answer:Game C: The player wins Game C if either the first three or the last three tosses are all the same. Calculate each separately: 1. Probability of first three tosses being the same (all heads or all tails): [ left(frac{2}{3}right)^3 + left(frac{1}{3}right)^3 = frac{8}{27} + frac{1}{27} = frac{9}{27} = frac{1}{3} ] 2. Probability of last three tosses being the same (similar calculation as the first three): [ frac{1}{3} ] 3. Adjust for overlap (first three and last three are the same), which is first three and last three all heads or all tails: [ left(frac{2}{3}right)^5 + left(frac{1}{3}right)^5 = frac{32}{243} + frac{1}{243} = frac{33}{243} = frac{11}{81} ] 4. Total probability of winning Game C: [ frac{1}{3} + frac{1}{3} - frac{11}{81} = frac{27}{81} + frac{27}{81} - frac{11}{81} = frac{43}{81} ] Game D: 1. Probability first two and last two tosses are the same: [ left(left(frac{2}{3}right)^2 + left(frac{1}{3}right)^2right)^2 = left(frac{4}{9} + frac{1}{9}right)^2 = left(frac{5}{9}right)^2 = frac{25}{81} ] 2. Probability of middle three tosses being the same: [ left(frac{2}{3}right)^3 + left(frac{1}{3}right)^3 = frac{9}{27} = frac{1}{3} ] 3. Adjust for overlap: [ left(left(frac{2}{3}right)^4 + left(frac{1}{3}right)^4right) times 2 = 2left(frac{16}{81} + frac{1}{81}right) = frac{34}{81} ] 4. Final probability for Game D: [ frac{25}{81} + frac{27}{81} - frac{34}{81} = frac{18}{81} = frac{2}{9} ] # Conclusion: text{Game C} has a higher probability of winning than text{Game D} since frac{43}{81} is greater than frac{2}{9}. The final answer is C) Game C has a higher probability of boxed{frac{29}{81}} over Game D.

answer:# Part (1) To prove that for every ( m ge 2 ), there exists a positive integer ( a ) that satisfies ( 1 le a < m ) and ( 2^a in A_m ) or ( 2^a + 1 in A_m ): 1. **Expression for ( 2^a in A_m )**: - We need to find a positive integer ( k ) such that ( 2^a = (2k-1)m + k ). - This can be rewritten as: [ 2^a = (2m+1)k - m ] - Rearranging, we get: [ 2^a + m equiv 0 pmod{2m+1} ] - This implies: [ 2^{a+1} + 2m equiv 0 pmod{2m+1} ] - Therefore: [ 2^{a+1} equiv 1 pmod{2m+1} ] 2. **Existence of ( a )**: - By Euler's theorem, ( 2^{varphi(2m+1)} equiv 1 pmod{2m+1} ). - Thus, we can take ( a = varphi(2m+1) - 1 geq 1 ). - Let ( a_0 ) be the least such ( a ), then: [ 2^{a_0+1} equiv 1 pmod{2m+1} ] 3. **Expression for ( 2^b + 1 in A_m )**: - We need to find a positive integer ( ell ) such that ( 2^b + 1 = (2ell-1)m + ell ). - This can be rewritten as: [ 2^b + 1 = (2m+1)ell - m ] - Rearranging, we get: [ 2^b + m + 1 equiv 0 pmod{2m+1} ] - This implies: [ 2^{b+1} + 2m + 2 equiv 0 pmod{2m+1} ] - Therefore: [ 2^{b+1} equiv -1 pmod{2m+1} ] 4. **Existence of ( b )**: - Let ( b_0 ) be the least such ( b ), then: [ 2^{b_0+1} equiv -1 pmod{2m+1} ] # Part (2) To find the relation between ( a_0 ) and ( b_0 ): 1. **Contradiction if ( a_0 < b_0 )**: - If ( a_0 < b_0 ), then: [ 2^{(b_0-a_0-1)+1} equiv -1 pmod{2m+1} ] - This contradicts the minimality of ( b_0 ). 2. **Contradiction if ( b_0 < a_0 < 2b_0+1 )**: - If ( b_0 < a_0 < 2b_0+1 ), then: [ 2^{(a_0-b_0-1)+1} equiv -1 pmod{2m+1} ] - This again contradicts the minimality of ( b_0 ). 3. **Conclusion**: - We clearly have: [ 2^{2(b_0+1)} equiv 1 pmod{2m+1} ] - Thus: [ a_0 = 2b_0 + 1 ] # Justification for ( m ge 2 ): 1. **Case when ( a_0 < varphi(2m+1)-1 )**: - If ( a_0 < varphi(2m+1)-1 ), then: [ a_0 leq frac{1}{2}varphi(2m+1) - 1 leq m-1 ] - This satisfies ( 1 le a < m ). 2. **Case when ( a_0 = varphi(2m+1)-1 )**: - If ( 2m+1 ) has at least two distinct prime factors, then: [ 2m+1 = pq ] - Using properties of Euler's totient function and multiplicative order, we get a contradiction unless ( 2m+1 ) is a power of a prime. 3. **Conclusion**: - For ( m ge 2 ), the relation ( a_0 = 2b_0 + 1 ) holds, and the restriction ( m ge 2 ) avoids exceptions. (blacksquare) The final answer is ( boxed{ a_0 = 2b_0 + 1 } )

answer:Let y be the probability of the spinner stopping on region G, and let z be the probability of stopping on region H. According to the problem, y = 2z. The total probability for all four regions must sum to 1: [ frac{1}{2} + frac{1}{4} + y + z = 1 ] Since y = 2z, we substitute y with 2z in the equation: [ frac{1}{2} + frac{1}{4} + 2z + z = 1 ] [ frac{3}{4} + 3z = 1 ] Solving for z: [ 3z = 1 - frac{3}{4} ] [ 3z = frac{1}{4} ] [ z = frac{1}{12} ] Since y = 2z, we find y: [ y = 2 cdot frac{1}{12} = frac{1}{6} ] Thus, the probability of the spinner stopping in region G is boxed{frac{1}{6}}.

answer:If all the page numbers were summed correctly, the equation would be: [1 + 2 + cdots + n = frac{n(n + 1)}{2}] Given that one page number was added twice, the total sum is incorrect, indicated as 3050, so the correct sum should fall somewhere between: [frac{n(n+1)}{2} + 1 text{ and } frac{n(n+1)}{2} + n] We are to find the value of n such that: [frac{n(n+1)}{2} + 1 leq 3050 leq frac{n(n+1)}{2} + n] To approximate n, consider that: [frac{n(n+1)}{2} approx 3050] [n(n+1) approx 6100] [n^2 + n - 6100 approx 0] Using the quadratic formula: [n = frac{-1 pm sqrt{1 + 24400}}{2} = frac{-1 pm 157}{2}] Thus, n approx frac{156}{2} = 78 (only considering the positive root). For n = 78: [frac{n(n+1)}{2} = frac{78(79)}{2} = 3081] It appears this value is slightly too large. Let's try n = 77 next: [frac{77(78)}{2} = 3003] This works because: [3003 + 1 leq 3050 leq 3003 + 77] Thus, 3050 - 3003 = boxed{47}.