Find the smallest number which leaves 22,35, 48 and 61 as remainders when divided by 26, 39, 52 and 65 respectively.

• (26-22,39-35,52-48,65-61) = (4,4,4,4)
  LCM of (26,39,52,65) =780
  Req. number= 780-4 =776
• 9 years agoHelpfull: Yes(9) No(1)
• since the common difference between remainder and divisor is constant i.e 26-22=4,39-35=4,52-48=4,65-61=4
  so the no will be =L.C.M of(26,39,52,65)-4
  =780-4
  =776
  
• 9 years agoHelpfull: Yes(2) No(0)
• 776 is ans
• 9 years agoHelpfull: Yes(0) No(0)
• the answer is 776
  26-22=4
  39-35=4
  52-48=4
  65-61=4
  LCM(26,39,52,65)=780
  780-4=776
  
• 9 years agoHelpfull: Yes(0) No(0)
• Lets take LCM of all four divisors= 780 is the LCM common multiple of all numbers. so, when these numbers divide 780 by themselves, the remainder is the divisor itself., means,
  780%26=26
  same goes for other numbers, so we need a number which leaves the given remainders in the questions, so we subtract 26-22=4, from 780, gives 776.
• 7 years agoHelpfull: Yes(0) No(0)