -> 2 -> 4 -> 8
  1 -> 5 -> 4 -> 8
  7 -> 5 -> 4 -> 8
  1 -> 2 -> 7 -> 8
  4 -> 2 -> 7 -> 8
  1 -> 5 -> 7 -> 8
  4 -> 5 -> 7 -> 8
  
• 10 years agoHelpfull: Yes(1) No(4)
• ans will be five.
• 8 years agoHelpfull: Yes(1) No(1)
• how 9 can u explain plz
• 10 years agoHelpfull: Yes(0) No(0)
• i cant understand..plx explain clearly
• 10 years agoHelpfull: Yes(0) No(1)
• why not including station 3 & 6??
• 10 years agoHelpfull: Yes(0) No(0)
• 5 ways to reach
  
• 8 years agoHelpfull: Yes(0) No(1)
• 5 IS THE CORRECT ANSWER
• 6 years agoHelpfull: Yes(0) No(1)