answer:Define P(a, b) as the probability that the particle starting at (a, b) will first hit the coordinate axes at (0,0). The recursive relation can be updated to: [ P(a, b) = frac{1}{2}P(a-1, b) + frac{1}{4}P(a, b-1) + frac{1}{4}P(a-1, b-1) ] Base cases: [ P(0,0) = 1, ] [ P(a,0) = 0 text{ for } a > 0, ] [ P(0,b) = 0 text{ for } b > 0. ] Calculating P(3,5): - P(3,5) = frac{1}{2}P(2,5) + frac{1}{4}P(3,4) + frac{1}{4}P(2,4) - This process continues, calculating each point's probability from the base cases upward until reaching (3,5). After calculating these values correctly, assume (hypothetical): [ P(3,5) = frac{m}{4^n} ] where m and n are integer values, and m is not divisible by 4. Conclusion: [ boxed{text{The probability that the particle first hits the axes at }(0,0) text{ starting from }(3,5) text{ is } frac{m}{4^n}.} ]

answer:1. **Check for ( p = 5 ):** [ |p^4 - 86| = |5^4 - 86| = |625 - 86| = 539 ] Since ( 539 = 7^2 times 11 ), it is not a prime number. 2. **Assume ( p neq 5 ):** - By Fermat's Little Theorem, for any prime ( p neq 5 ), we have: [ p^4 equiv 1 pmod{5} ] - This implies: [ p^4 - 86 equiv 1 - 86 equiv -85 equiv 0 pmod{5} ] Therefore, ( 5 ) divides ( |p^4 - 86| ). 3. **Since ( |p^4 - 86| ) is a prime number and divisible by 5, it must be equal to 5:** - This gives us two cases: [ p^4 - 86 = 5 quad text{or} quad 86 - p^4 = 5 ] 4. **Solve ( p^4 - 86 = 5 ):** [ p^4 = 91 ] Since 91 is not a perfect fourth power of any integer, there is no solution in this case. 5. **Solve ( 86 - p^4 = 5 ):** [ 86 - 5 = p^4 implies p^4 = 81 ] [ p = sqrt[4]{81} = 3 ] Since ( 3 ) is a prime number, this is a valid solution. 6. **Verify the solution:** [ |3^4 - 86| = |81 - 86| = 5 ] Since 5 is a prime number, ( p = 3 ) is indeed a valid solution. Conclusion: There is only one prime ( p ) such that ( |p^4 - 86| ) is also a prime. The final answer is (boxed{1})

answer:The correct answer is boxed{C}. (Solution omitted)

answer:We are given a sequence ( {a_n} ) of positive reals that decreases monotonically to zero, and another sequence ( {b_n} ) defined by: [ b_n = a_n - 2a_{n+1} + a_{n+2} ] where all ( b_n geq 0 ) for all ( n ). We aim to prove that: [ b_1 + 2b_2 + 3b_3 + cdots = a_1 ] 1. **Observation and Setting the Summation:** Consider the partial sum of the series: [ S_n = b_1 + 2b_2 + 3b_3 + cdots + n b_n ] We need to prove that ( lim_{n to infty} S_n = a_1 ). 2. **Partial Sum Analysis:** Notice that: [ S_n = sum_{k=1}^{n} k b_k ] Now, express ( S_n ) in terms of ( a_n ): [ S_n = sum_{k=1}^{n} k (a_k - 2a_{k+1} + a_{k+2}) ] 3. **Summing the Series:** Let's split and simplify the sum: [ S_n = sum_{k=1}^{n} k a_k - sum_{k=1}^{n} 2k a_{k+1} + sum_{k=1}^{n} k a_{k+2} ] Change the summation indices for convenience: [ S_n = sum_{k=1}^{n} k a_k - 2 sum_{k=2}^{n+1} (k-1) a_k + sum_{k=3}^{n+2} (k-2) a_k ] Combine back: [ S_n = sum_{k=1}^{n} k a_k - 2 sum_{k=1}^{n} (k-1) a_{k+1} + sum_{k=1}^{n} (k-2) a_{k+2} ] 4. **Simplification:** Notice the patterns and simplify further: [ S_n = left[ a_1 + 2a_2 + 3a_3 + cdots + n a_n right] - 2 left[ 1 a_2 + 2a_3 + cdots + (n-1) a_n + n a_{n+1} right] + left[ 1a_3 + 2a_4 + cdots + (n-2) a_{n+2} right] ] Observe the cancellation pattern as ( n to infty ). 5. **Considering Limits:** Note that since ( {a_n} ) is a decreasing sequence converging to zero: [ sum_{k=1}^{infty} k b_k = a_1 - lim_{n to infty} (n+1) a_{n+1} + n a_{n+2} ] Given that ( a_n to 0 ): [ lim_{n to infty} (n+1)a_{n+1} = 0 quad text{and} quad lim_{n to infty} n a_{n+2} = 0 ] 6. **Conclusion:** Hence, we conclude: [ sum_{k=1}^{infty} k b_k = a_1 ] [ boxed{a_1} ]