When two different numbers are divided by a certain divisor the remainders are 147 and 192. When the sum of the two
numbers is divided by the same divisor the ‘remainder is 38: Find the divisor.

• Let divisior be x.so 1st no.=a+147
  2nd no=b+192.
  After adding:(a+147+b+192)/x=>a+b+339/x
  Remainder of above expression is 38(given).so rem[a+b+(301+38)]/x=>38 therefore x would be 301.
  
• 9 years agoHelpfull: Yes(10) No(0)
• m*x = y-147
  and n*x = z-192
  
  (m+n+1)*x = y+z+38
  => mx + nx+ x = y+z-38
  => y-147+z-192+x = y+z-38
  => x-339 = -38
  => x = 301 ANS
• 9 years agoHelpfull: Yes(5) No(7)
• why did u write (m+n+1)*x please explain
  
• 9 years agoHelpfull: Yes(3) No(0)
• use remainder rule to solve this problem
  (147+192)/d=38
  therefore d should be either 301 or any of its factors like 43,7,147 but they are not possible
  therfore ans is 301
• 9 years agoHelpfull: Yes(3) No(0)
• Look remainder for firsthandno=147
  Remainder for second no=192
  When we divided addition of first & second no then add the remainder of first & second no
  147+192=339
  Substract the remainder 38 and find the divisor
  339-38=301
  
• 9 years agoHelpfull: Yes(2) No(0)
• why m+n+1???
• 9 years agoHelpfull: Yes(0) No(0)