answer：To determine how many ways 52 can be expressed as the sum of two primes, we start by noting that since 52 is even, one of the primes must be 2 (the only even prime). Subtracting 2 from 52 gives 50, which is not a prime number. Therefore, we need to consider other pairs of odd primes that add up to 52.We systematically check pairs of primes:1. (3 + 49): 49 is not prime (since (49 = 7 times 7)).2. (5 + 47): Both 5 and 47 are primes.3. (7 + 45): 45 is not prime (since (45 = 5 times 9)).4. (11 + 41): Both 11 and 41 are primes.5. (13 + 39): 39 is not prime (since (39 = 3 times 13)).6. (17 + 35): 35 is not prime (since (35 = 5 times 7)).7. (19 + 33): 33 is not prime (since (33 = 3 times 11)).8. (23 + 29): Both 23 and 29 are primes.The valid pairs are ((5, 47)), ((11, 41)), and ((23, 29)). Therefore, there are 3 ways to write 52 as the sum of two primes.

answer：Alright, so I have this problem here about two triangular regions. The first one is bounded by the x-axis, y-axis, and the line y = 2x - 6. The second one is bounded by the x-axis, the same line y = 2x - 6, and another vertical line x = d, where d is greater than 0. The problem says that the area of the first triangle is one-quarter of the area of the second triangle. I need to find the value of d.Okay, let me start by understanding the first triangle. It's bounded by the x-axis, y-axis, and the line y = 2x - 6. To find its area, I should probably find the intercepts of the line with the axes because those will give me the base and height of the triangle.First, let's find the x-intercept. The x-intercept occurs where y = 0. So, I'll set y to 0 in the equation and solve for x:0 = 2x - 6 2x = 6 x = 3So, the x-intercept is at (3, 0). That means the base of the triangle along the x-axis is from (0, 0) to (3, 0), which is 3 units long.Next, the y-intercept occurs where x = 0. Plugging x = 0 into the equation:y = 2(0) - 6 y = -6So, the y-intercept is at (0, -6). That means the height of the triangle along the y-axis is from (0, 0) to (0, -6), which is 6 units long.Now, the area of a triangle is (1/2)*base*height. So, the area of the first triangle is:Area1 = (1/2) * 3 * 6 = 9Okay, so the area of the first triangle is 9.Now, the problem says that this area is one-quarter of the area of the second triangle. So, the area of the second triangle must be 4 times 9, which is 36.So, Area2 = 36.Now, let's figure out the second triangle. It's bounded by the x-axis, the line y = 2x - 6, and the vertical line x = d. So, this triangle is going to have a base along the x-axis from (3, 0) to (d, 0), and its height will be the y-value of the line y = 2x - 6 at x = d.Wait, let me visualize this. The line y = 2x - 6 crosses the x-axis at (3, 0) and goes downward into the fourth quadrant. But since we're talking about the area bounded by the x-axis, the line, and x = d, I think we need to consider the region where the line is above the x-axis or below?Wait, at x = d, if d is greater than 3, then y = 2d - 6 will be positive because 2d - 6 will be greater than 0 when d > 3. So, the line will be above the x-axis for x > 3. Therefore, the triangle will be above the x-axis, with vertices at (3, 0), (d, 0), and (d, y), where y is the value of the line at x = d.So, the base of the second triangle is from (3, 0) to (d, 0), which is (d - 3) units long. The height is the y-value at x = d, which is y = 2d - 6.Therefore, the area of the second triangle is:Area2 = (1/2) * (d - 3) * (2d - 6)We know that Area2 is 36, so:(1/2) * (d - 3) * (2d - 6) = 36Let me simplify this equation step by step.First, multiply both sides by 2 to eliminate the fraction:(d - 3) * (2d - 6) = 72Now, let's expand the left side:(d - 3)(2d - 6) = 2d(d - 3) - 6(d - 3) = 2d^2 - 6d - 6d + 18 = 2d^2 - 12d + 18So, the equation becomes:2d^2 - 12d + 18 = 72Subtract 72 from both sides to set the equation to zero:2d^2 - 12d + 18 - 72 = 0 2d^2 - 12d - 54 = 0Now, let's simplify this quadratic equation. First, I can divide all terms by 2 to make it simpler:d^2 - 6d - 27 = 0Now, I need to solve for d. This is a quadratic equation, so I can use the quadratic formula:d = [6 ± sqrt( (-6)^2 - 4*1*(-27) )]/(2*1) = [6 ± sqrt(36 + 108)]/2 = [6 ± sqrt(144)]/2 = [6 ± 12]/2So, we have two solutions:d = (6 + 12)/2 = 18/2 = 9 d = (6 - 12)/2 = (-6)/2 = -3But since d > 0, we discard the negative solution. Therefore, d = 9.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Found the intercepts correctly: x-intercept at (3, 0), y-intercept at (0, -6).2. Calculated Area1 as (1/2)*3*6 = 9. That seems correct.3. Then, since Area1 is one-quarter of Area2, so Area2 = 4*9 = 36. That makes sense.4. For the second triangle, identified the base as (d - 3) and the height as (2d - 6). That seems right because at x = d, y = 2d - 6, which is positive since d > 3.5. Set up the equation (1/2)*(d - 3)*(2d - 6) = 36. Multiplied both sides by 2: (d - 3)*(2d - 6) = 72.6. Expanded it correctly to 2d^2 - 12d + 18 = 72.7. Subtracted 72: 2d^2 - 12d - 54 = 0.8. Divided by 2: d^2 - 6d - 27 = 0.9. Applied quadratic formula: d = [6 ± sqrt(36 + 108)]/2 = [6 ± sqrt(144)]/2 = [6 ± 12]/2.10. Got d = 9 and d = -3. Discarded negative, so d = 9.Everything seems to check out. So, the value of d is 9.**Final Answer**boxed{A}

answer：First, I recognize that the equation involves a cube, so I'll need to take the cube root to solve for ( x ).I start by isolating the term with the cube by taking the cube root of both sides of the equation.Next, I simplify the cube root of 64, which I know is 4 because ( 4^3 = 64 ).After that, I solve for ( x ) by adding 1 to both sides of the equation.Finally, I arrive at the solution ( x = 5 ).

answer：Okay, so I need to find the value of 15! divided by 7! times 8!. Hmm, factorials can get really big, so calculating them directly might not be the best approach. Let me think about how to simplify this expression first.I remember that factorials involve multiplying a series of descending natural numbers. So, 15! is 15 × 14 × 13 × ... × 1, and similarly for 7! and 8!. But calculating each of these separately would result in very large numbers, which could be cumbersome and prone to errors. There must be a smarter way to simplify this fraction before multiplying everything out.Let me write down the expression:15! / (7! × 8!)I notice that 15! can be written as 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8!. That's because 15! is the product of all integers from 1 to 15, and 8! is the product from 1 to 8. So, if I factor out 8! from 15!, I'm left with the product of 15 down to 9.So, substituting that back into the original expression:(15 × 14 × 13 × 12 × 11 × 10 × 9 × 8!) / (7! × 8!)Now, I can cancel out the 8! in the numerator and the denominator since they appear in both. That simplifies the expression to:(15 × 14 × 13 × 12 × 11 × 10 × 9) / 7!Okay, now I need to compute 7! to see what I'm dealing with. 7! is 7 × 6 × 5 × 4 × 3 × 2 × 1, which equals 5040. So, the expression becomes:(15 × 14 × 13 × 12 × 11 × 10 × 9) / 5040Now, I need to compute the numerator. Let's do that step by step:15 × 14 = 210210 × 13 = 27302730 × 12 = 3276032760 × 11 = 360,360360,360 × 10 = 3,603,6003,603,600 × 9 = 32,432,400So, the numerator is 32,432,400.Now, divide that by 5040:32,432,400 / 5040Hmm, that's still a bit large. Maybe I can simplify this division by breaking it down. Let's see if both numbers are divisible by 10 first:32,432,400 ÷ 10 = 3,243,2405040 ÷ 10 = 504So now, it's 3,243,240 / 504I can try dividing both numerator and denominator by 12 to see if that helps:3,243,240 ÷ 12 = 270,270504 ÷ 12 = 42So now, it's 270,270 / 42Let me divide 270,270 by 42:42 × 6,000 = 252,000Subtracting that from 270,270 gives 18,270Now, 42 × 435 = 18,270So, total is 6,000 + 435 = 6,435Wait, that doesn't seem right because 42 × 6,435 should be 270,270, but let me check:42 × 6,435First, 42 × 6,000 = 252,00042 × 400 = 16,80042 × 35 = 1,470Adding those up: 252,000 + 16,800 = 268,800; 268,800 + 1,470 = 270,270Yes, that's correct. So, 270,270 / 42 = 6,435Therefore, the original expression simplifies to 6,435.Wait a minute, but earlier I thought it was 181.5. That can't be right because 6,435 is much larger. Did I make a mistake somewhere?Let me go back and check my steps.Starting from the beginning:15! / (7! × 8!) = (15 × 14 × 13 × 12 × 11 × 10 × 9 × 8!) / (7! × 8!) = (15 × 14 × 13 × 12 × 11 × 10 × 9) / 7!7! = 5040So, numerator is 15 × 14 × 13 × 12 × 11 × 10 × 9Let me compute that again step by step:15 × 14 = 210210 × 13 = 27302730 × 12 = 32,76032,760 × 11 = 360,360360,360 × 10 = 3,603,6003,603,600 × 9 = 32,432,400So, numerator is 32,432,400Divide by 5040:32,432,400 / 5040Let me try another approach. Maybe factor both numerator and denominator into their prime factors to cancel out terms.Prime factors of numerator:15 = 3 × 514 = 2 × 713 = 1312 = 2² × 311 = 1110 = 2 × 59 = 3²So, numerator factors:3 × 5 × 2 × 7 × 13 × 2² × 3 × 11 × 2 × 5 × 3²Combine like terms:2^(1+2+1) = 2^43^(1+1+2) = 3^45^(1+1) = 5^27, 11, 13So, numerator is 2^4 × 3^4 × 5^2 × 7 × 11 × 13Denominator is 7! = 5040Prime factors of 5040:5040 = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1Breaking down each:7 = 76 = 2 × 35 = 54 = 2²3 = 32 = 2So, prime factors:2^(1+2+1) = 2^43^(1+1) = 3^25 = 57 = 7So, denominator is 2^4 × 3^2 × 5 × 7Now, divide numerator by denominator:(2^4 × 3^4 × 5^2 × 7 × 11 × 13) / (2^4 × 3^2 × 5 × 7)Cancel out common factors:2^4 cancels out3^2 cancels out, leaving 3^(4-2) = 3^25 cancels out one 5, leaving 5^(2-1) = 5^17 cancels outSo, remaining factors:3^2 × 5 × 11 × 13Compute that:3^2 = 99 × 5 = 4545 × 11 = 495495 × 13Let me compute 495 × 10 = 4,950495 × 3 = 1,485Add them together: 4,950 + 1,485 = 6,435So, the result is 6,435.Wait, earlier I thought it was 181.5, but that must have been a mistake. Let me check where I went wrong.Looking back, I see that in the initial calculation, I had:(15 × 14 × 13 × 12 × 11 × 10 × 9) / 5040 = 32,432,400 / 5040Then I tried dividing by 10 to get 3,243,240 / 504Then divided by 12 to get 270,270 / 42Then got 6,435But in the first attempt, I thought it was 181.5, which is clearly wrong because 6,435 is much larger.So, the correct answer should be 6,435.I think I must have made a mistake in the initial calculation when I thought it was 181.5. Probably, I divided incorrectly or missed some steps.Therefore, the correct value of 15! / (7! × 8!) is 6,435.

answer：Let the digits of the three-digit number be ( A ), ( B ), and ( C ) from left to right.1. **Middle Digit Average Condition**: [ B = frac{A + C}{2} ] For ( B ) to be an integer, ( A + C ) must be even.2. **Sum Condition**: [ A + B + C = 12 ] Substitute ( B ) from the first condition: [ A + frac{A + C}{2} + C = 12 ] Simplifying this equation: [ 2A + A + C + 2C = 24 ] [ 3A + 3C = 24 ] [ A + C = 8 ] Since ( A + C = 8 ) and needs to be even, this aligns perfectly with the previous condition.3. **Counting Valid Combinations**: - ( A ) can range from 1 to 9. - ( C ) can range from 0 to 9 since it is the last digit of a three-digit number. However, since ( A + C = 8 ), the possible values for ( C ) are restricted. - For each ( A ) from 1 to 7, there is exactly one value of ( C ) such that ( A + C = 8 ) and they follow the even/odd pattern needed for an integer ( B ). Valid pairs are: - ( (1, 7) ) - ( (2, 6) ) - ( (3, 5) ) - ( (4, 4) ) - ( (5, 3) ) - ( (6, 2) ) - ( (7, 1) ) Each of these pairs provides a unique ( B = frac{A + C}{2} ). **Total Count**: There are 7 valid pairs, each leading to a unique three-digit number.Conclusion:There are 7 valid three-digit numbers that meet the criteria.

answer：First, I recognize that the expression (635^2 - 365^2) is a difference of squares. I can apply the difference of squares formula, which states that (a^2 - b^2 = (a + b)(a - b)).Next, I identify (a) as 635 and (b) as 365. I then calculate the sum of (a) and (b), which is (635 + 365 = 1000).After that, I find the difference between (a) and (b), which is (635 - 365 = 270).Finally, I multiply the sum and the difference together: (1000 times 270 = 270,000).Therefore, the value of (635^2 - 365^2) is 270,000.