There are a number of chocolates in a bag. If they were to be equally divided among 14 children, there are 10 chocolates left. If they were to be equally divided among 15 children, there are 8 chocolates left. Obviously, this can be satisfied if any multiple of 210 chocolates are added to the bag. What is the remainder when the minimum feasible number of chocolates in the bag is divided by 9?

• Let there will be X chocolates is given to each children.
  from first condition total number of chocolates=14*X+10
  from second condition total number of chocolates=15*X+8
  14*X+10=15*X+8
  X=2
  so total number of chocolates=14*2+10=38
  when multiple of 210 chocolates added with total number of chocolates then number of chocolates=38+210=248
  when this is divided by 9 then remainder=5
• 10 years agoHelpfull: Yes(122) No(13)
• from the queston n=14x+10 and n= 15x+8. by solving these two equations we get x=2
  by substitung this x in any one of the equation we get n=38 nd acc to question n=38+210=249 and ans is 249%9=5
• 10 years agoHelpfull: Yes(77) No(25)
• let x be the total number of chocolates.
  According to the problem,
  x= 14*m+10 and x= 15*m+8. Solving we get m=2. So number of chocolates is 38. When divided by 9 ans is 2.
• 10 years agoHelpfull: Yes(35) No(36)
• let N = no of choclates
  .
  accrding to GIVEN cond.
  N=14x+10 and N= 15y+8.....(no need of x & y to be same)
  
  now equate both 14x+10 = 15y + 8
  ==> 15y - 14x = 2
  it can have many solutions like(for (y,x)) (2,2) , (16,17) , (30,32)...etc
  bt the nearest solution is (2,2)
  by this we can calculte no of choclates and the remainder thus solved by adding 120 (feasible) is 5
  
• 10 years agoHelpfull: Yes(32) No(2)
• There are minimum 38 chocolates in the bag..
  38/14= 10( remainder)
  38/15= 8 (reminder)
  any multiple this with 210,satisfies this..
  so,when the numbers divided by 9,
  the remaining will be,38/2=2(reminder)..
  so the answer is 2..
• 10 years agoHelpfull: Yes(6) No(9)
• from the queston n=14x+10 and n= 15x+8. by solving these two equations we get x=2
  by substitung this x in any one of the equation we get n=38 nd acc to question n=38+210=248 and ans is 248%9=5
• 10 years agoHelpfull: Yes(3) No(3)
• c=14x+10 and c= 15x+8. by solving these two equations we get x=2
  by substitung this x in any one of the equation we get n=38 nd acc to question n=38+210=249 and ans is 249/9=5
• 10 years agoHelpfull: Yes(2) No(3)
• 14x1+10=N
  15x2+8=N.
  simplest and least soln of N is 38 by hit and try method....so when divided among 9 children..there will be 2 choc left
• 10 years agoHelpfull: Yes(2) No(0)
• 14*m+10=15*m+8??????????
  here it iss written as equally divided... where from u get that, both the times, m no of chocolates are given..????
• 10 years agoHelpfull: Yes(0) No(0)
• So, total number of chocolates = 38 Ans.
• 10 years agoHelpfull: Yes(0) No(4)
• total chocolates in first case = 14*x+10
  in second case= 15*x +8
  equating both we get x=2
  so total chocolates =38 which when divided by four gives 2 remainder
• 9 years agoHelpfull: Yes(0) No(1)
• WHY 14x+10=15x+8
• 8 years agoHelpfull: Yes(0) No(0)
• 8*15=120+210=330%9=6 remainder is 6.
  
• 8 years agoHelpfull: Yes(0) No(0)
• Division of m+n+p objects into three groups is given by (m+n+p)!m!×n!×p!
  But 7 = 1 + 3 + 3 or 1 + 2 + 4 or 1 + 1 + 5
  So The number of ways are (7)!/1!×3!×3!×2! + (7)!/1!×2!×4! + (7)!/1!×1!×5!×2! = 70 + 105 + 21 = 196
• 8 years agoHelpfull: Yes(0) No(0)
• given
  n=14x+10
  n= 15x+8.
  by solving these two equations we get x=2
  and n=14*2+10=38
  acc to question n=38+210=248 and
  if we divide 248 by 9 then we get remainder 5 which is the answer
  
• 8 years agoHelpfull: Yes(0) No(1)
• 630 divided by 9 so ans is 630
• 8 years agoHelpfull: Yes(0) No(0)
• Ans 0
  38 total chocolate..
  Multiple of 210 = 3* 7* 5* 2
  So when 38 + 7(multiple of 210) / 9 answer is 0
• 8 years agoHelpfull: Yes(0) No(1)
• 14*X+10=15*X+8
  x=2.
  total Chocolate=38
  
  if 210 is added to total chocolate then the bag's chocolate becomes .....248.
  now 248/9 is remainder 5.
  so ans =5.
  
• 7 years agoHelpfull: Yes(0) No(0)
• The given condition of remainder of 10 and 8 on dividing the no. of chocolates by 14 and 15 resp.
  can only be satisfied when no. of chocolates is of the form
  N = 210*a + 38
  hence put a=0, we get minimum no. of chocolates = 38
  hence 38/9 gives a remainder of 2.
  therefore answer is 2.
• 7 years agoHelpfull: Yes(0) No(0)
• bviously, this can be satisfied if any multiple of 210 chocolates are added to the bag. can anyone tell me in the above statement "this can be satisfied "refers to what none of you use this constrain given in the problem
• 7 years agoHelpfull: Yes(0) No(0)
• Why the heck we added 210.
  
• 7 years agoHelpfull: Yes(0) No(0)
• Why 38 is not the feasible number of chocolates. It asked for minimum feasible number and 38 satisfies this.
• 6 years agoHelpfull: Yes(0) No(0)