Assume n is even:
if (a0+a2+...+an)-(a1+a3+...+an-1)=13N
then (a0+a2+...+an)-(a1+a3+...+an-1)+1001a1+999999a2+...+(103n-1)an=13N+13M
a0+101a1+102a2+...+10nan=13(M+N)
n is odd will have the similar result.
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Let a0, a1, a2,...,an are the 3-digit number:
The long number will be a0+103a1+106a2+...+103nan
106K-3+1=13A
106L-1=13B
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