(123456789101112131415161718..........4041424344).what is the remainder when the number in brackets is divided by 45

• let (123456789101112131415161718..........4041424344) =X
  and (123456789101112131415161718..........4041424200)=Y
  then
  X=Y+144
  
  Now Y is divisible by 5 .... as last digit is zero.
  Y is also divisible by 9
  as sum of digits of Y is divisible by 9.
  when 144 is divided by 45, we get 9 as remainder.
  
  so 9 is remainder when (123456789101112131415161718..........4041424344) is divided by 45.
• 12 years agoHelpfull: Yes(11) No(0)
• If the last digit is 0 or 5, the number is divisible by 5.
  If sum of digits is divisible by 9, the number is divisible by 9.
  So, 123456789101112131415161718..........404142434445 is perfectly divisible by 45.
  Now, we have number 123456789101112131415161718..........4041424344, it is divisible by 9 (as we have removed 45 whose sum of 2 digits is 9).
  Now, we can write another number 123456789101112131415161718..........4041424335 ( 1 to 43 and then 35). Its sum is same as previous number and it is also divisible by 5. So, this is divisible by 45.
  On subtracting second number from first, we get 9. So, the remainder is 9 when the original number is divided by 45.
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• can u send me the materials pls my id is Dharaniedu@gmail.com
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• hi i am aathi...
  I am seriously preparing for placement next month..
  Please send any materials regarding TCS plsssss..
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