[sp]$2019$ has digital sum $12$. Twice $2019$ is $4038$, which has digital sum $15$. Also, $$2019 = 15 + 2004 = 15 + 12\cdot167.$$ So the number $$4038\;\overbrace{2019\;2019\;\ldots\;2019}^{167\text{ blocks}},$$ whose decimal expansion consists of $4038$ followed by $167$ blocks of $2019$, has decimal sum $2019$. It is clearly a multiple of $2019$, the quotient being $$2\;\overbrace{0001\;0001\;\ldots\;0001}^{167\text{ blocks}}.$$lfdahl said:Prove the existence of a positive integer divisible by $2019$ the sum of whose decimal digits is $2019$.Source: Nordic Math. Contest
The smallest positive integer that meets these criteria is 2019.
No, there are infinitely many integers that meet these conditions. Some examples include 4038, 6057, and 8076.
This is because the divisibility rule for 2019 states that the sum of the first three digits must be divisible by 2019. Since 2019 is a four-digit number, the sum of its digits must also be divisible by 2019.
Yes, there is a pattern. The integers that meet these criteria can be written as 2019n, where n is any positive integer. For example, 2019, 4038, 6057, etc. are all divisible by 2019 and have a sum of decimal digits equal to 2019.
Yes, it is possible. For example, the integer 20190 is divisible by 2019 and has a sum of decimal digits equal to 12. However, this is not as common as having a sum of decimal digits equal to the number itself (2019).