1?
\\\texttt{ Given the three digit number } N= a_1a_2a_3.. m_1,m_2,m_3 \\\texttt{ are the smallest possible natural number such that N is divisible by 7 whenever } \\ m_1a_1+m_2a_2+m_3a_3 \texttt{ is divisible by 7 then} \\ \texttt{find}...\\ 1)m_1+m_2\\ 2)m_1m_2\\ 3)m_1m_2m_3-1\\ 4)m_{1}^{m_2}
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5 Answers
341Sorry, my posts are not going through for some reason.
Clarification: is the 1st question asking for m1+m3?
Otherwise m1+m2 = 3 and m1m2=6 is not possible.
My reckoning was:
N = 100a_1 + 10a_2+a_3 = 98a_1+7a_2 + (2a_1+3a_2+a_3)
But I really dont know what is meant by "smallest natural numbers" when we are talking of a triplet like this. In any case it is easy to prove that it does not hold for coefficients lesser than those given here.