Find the smallest positive integer which can be written both as (i) a sum of $2002$ positive integers (not necessarily distinct), each of which has the same sum of digits and (ii) as a sum of $2003$ positive integers (not necessarily distinct), each of which has the same sum of digits.
The book's solution is to 'observe' the answer is $10010$:$$ 10010 = 2002·5 = 1781·4+222·13$$ and prove it is the minimal integer.
The answer is $10010$. First observe that this is indeed a solution: $$10010 = 2002·5 = 1781·4+222·13$$ so we may express $10010$ as the sum of $2002$ fives or of $1781$ fours and $222$ thirteens, where $1781+222 = 2003$. To prove minimality, observe that a number is congruent modulo $9$ to the sum of its digits, so two positive integers with the same digit sum are in the same residue class modulo $9$. Let $k_1$ be the digit sum of the $2002$ numbers and $k_2$ the digit sum of the $2003$ numbers. Then $$4k_1 ≡ 2002k_1 ≡ 2003k_2 ≡ 5k_2 \pmod 9$$ If $k_1 ≥ 5$, the sum of the $2002$ numbers is at least $10010$; if $k_2 ≥ 5$, the sum of the $2003$ numbers is greater than $10010$. However, the solutions $$k_1 ≡ 1, 2, 3, 4 \pmod 9$$ give $k_2 ≡ 8, 7, 6, 5$, respectively, so that at least one of $k_1$ or $k_2$ is greater than or equal to $5$, and the minimal integer is $10010$.
My question: Is there any other way to solve this? If it isn't, what is the quickest way to find that $10010$ is the minimal integer which can be written both as (i) a sum of $2002$ positive integers (not necessarily distinct), each of which has the same sum of digits and (ii) as a sum of $2003$ positive integers (not necessarily distinct).
Any help?Thanks in advance:)