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Answers ( 5 )
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    We have:

    \(\overline{a_na_{n-1}...a_3a_2a_1}=\overline{a_na_{n-1}..a_4}.1000+\overline{a_3a_2a_1}\)

    Since \(1000⋮4\) => \(\overline{a_na_{n-1}..a_4}.1000\) \(⋮\) \(4\).

    \(\overline{a_na_{n-1}...a_3a_2a_1}\)  \(⋮\) \(4\)  iff   \(\overline{a_3a_2a_1}\) \(⋮\)  \(4\)

    In other word: a number is divisibility by 4 if only if the number formed by its three right digits is divisibility by 4.

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    Factors of N: {pi qj | i = 0, .., m ; j = 0, .., n}

    => Total: (m + 1)(n + 1) factors.

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    Factors of N:  {1, p, p2, ...., pk-1, pk }

    => There are k + 1 factors.

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    five 5-student committees and one 7-student committees.

    Total: 6 committees.

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    Each year has 365 or 366 days, so the least number of students should be 367 to be sure that there are at leat two students with the same birthday. 

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Questions ( 1 )
  • Of the three figures shown below, which ones can be drawn without going over any lines twice and without crossing any lines? 

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