Let A be a natural number consisting only 1. B is another natural number which is equal to quotient when A is divided by 13. C is yet another natural number equal to the quotient when B is divided by 7. Find B-C.

• a/c to question A contain only 1
  let A=111111 when which is divide by 13 we get quotient 8547 and remainder=0 {111111 is 1st no. which is divisible by 13}.
  this quotient is equal to i.e B=8547
  and c equal to quotient of B when it is divided by 7
  then we get C=1221
  now B-C=8547-1221=7326. ans
• 8 years agoHelpfull: Yes(24) No(0)
• In above problem reminder should be zero. Otherwise each one will get different answer. Also A should be least possible number (that is not mentioned in question, this should be understood).
  
  111111/13=8547=B
  8547/7=1221=C
  
  B-C=7326
• 8 years agoHelpfull: Yes(6) No(0)
• a=111,b=8, c=1
  ans 7
• 8 years agoHelpfull: Yes(2) No(6)
• To proof: x^2=a*8(multiple of 8)
  
  checking i). x^3=b*16
  take cube root and then square both sides
  x^2=2^(8/3) * b^(2/3)
  which is not a multiple of 8 certainly.
  
  checking ii). x=c*18=2*3^2*c
  square both sides
  x^2=4*3^4*c^2
  which is not always a multiple of 8 .for it c should be even.
  taking both
  x^3/x= x^2=8*(b/9c)=8*k
  
  BOTH statements (1) and (2)
  TOGETHER are sufficient to
  answer the question asked,
  but NEITHER statement
  ALONE is sufficient to answer
  the question asked
• 8 years agoHelpfull: Yes(0) No(7)
• B-C would be = 6
  here a= 11, b=1,c=7
• 8 years agoHelpfull: Yes(0) No(4)