answer:When the sum of two integers a and b is even, it implies that either both a and b are even or both are odd. Given that a and b have no common factors besides 1, the case where both are even is impossible, as they would both be divisible by 2. Therefore, both a and b must be odd. For an odd integer to be prime, it needs to be only divisible by 1 and itself. Considering the distribution of prime numbers, we can see that this is less likely to happen as the numbers get larger due to the increasing number of potential divisors. However, in this problem, we are concerned with the probability of a and b both being prime, given that their sum is even. Since both a and b are odd, we can express them as a = 2n + 1 and b = 2m + 1, where n and m are non-negative integers. The sum of a and b would be a + b = 2n + 2m + 2 = 2(n + m + 1), which is even. For both a and b to be prime, they would need to be two distinct odd prime numbers. We can consider the probability of this happening by looking at the distribution of odd prime numbers among all odd numbers. However, we have to note that, while we can derive the probability for specific ranges or values of a and b using exact counts of prime numbers, the general probability question asked here does not have a direct answer because the probability of being prime decreases as the integers increase without bound. But if we look at the initial values, such as a and b up to some reasonable upper bound N, the number of odd primes is approximately given by (N / (ln N))/2 since N / (ln N) is an approximation of the prime-counting function for large N, and roughly half of those primes would be odd. However, the total number of odd numbers up to N is (N+1)/2 (rounding down to disregard the even number N/2). Now if we have an even sum, such as a and b both odd, we want to know the number of pairs that both are prime. This is the square of the probability that an odd number is prime. This odd prime probability would be roughly ((N / (ln N))/2) / ((N+1)/2) for one number, and thus the square of this fraction for two numbers. The general answer can thus be described as follows: the probability decreases roughly as N^2 / (ln N)^2 divided by (N^2), which simplifies to (1 / (ln N))^2 for larger N values. As the values increase, the probability decreases towards zero since it is inversely proportional to (ln N)^2, though not according to a simple specific formula, and is highly dependent on the exact value of N considered.

answer:We can extend the problem to higher numbers and represent the probability mathematically using the theory of numbers by applying the Prime Number Theorem (PNT). The PNT states that the number of primes less than or equal to x, denoted by π(x), is asymptotically equal to x / ln(x) as x approaches infinity. To express the probability that two odd integers a and b are both prime when their sum is even, we can use the following approach: Let P(a, b) be the probability that a and b are both prime. We want to find P(a, b | a + b = even). Using the fact that a and b are odd, we can express them as a = 2n + 1 and b = 2m + 1, where n and m are non-negative integers. The sum of a and b would be a + b = 2n + 2m + 2 = 2(n + m + 1), which is even. The probability that a is prime can be expressed as P(a) = π(a) / (a / 2), since a is odd. Similarly, the probability that b is prime is P(b) = π(b) / (b / 2). Assuming independence of primality, which is a reasonable assumption for large values of a and b, we can write the joint probability as: P(a, b) = P(a) * P(b) = (π(a) / (a / 2)) * (π(b) / (b / 2)) Using the Prime Number Theorem, we can replace π(a) and π(b) with a / ln(a) and b / ln(b), respectively, as a and b approach infinity. P(a, b) ≈ (a / ln(a)) / (a / 2) * (b / ln(b)) / (b / 2) = 4 / (ln(a) * ln(b)) Since a + b = 2n + 2m + 2, we can write b ≈ 2N - a, where N = n + m + 1. Substituting this into the expression for P(a, b), we get: P(a, b) ≈ 4 / (ln(a) * ln(2N - a)) To find the probability that a and b are both prime given that their sum is even, we need to average this expression over all possible values of a and b that satisfy a + b = even. Using the average order of the reciprocal logarithm function, we can simplify the expression to: P(a, b | a + b = even) ≈ C / (ln N)^2 where C is a constant. This expression shows that the probability that two odd integers a and b are both prime when their sum is even decreases like 1 / (ln N)^2 as N approaches infinity. This result is consistent with our earlier heuristic argument and provides a more rigorous mathematical representation of the probability using the theory of numbers.

answer:Let's break down the problem using the properties of prime numbers. We know that for a and b to be prime, they must be odd (except for 2, which is the only even prime number). Since a + b is even, we can write a = 2n + 1 and b = 2m + 1, where n and m are non-negative integers. Now, we can use the properties of prime numbers to find the probability that a and b are both prime. A prime number p has the following properties: 1. p is divisible only by 1 and itself (p). 2. p is a positive integer greater than 1. Using these properties, we can write the probability that a is prime as: P(a) = (number of prime numbers less than or equal to a) / (number of odd integers less than or equal to a) Since a is odd, the number of odd integers less than or equal to a is approximately a / 2. Using the Prime Number Theorem, we can approximate the number of prime numbers less than or equal to a as a / ln(a). So, P(a) ≈ (a / ln(a)) / (a / 2) = 2 / ln(a) Similarly, P(b) ≈ 2 / ln(b) Assuming independence of primality, we can write the joint probability as: P(a, b) = P(a) * P(b) ≈ (2 / ln(a)) * (2 / ln(b)) = 4 / (ln(a) * ln(b)) Now, let's find the probability that a and b are both prime given that their sum is even. We can write the conditional probability as: P(a, b | a + b = even) = P(a, b) / P(a + b = even) Since a and b are odd, a + b is always even. Therefore, P(a + b = even) = 1. So, P(a, b | a + b = even) ≈ P(a, b) = 4 / (ln(a) * ln(b)) To find the probability in terms of the sum N = n + m + 1, we can write b ≈ 2N - a. Substituting this into the expression for P(a, b), we get: P(a, b | a + b = even) ≈ 4 / (ln(a) * ln(2N - a)) Using the average order of the reciprocal logarithm function, we can simplify the expression to: P(a, b | a + b = even) ≈ C / (ln N)^2 where C is a constant. This expression shows that the probability that two odd integers a and b are both prime when their sum is even decreases like 1 / (ln N)^2 as N approaches infinity. To find the value of C, we can use numerical methods or the following approximation: C ≈ 4 / (e^(-γ))^2 where γ is the Euler-Mascheroni constant. Using this approximation, we get: P(a, b | a + b = even) ≈ (4 / (e^(-γ))^2) / (ln N)^2 = (4e^(2γ)) / (ln N)^2 This expression provides an approximate value of the probability in terms of the sum N. Note that this result is consistent with our earlier heuristic argument and provides a more rigorous mathematical representation of the probability using the theory of numbers.

answer:We know that one machine can produce 2 kg of material every hour. There are 3 machines working 23 hours a day, so they can produce 3 * 2 * 23 = 138 kg of material in one day. The fourth machine works only 12 hours a day, so it can produce 1 * 2 * 12 = 24 kg of material in one day. The total amount of material produced in one day is 138 + 24 = 162 kg. The factory sells the produced material for x per 1 kg, so the total earnings in one day is 162 * x. We are given that the total earnings in one day is 8100, so we can write: 162 * x = 8100. Dividing both sides by 162, we get: x = 50. The value of x is 50. 50 The answer is: 50