answer:1. **Set up the equations:** Let t be the length of each side of the equilateral triangle, and s the length of each side of the square. [ 3t - 4s = 108 ] [ t - s = d + 2 ] 2. **Express t in terms of s and substitute:** [ t = s + d + 2 ] Substitute into the perimeter equation: [ 3(s + d + 2) - 4s = 108 ] Simplify: [ 3s + 3d + 6 - 4s = 108 ] [ -s + 3d = 102 ] [ s = 3d - 102 ] 3. **Determine the condition for s:** Since the perimeter of the square must be greater than 0 and s > 0: [ 3d - 102 > 0 ] [ 3d > 102 ] [ d > frac{102}{3} = 34 ] 4. **Conclusion on the values of d:** Since d must be greater than 34, the first 34 positive integers (1 through 34) are not possible values for d. Therefore, there are 34 positive integers that are not possible values for d. The number of positive integers that are NOT possible values for d is 34. The final answer is boxed{C text{ (34)}}

answer:1. **Identify the groups**: There are two distinct groups: - Group 1: 25 people who all know each other. - Group 2: 15 people who know no one. 2. **Understanding interactions**: - Members of Group 1 only hug each other, so no handshakes occur within this group. - Members of Group 2 do not know anyone, so they only shake hands. 3. **Counting handshakes involving Group 2**: - Each person in Group 2 will shake hands with every other person at the event (both from Group 1 and Group 2), except themselves. - For each person in Group 2, the number of handshakes is 40 - 1 = 39 (total people at the event minus themselves). 4. **Calculating total handshakes**: - Since there are 15 people in Group 2, and each initiates 39 handshakes, the total number of initiated handshakes is 15 times 39 = 585. - However, this count considers each handshake twice (once for each participant in the handshake). Therefore, we need to divide by 2 to avoid double-counting: [ text{Total handshakes} = frac{15 times 39}{2} = 292.5 ] 5. **Conclusion**: - The correct number of handshakes, considering the avoidance of double-counting, is 292.5. However, since the number of handshakes cannot be a fraction, we need to reassess the problem or solution approach. There might be an oversight in conceptualizing how the interactions are counted, or the problem set-up might be inconsistent with realistic social interactions. The final answer is boxed{292}

answer:1. First, we need to convert the given sequence ( text{"оуавоуав"} ) from base-4 to base-10. 2. Assign the digits according to the given sounds: - ("о") corresponds to (0) - ("у") corresponds to (1) - ("в") corresponds to (2) - ("а") corresponds to (3) So, the sequence "оуавоуав" translates to the base-4 number (01320132). 3. Next, we write down the base-4 representation: [ (01320132)_4 ] 4. To convert this to base-10, we use the formula for base conversion: [ n = a_k cdot b^k + a_{k-1} cdot b^{k-1} + ldots + a_1 cdot b^1 + a_0 cdot b^0 ] where ( b ) is the base (which is 4 in this case), and ( a_i ) are the digits. 5. Calculating each term: [ (01320132)_4 = 0 cdot 4^7 + 1 cdot 4^6 + 3 cdot 4^5 + 2 cdot 4^4 + 0 cdot 4^3 + 1 cdot 4^2 + 3 cdot 4^1 + 2 cdot 4^0 ] - ( 0 cdot 4^7 = 0 ) - ( 1 cdot 4^6 = 1 cdot 4096 = 4096 ) - ( 3 cdot 4^5 = 3 cdot 1024 = 3072 ) - ( 2 cdot 4^4 = 2 cdot 256 = 512 ) - ( 0 cdot 4^3 = 0 ) - ( 1 cdot 4^2 = 1 cdot 16 = 16 ) - ( 3 cdot 4^1 = 3 cdot 4 = 12 ) - ( 2 cdot 4^0 = 2 ) 6. Summing these values: [ 0 + 4096 + 3072 + 512 + 0 + 16 + 12 + 2 = 7710 ] 7. Therefore, the number represented by the base-4 sequence "оуавоуав" in base-10 is: [ boxed{7710} ]

answer:1. Let ( x = k + t ) where ( k ) is an integer and ( 0 leq t < 1 ). Then, the integer part of ( x ) is ( k ) and the fractional part of ( x ) is ( t ). Therefore, the given equation can be rewritten as: [ [x]^5 + {x}^5 = x^5 implies k^5 + t^5 = (k + t)^5 ] 2. Expanding ( (k + t)^5 ) using the binomial theorem, we get: [ (k + t)^5 = k^5 + 5k^4t + 10k^3t^2 + 10k^2t^3 + 5kt^4 + t^5 ] 3. Substituting this back into the equation, we have: [ k^5 + t^5 = k^5 + 5k^4t + 10k^3t^2 + 10k^2t^3 + 5kt^4 + t^5 ] 4. Simplifying, we get: [ 0 = 5k^4t + 10k^3t^2 + 10k^2t^3 + 5kt^4 ] 5. Factoring out ( t ) from the right-hand side, we obtain: [ 0 = t(5k^4 + 10k^3t + 10k^2t^2 + 5kt^3) ] 6. This implies: [ t = 0 quad text{or} quad 5k^4 + 10k^3t + 10k^2t^2 + 5kt^3 = 0 ] 7. If ( t = 0 ), then ( x = k ), which means ( x ) is an integer. Therefore, ( x in mathbb{Z} ). 8. If ( t neq 0 ), then we have: [ 5k^4 + 10k^3t + 10k^2t^2 + 5kt^3 = 0 ] 9. Dividing the entire equation by ( t ), we get: [ 5k^4 + 10k^3t + 10k^2t^2 + 5kt^3 = 0 implies 5k^4 + 10k^3t + 10k^2t^2 + 5kt^3 = 0 ] 10. Factoring out ( 5k ) from the equation, we get: [ k(5k^3 + 10k^2t + 10kt^2 + 5t^3) = 0 ] 11. This implies: [ k = 0 quad text{or} quad 5k^3 + 10k^2t + 10kt^2 + 5t^3 = 0 ] 12. If ( k = 0 ), then ( x = t ) where ( 0 leq t < 1 ). Therefore, ( x in [0, 1) ). 13. If ( k neq 0 ), then we have: [ 5k^3 + 10k^2t + 10kt^2 + 5t^3 = 0 ] 14. Dividing the entire equation by 5, we get: [ k^3 + 2k^2t + 2kt^2 + t^3 = 0 ] 15. Factoring the left-hand side, we get: [ (k + t)(k^2 + kt + t^2) = 0 ] 16. This implies: [ k + t = 0 quad text{or} quad k^2 + kt + t^2 = 0 ] 17. If ( k + t = 0 ), then ( k = -t ). Since ( k ) is an integer and ( 0 leq t < 1 ), ( k ) must be 0. Therefore, ( x = t ) where ( 0 leq t < 1 ). Hence, ( x in [0, 1) ). 18. If ( k^2 + kt + t^2 = 0 ), then since ( k ) and ( t ) are real numbers, this equation has no real solutions for ( k neq 0 ) and ( 0 leq t < 1 ). 19. Therefore, the only solutions are ( x in [0, 1) cup mathbb{Z} ). The final answer is ( boxed{ [0, 1) cup mathbb{Z} } ).