Find the greatest no. of 4 digit which when divided by 10,11,15 and 22 leaves 3,4,8 and 15 as remainder respt.
a) 9907
b)9903
c)9893
d)None
correct answer: c)

• Largest 4 digit no. is 9999. We see that,each (divisor - remainder) = 7. (10-3)=7,(11-4)=7,(15-8)=7,(22-15)=7. Now take lcm of 10,11,15,22..which is 330. So, lcm + common difference i.e., 330+7= 337. now by dividing 9999 by 337 we get remainder as 106, on subtracting 106 from 9999 we get 9893 which is the required largest four digit no.. Hence, option C is correct.
  
• 9 years agoHelpfull: Yes(1) No(0)
• LCM(10,11,15,22)=330
  GREATEST 4 DIGIT NO=9999
  DIVISIBLE BY 330 WILL BE 9900
  HERE 10-3=7,11-4=7,15-8=7,22-15=7
  THUS 9900-7=9893
• 9 years agoHelpfull: Yes(1) No(0)
• take lcm of 10,11,15,and 22
  which comes out to be 330
  and 4 digit maximun number using this lcm is 9900 which is divided by all these i.e 10,11,15,and 22
  
  
  n now we have to just check out the desired option
  n only option c gives us the right answer.
  
• 9 years agoHelpfull: Yes(0) No(1)
• >At first ... take a LCM of (10,11,15,22) = 330
  >Now as we know largest 4 digit number is (9999)
  >so divide (9999/330) = 30 #reminder we got is 99
  > now minus 99 from 9999= 9900 ( here we got highest 4 digit multiple of 330)
  
  #2 step > take a difference of divisor - reminder as follow :-
  10-3= 7
  11-4= 7
  15-8= 7
  22-15= 7
  as we see above 7 is common difference
  
  #3 step > now minus 7 from 9900 = {9893 } is your answer.
  
  # ( C) 9893
• 6 years agoHelpfull: Yes(0) No(0)