answer:1. **Understanding the Initial Observation**: - Each of the three wise men sees that the other two have soot on their faces. - Initially, everyone starts laughing because they can see soot on their companions' faces. 2. **Formulating the Key Insight**: - Let us denote the wise men as ( A ), ( B ), and ( C ). - Suppose that wise man ( A ) is the one who figures out that he also has soot on his face. 3. **Wise Man ( A ) Reasoning**: - Wise man ( A ) thinks: "If my face were not dirty, then wise men ( B ) and ( C ) are laughing because they see soot on the face of the other one only." - Now, if wise man ( A )'s face were clean, here is what the situation for wise man ( B ) would be: - Wise man ( B ) sees soot on ( C )'s face. - ( B ) would reason: "If my face were clean, then ( C ) would be the only one with a dirty face and should stop laughing as he realizes why we laugh." - However, ( C ) is laughing, which means ( C ) sees soot on another face aside from ( B ). 4. **Concluding the Clean Face Hypothesis is False**: - Since ( B ) continues to laugh, it means ( B ) saw soot on another face apart from ( C ), indicating there’s soot on ( A )’s face because ( A )’s clean face hypothesis does not hold true. 5. **Reaching the Final Conclusion**: - Therefore, from ( A )’s perspective: "Since both ( B ) and ( C ) continue to laugh, they must see soot on my face too." - This reasoning leads wise man ( A ) to logically deduce that his face must also be covered in soot. 6. **Final Thought Process**: - All the wise men continue laughing, leading ( A ) to the realization of his own dirty face through this chain of logical deduction based on the persistence of laughter indicating unclean faces amongst them. [ boxed{text{Wise man } A text{ deduced that his own face was dirty.}} ]

answer:1. **Calculate AC in triangle ABC**: Since triangle ABC is isosceles and right-angled at B: [ AC = sqrt{AB^2 + BC^2} = sqrt{2^2 + 2^2} = sqrt{8} = 2sqrt{2} ] 2. **Determine the Perimeter of triangle ABC**: [ AB + BC + AC = 2 + 2 + 2sqrt{2} = 4 + 2sqrt{2} ] 3. **Set Equal Perimeters for triangle ACD**: Since AC = 2sqrt{2}, perimeters equate thus: [ AC + CD + DA = 4 + 2sqrt{2} ] [ 2sqrt{2} + CD + DA = 4 + 2sqrt{2} implies CD + DA = 4 ] 4. **Using the Angle Given, Calculate CD**: Considering angle CAD = 30^circ, and if we assume DA = d, [ sin(30^circ) = frac{CD}{d} = frac{1}{2} implies CD = frac{d}{2} ] and from the perimeter calculation, [ frac{d}{2} + d = 4 implies frac{3d}{2} = 4 implies d = frac{8}{3}, , CD = frac{4}{3} ] 5. **Angle Sum in triangle ACD **: [ angle BAC = 45^circ, , angle CAD = 30^circ, , angle BAD = angle BAC + angle CAD = 45^circ + 30^circ = 75^circ ] 6. **Finding sin(2angle BAD)**: [ sin(2 times 75^circ) = sin(150^circ) = frac{1}{2} ] Conclusion: [ frac{1{2} } ] The correct answer is boxed{textbf{(E)} frac{1}{2}}.

answer:Let's call the amount of money Oliver spent by March "S". In January, Oliver had 33. By March, he had spent some money, so he had 33 - S. Then, he got another 32 from his mom, which means he had 33 - S + 32. According to the information given, after receiving the 32 from his mom, Oliver had 61. So we can set up the following equation: 33 - S + 32 = 61 Now, let's solve for S: 33 + 32 - S = 61 65 - S = 61 Subtract 65 from both sides to isolate S: -S = 61 - 65 -S = -4 Since we're looking for the amount spent, which is a positive number, we take the absolute value: S = 4 So, Oliver spent boxed{4} by March.

answer:1. **Identify the set and the condition:** We need to choose distinct numbers (a) and (b) from the set ({1, 2, 3, ldots, 2005}) such that (a + b) is a multiple of 5. 2. **Consider the numbers modulo 5:** The numbers in the set ({1, 2, 3, ldots, 2005}) can be classified based on their residues modulo 5. The possible residues are (0, 1, 2, 3,) and (4). 3. **Count the numbers with each residue:** Since (2005 div 5 = 401) with no remainder, there are exactly 401 numbers for each residue class modulo 5. 4. **Pairing residues to satisfy the condition (a + b equiv 0 pmod{5}):** - If (a equiv 0 pmod{5}), then (b equiv 0 pmod{5}). - If (a equiv 1 pmod{5}), then (b equiv 4 pmod{5}). - If (a equiv 2 pmod{5}), then (b equiv 3 pmod{5}). - If (a equiv 3 pmod{5}), then (b equiv 2 pmod{5}). - If (a equiv 4 pmod{5}), then (b equiv 1 pmod{5}). 5. **Calculate the number of ways to choose (a) and (b) for each case:** - For (a equiv 0 pmod{5}) and (b equiv 0 pmod{5}): There are 401 numbers that are multiples of 5. We need to choose 2 out of these 401 numbers: [ binom{401}{2} = frac{401 times 400}{2} = 80200 ] - For (a equiv 1 pmod{5}) and (b equiv 4 pmod{5}): There are 401 numbers that are (1 pmod{5}) and 401 numbers that are (4 pmod{5}). We can pair any number from the first group with any number from the second group: [ 401 times 401 = 160801 ] - For (a equiv 2 pmod{5}) and (b equiv 3 pmod{5}): Similarly, there are 401 numbers that are (2 pmod{5}) and 401 numbers that are (3 pmod{5}): [ 401 times 401 = 160801 ] 6. **Sum the number of ways for all cases:** [ binom{401}{2} + 401^2 + 401^2 = 80200 + 160801 + 160801 = 401802 ] Conclusion: [ boxed{401802} ]