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A positive integer divisible by 2019 the sum of whose decimal digits is 2019.

lfdahl

Well-known member
Nov 26, 2013
719
Prove the existence of a positive integer divisible by $2019$ the sum of whose decimal digits is $2019$.


Source: Nordic Math. Contest
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,684
Prove the existence of a positive integer divisible by $2019$ the sum of whose decimal digits is $2019$.


Source: Nordic Math. Contest
$2019$ 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

Well-known member
Nov 26, 2013
719
Thankyou, Opalg , for your participation and - as always - for a clever answer! (Yes)