Let $A=1 \cdot 4 \cdot 7 \cdot \ldots \cdot 2014$ be the product of the numbers less or equal to 2014 that give remainder 1 when divided by 3. Find the last non-zero digit of $A$.
Grouping the elements of the product by ten we get:

$$
\begin{aligned}
& (30 k+1)(30 k+4)(30 k+7)(30 k+10)(30 k+13)(30 k+16) \\
& (30 k+19)(30 k+22)(30 k+25)(30 k+28)= \\
& =(30 k+1)(15 k+2)(30 k+7)(120 k+40)(30 k+13)(15 k+8) \\
& (30 k+19)(15 k+11)(120 k+100)(15 k+14)
\end{aligned}
$$

(We divide all even numbers not divisible by five, by two and multiply all numbers divisible by five with four.)

We denote \( P_{k}=(30 k+1)(15 k+2)(30 k+7)(30 k+13)(15 k+8)(30 k+19)(15 k+11)(15 k+14) \). For all the numbers not divisible by five, only the last digit affects the solution, since the power of two in the numbers divisible by five is greater than the power of five. Considering this, for even \( k \), \( P_{k} \) ends with the same digit as \( 1 \cdot 2 \cdot 7 \cdot 3 \cdot 8 \cdot 9 \cdot 1 \cdot 4 \), i.e., six, and for odd \( k \), \( P_{k} \) ends with the same digit as \( 1 \cdot 7 \cdot 7 \cdot 3 \cdot 3 \cdot 9 \cdot 6 \cdot 9 \), i.e., six. Thus \( P_{0} P_{1} \ldots P_{66} \) ends with six. If we remove one zero from the end of all numbers divisible by five, we get that the last nonzero digit of the given product is the same as the one from \( 6 \cdot 2011 \cdot 2014 \cdot 4 \cdot 10 \cdot 16 \cdot \ldots \cdot 796 \cdot 802 \). Considering that \( 4 \cdot 6 \cdot 2 \cdot 8 \) ends with four and removing one zero from every fifth number we get that the last nonzero digit is the same as in \( 4 \cdot 4^{26} \cdot 784 \cdot 796 \cdot 802 \cdot 1 \cdot 4 \cdot \ldots \cdot 76 \cdot 79 \). Repeating the process we did for the starting sequence we conclude that the last nonzero number will be the same as in \( 2 \cdot 6 \cdot 6 \cdot 40 \cdot 100 \cdot 160 \cdot 220 \cdot 280 \cdot 61 \cdot 32 \cdot 67 \cdot 73 \cdot 38 \cdot 79 \), which is two.

Let \( ABC \) be a triangle with \( \measuredangle B = \measuredangle C = 40^{\circ} \). The bisector of the \( \measuredangle B \) meets \( AC \) at the point \( D \). Prove that \( \overline{BD} + \overline{DA} = \overline{BC} \).

Solution. Since \( \measuredangle BAC = 100^{\circ} \) and \( \measuredangle BDC = 120^{\circ} \) we have \( \overline{BD} < \overline{BC} \). Let \( E \) be the point on \( \overline{BC} \) such that \( \overline{BD} = \overline{BE} \). Then \( \measuredangle DEC = 100^{\circ} \) and \( \measuredangle EDC = 40^{\circ} \), hence \( \overline{DE} = \overline{EC} \), and \( \measuredangle BAC + \measuredangle DEB = 180^{\circ} \). So \( A, B, E \) and \( D \) are concyclic, implying \( \overline{AD} = \overline{DE} \) (since \( \measuredangle ABD = \measuredangle DBC = 20^{\circ} \) ), which completes the proof.