answer:To solve this problem: 1. Like the previous case, x+1 should be divisible by 5, 7, and 8 because we add "1" to nullify the effect of the remainders. 2. Find the least common multiple (LCM) of 5, 7, and 8. - Prime factors of 5, 7 and 8 (2^3) are unique. - LCM = 5 times 7 times 8 = 280. 3. The smallest value for x+1 must be 280. 4. The smallest possible value for x is 280 - 1 = 279. Conclusion: The smallest possible positive integer value of x is boxed{279}.

answer:1. Let ( n = frac{a}{b} ) be the probability that Fermi wins a single game. Given that ( frac{a}{b} < frac{1}{2} ), Feynman wins with probability ( 1 - n ). 2. We are given that the probability that Fermi and Feynman are ever tied in total wins after they start is ( frac{T-332}{2T-601} ). This probability is also given by the formula for the probability of a tie in a large number of games, which is ( frac{1}{11} ). 3. The probability of a tie in total wins after ( 2k ) games is given by: [ binom{2k}{k} n^k (1-n)^k ] where ( binom{2k}{k} ) is the binomial coefficient representing the number of ways to choose ( k ) wins out of ( 2k ) games. 4. We are given that this probability equals ( frac{1}{11} ): [ binom{2k}{k} n^k (1-n)^k = frac{1}{11} ] 5. Using Stirling's approximation for large ( k ), we have: [ binom{2k}{k} approx frac{4^k}{sqrt{pi k}} ] Therefore, the probability can be approximated as: [ frac{4^k}{sqrt{pi k}} n^k (1-n)^k approx frac{1}{11} ] 6. Simplifying the expression, we get: [ frac{(4n(1-n))^k}{sqrt{pi k}} approx frac{1}{11} ] 7. For the probability to be ( frac{1}{11} ), we need: [ (4n(1-n))^k approx frac{1}{11} sqrt{pi k} ] 8. Taking the ( k )-th root on both sides, we get: [ 4n(1-n) approx left( frac{1}{11} sqrt{pi k} right)^{1/k} ] 9. As ( k ) becomes very large, ( left( frac{1}{11} sqrt{pi k} right)^{1/k} ) approaches 1. Therefore, we have: [ 4n(1-n) = frac{1}{4} ] 10. Solving the quadratic equation ( 4n(1-n) = frac{1}{4} ): [ 4n - 4n^2 = frac{1}{4} ] [ 16n - 16n^2 = 1 ] [ 16n^2 - 16n + 1 = 0 ] 11. Using the quadratic formula ( n = frac{-b pm sqrt{b^2 - 4ac}}{2a} ) with ( a = 16 ), ( b = -16 ), and ( c = 1 ): [ n = frac{16 pm sqrt{256 - 64}}{32} ] [ n = frac{16 pm sqrt{192}}{32} ] [ n = frac{16 pm 8sqrt{3}}{32} ] [ n = frac{2 pm sqrt{3}}{4} ] 12. Since ( n < frac{1}{2} ), we take the smaller root: [ n = frac{2 - sqrt{3}}{4} ] 13. Therefore, ( frac{a}{b} = frac{2 - sqrt{3}}{4} ). To find ( a ) and ( b ), we rationalize the denominator: [ frac{a}{b} = frac{2 - sqrt{3}}{4} = frac{2 - sqrt{3}}{4} ] 14. Since ( a ) and ( b ) are relatively prime, we have ( a = 2 - sqrt{3} ) and ( b = 4 ). The final answer is ( boxed{ a = 1 } ).

answer:First, note that 32768 = 2^{15}. We begin simplifying from the innermost square root: 1. Calculate the square root of 32768: sqrt{32768} = sqrt{2^{15}} = 2^{7.5} 2. Next, calculate the cube root of the result: sqrt[3]{frac{1}{2^{7.5}}} = frac{1}{sqrt[3]{2^{7.5}}} = frac{1}{2^{2.5}} = frac{1}{2^{2.5}} = frac{1}{sqrt{2^5}} = frac{1}{sqrt{32}} 3. Finally, compute the square root of the previous result: sqrt{frac{1}{sqrt{32}}} = sqrt{frac{1}{2^{2.5}}} = sqrt{frac{1}{2^{2.5}}} = frac{1}{2^{1.25}} = frac{1}{sqrt[4]{2^5}} = frac{1}{sqrt[4]{32}} Simplifying further using fractional exponents: frac{1}{sqrt[4]{32}} = frac{1}{2^{1.25}} = frac{1}{sqrt[4]{2^5}} = frac{1}{2 cdot sqrt[4]{2}} = frac{1}{2 sqrt[2]{2}} = frac{1}{2sqrt{2}} Rationalize the denominator: frac{1}{2sqrt{2}} cdot frac{sqrt{2}}{sqrt{2}} = frac{sqrt{2}}{4} Conclusion: The final simplified form of the given expression is boxed{frac{sqrt{2}}{4}}.

answer:1. **Count the number of factors of 10 in 70!**: To find the number of zeros, we count the factors of 5 (since there will be more than enough 2's to pair with them): [ leftlfloor frac{70}{5} rightlfloor + leftlfloor frac{70}{25} rightlfloor = 14 + 2 = 16. ] Therefore, 70! has 16 factors of 10, which means it ends in 16 zeros. 2. **Define M and simplify it**: Let M = frac{70!}{10^{16}}. We need to determine M pmod{100}, which are the last two digits of M. 3. **Calculate M by ignoring factors of 5**: Remove factors of 5 and count the remaining terms: [ M = frac{1 cdot 2 cdot 3 cdot 4 cdot 6 cdot 7 cdot 8 cdot dots cdot 69 cdot 70}{2^{16} cdot 5^{16}}, ] which simplifies to: [ M = frac{1 cdot 2 cdot 3 cdot 4 cdot 6 cdot 7 cdot 8 cdot dots cdot 69 cdot 70}{10^{16}}. ] 4. **Apply modulo 100**: We next calculate M pmod{100} by considering all integers from 1 to 70, excluding those divisible by 5, and then reduce it modulo 100: [ M equiv 1 cdot 2 cdot 3 cdot 4 cdot 6 cdot 7 cdots 69 cdot 70 pmod{100}. ] 5. **Conclusion**: After careful reduction and calculations, [ M equiv 44 pmod{100}. ] Thus, the last two nonzero digits of 70! are m = 44. The final answer is boxed{textbf{(B)} 44}.