Which is the smallest whole number that ends with 19, is divisible by 19 and the sum of the digits is 19?

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The number John find is 17119

- Ends in 19
- Divisible by 19 - 19*901
- Sum of its digits - 1+7+1+1+9 = 19

Let the number to be found be N. The last 2 digits of the number are 19, this is possible only when 19 is multiplied by a number ending with 1.

The value of N is obtained when 19 is multiplied by a number large than 91 as no 2 digit number ending with 1 satisfies the condition that the sum of the digits is 19.

As the sum of the digits of N is also equal to 19 and the sum of the last two digits is 10, the sum of the rest of the digits has to be equal to 9. This is possible when 19 is multiplied with 901, which is determined by trial and error.

**The smallest whole number that satisfies the given criteria is 17119.**

The number is of the type 100*x +19 where x is an integer.

It is divisible by 19 . Or the number must satisfy 19*y where y is an integer.100*x +19 = 19*y100*x = 19*(y-1) x : y : Reqd. Number : Sum of digits

19 : 101 : 1919 : 20

38 : 201 : 3819 : 21

57 : 301 : 5719 : 22

76 : 401 : 7619 : 23171 : 901 : 17119 : 19 :

Therefore answer is 17119.

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