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\\\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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Hari Shankar ·2010-03-06 02:57:47

Are they:

1) 5

2) 6

3) 5

4) 8

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Che ·2010-03-06 03:17:28

1 st is 3

rest r rit [1]

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Che ·2010-03-07 02:23:01

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341

Hari Shankar ·2010-03-07 02:40:49

Sorry, 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.

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Che ·2010-03-07 23:35:18

no first is m₁+m₂ only.

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