Answer this question.
Q. How many 3-digit numbers will be there which are divisible by 19?
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  • Priya
  • 08 jul.
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  • Let's solve this problem without using any mathematical formula of Arithmetic Progression. This gives us the speedy answer which is so required in the competitive exams where there is a constraint of time.
    The greatest three digit number is 999.
    And 999/19 = 52.57.
    This implies upto 999, a total of 52 numbers are completely divisible.
    Subtracting from this, number of two- digit numbers also divisible by 19 taken into account in the previous calculation, i.e 19,38,57,76, and 95, we get :
    52-5 = 47 
    Thus there are 47 such 3 digit numbers in all.


  • hmm 47 numbers


  • There are 47 such numbers.


  • There should be 3 digit number,So First Digit=114 
    Last digit which is divisible by 19 is 19*52=988 
      
    Apply the formula of A.P....An=a+(n-1)d---------------Eq(1) 
      
    Where An=988, a=114, common difference d= 19 
      
    Applying above value in Eq(1) we get, 
      
    988=114+(n-1)19 
      
    874=19n-19 
      
    893=19n 
      
    n=893/19 
      
    n=47


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