💡First, I need to determine the most cost-effective way to purchase at least 65 pounds of grass seed without exceeding 80 pounds. The garden center offers three bag sizes: 5-pound, 10-pound, and 25-pound bags, each with different prices.I'll start by calculating the price per pound for each bag size to identify which offers the best value:- **5-pound bag**: 13.85 ÷ 5 = 2.77 per pound- **10-pound bag**: 20.42 ÷ 10 = 2.042 per pound- **25-pound bag**: 32.25 ÷ 25 = 1.29 per poundThe 25-pound bag has the lowest price per pound, so I'll prioritize buying as many of these as possible. Since the customer needs at least 65 pounds, I'll start by purchasing two 25-pound bags:- **2 × 25 pounds = 50 pounds**This leaves me needing an additional **15 pounds** to meet the minimum requirement of 65 pounds. To efficiently reach this, I'll purchase one 10-pound bag:- **50 pounds + 10 pounds = 60 pounds**Now, I still need **5 more pounds** to reach 65 pounds. I'll buy one 5-pound bag:- **60 pounds + 5 pounds = 65 pounds**Finally, I'll calculate the total cost:- **2 × 25-pound bags = 2 × 32.25 = 64.50**- **1 × 10-pound bag = 1 × 20.42 = 20.42**- **1 × 5-pound bag = 1 × 13.85 = 13.85**Adding these together:- **Total cost = 64.50 + 20.42 + 13.85 = 98.77**Therefore, the least possible cost for purchasing at least 65 pounds of grass seed is **98.77**.

💡First, I need to understand what the least common multiple (LCM) represents. The LCM of several numbers is the smallest number that is a multiple of each of the numbers.Given the numbers ( x ), 15, and 21, and their LCM is 105, I can start by prime factorizing these numbers:[15 = 3 times 5, quad 21 = 3 times 7, quad 105 = 3 times 5 times 7]The LCM of 15 and 21 is already 105, which means that ( x ) must be a divisor of 105 to ensure that the LCM of ( x ), 15, and 21 remains 105.Next, I list the divisors of 105:[1, 3, 5, 7, 15, 21, 35, 105]Among these divisors, the largest value is 105. Therefore, the greatest possible value of ( x ) that satisfies the condition is 105.

💡First, I need to simplify the expression ( dfrac{10!}{210} ). I know that ( 10! ) is the product of all integers from 1 to 10, and 210 can be factored into its prime components.I'll start by writing out ( 10! ) and 210:[10! = 10 cdot 9 cdot 8 cdot 7 cdot 6 cdot 5 cdot 4 cdot 3 cdot 2 cdot 1][210 = 2 cdot 3 cdot 5 cdot 7]Next, I'll divide ( 10! ) by 210 by canceling out the common factors in the numerator and the denominator:[dfrac{10!}{210} = dfrac{10 cdot 9 cdot 8 cdot 7 cdot 6 cdot 5 cdot 4 cdot 3 cdot 2 cdot 1}{2 cdot 3 cdot 5 cdot 7}]After canceling, I'm left with:[dfrac{10!}{210} = 10 cdot 9 cdot 8 cdot 6 cdot 4]Now, I'll simplify this further by multiplying the remaining numbers step by step:[10 cdot 9 = 90][8 cdot 6 = 48][90 cdot 48 = 4320][4320 cdot 4 = 17280]So, ( dfrac{10!}{210} = 17280 ).Next, I need to find the square root of 17280. I'll factorize 17280 into its prime factors to identify perfect squares:[17280 = 2^7 cdot 3^3 cdot 5]I'll separate the perfect squares from the remaining factors:[sqrt{17280} = sqrt{2^6 cdot 3^2 cdot 2 cdot 3 cdot 5}]Taking the square root of the perfect squares:[sqrt{2^6 cdot 3^2} = 2^3 cdot 3 = 8 cdot 3 = 24]The remaining factors under the square root are ( 2 cdot 3 cdot 5 = 30 ). Therefore:[sqrt{17280} = 24 cdot sqrt{30}]So, the square root of ( dfrac{10!}{210} ) is ( 24sqrt{30} ).

💡First, I need to evaluate the ceiling of the square root of 165, which is written as ⎡√165⎤.To do this, I'll start by identifying the perfect squares that are closest to 165. I know that 12 squared is 144 and 13 squared is 169. This means that 144 is less than 165, and 165 is less than 169.Since 144 < 165 < 169, taking the square root of each part of the inequality gives me 12 < √165 < 13. This tells me that the square root of 165 is somewhere between 12 and 13.Because √165 is not an integer, the ceiling function will round it up to the next integer. Therefore, the smallest integer greater than √165 is 13.So, the value of ⎡√165⎤ is 13.

💡Okay, so I have this problem here: "What is the product of all real numbers that are tripled when added to their reciprocals?" Hmm, let me try to understand what this means.First, I need to translate the words into a mathematical equation. The phrase "triple when added to their reciprocals" suggests that if I take a number, add its reciprocal to it, the result is three times the original number. So, if I let the number be x, then the equation would be:x + (1/x) = 3xAlright, that seems right. Now, I need to solve this equation for x and then find the product of all real solutions.Let me write that equation again:x + (1/x) = 3xHmm, maybe I can rearrange this equation to make it easier to solve. Let me subtract x from both sides to get:(1/x) = 3x - xSimplifying the right side:(1/x) = 2xOkay, so now I have 1/x equals 2x. To solve for x, I can multiply both sides by x to eliminate the denominator:1 = 2x^2That gives me 2x^2 = 1. To solve for x^2, I'll divide both sides by 2:x^2 = 1/2Taking the square root of both sides, I get:x = ±√(1/2)Which simplifies to:x = ±(√2)/2So, the two real solutions are (√2)/2 and -(√2)/2.Now, the problem asks for the product of all these real numbers. So, I need to multiply these two solutions together:(√2)/2 * -(√2)/2Let me compute that step by step. First, multiply the numerators:√2 * -(√2) = - (√2 * √2) = -2Then, multiply the denominators:2 * 2 = 4So, putting it together, the product is:-2/4 = -1/2Therefore, the product of all real numbers that satisfy the given condition is -1/2.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the equation:x + (1/x) = 3xSubtract x:1/x = 2xMultiply both sides by x:1 = 2x^2Divide by 2:x^2 = 1/2Square root:x = ±√(1/2) = ±(√2)/2Product:(√2)/2 * -(√2)/2 = - ( (√2)^2 ) / (2*2) = -2/4 = -1/2Yep, that all checks out. So, I'm confident that the product is indeed -1/2.

💡Okay, so I have this problem to prove that if the sum or difference of two natural numbers is divisible by 50, then the last two digits of the squares of these numbers will be the same. Hmm, that sounds interesting. Let me try to break it down step by step.First, let me understand what it means for the sum or difference of two numbers to be divisible by 50. If two numbers, say a and b, have a sum or difference that's divisible by 50, that means either a + b is a multiple of 50 or a - b is a multiple of 50. So, mathematically, that would be:- If a + b is divisible by 50, then a + b = 50k for some integer k.- If a - b is divisible by 50, then a - b = 50m for some integer m.Now, the goal is to show that the last two digits of a² and b² are the same. In other words, a² ≡ b² mod 100. Because the last two digits of a number are determined by its remainder when divided by 100.So, how can I relate a² and b²? Maybe I can use the difference of squares formula. I remember that a² - b² = (a - b)(a + b). That's a useful identity. Let me write that down:a² - b² = (a - b)(a + b)Now, if either a + b or a - b is divisible by 50, what does that tell me about a² - b²?Case 1: Suppose a + b is divisible by 50. So, a + b = 50k. Then, a² - b² = (a - b)(50k). If a - b is even, meaning a - b = 2m, then a² - b² = 2m * 50k = 100mk. So, a² - b² is divisible by 100. That means a² ≡ b² mod 100, which implies the last two digits of a² and b² are the same.Wait, but why is a - b necessarily even? Hmm, let me think. If a + b is divisible by 50, which is an even number, then a and b must be both even or both odd. Because if one is even and the other is odd, their sum would be odd, which can't be divisible by 50. So, if a and b are both even or both odd, then a - b would also be even. So, yes, a - b is even in this case.Case 2: Suppose a - b is divisible by 50. So, a - b = 50m. Then, a² - b² = (50m)(a + b). If a + b is even, then a + b = 2n, so a² - b² = 50m * 2n = 100mn. Again, a² - b² is divisible by 100, so a² ≡ b² mod 100.But wait, why is a + b even in this case? If a - b is divisible by 50, which is even, then similar to before, a and b must be both even or both odd. Therefore, their sum a + b is also even. So, a + b = 2n holds true.So, in both cases, whether the sum or the difference is divisible by 50, we end up with a² - b² being divisible by 100, which means their squares have the same last two digits.Let me test this with some examples to make sure I'm not missing something.Example 1: Let a = 25 and b = 25. Then, a + b = 50, which is divisible by 50. a² = 625 and b² = 625. The last two digits are both 25, so they are the same.Example 2: Let a = 75 and b = 25. Then, a - b = 50, which is divisible by 50. a² = 5625 and b² = 625. The last two digits are 25 in both cases.Wait, hold on. In the second example, a = 75 and b = 25, a - b = 50, which is divisible by 50. a² = 5625 and b² = 625. The last two digits are 25 for both. That works.Another example: Let a = 100 and b = 50. Then, a - b = 50, which is divisible by 50. a² = 10000 and b² = 2500. The last two digits are 00 for both. So that works too.Wait, but what if a and b are both even or both odd? Let me try a = 30 and b = 20. Then, a + b = 50, which is divisible by 50. a² = 900 and b² = 400. The last two digits are 00 and 00, which are the same.Another example: a = 35 and b = 15. Then, a + b = 50, which is divisible by 50. a² = 1225 and b² = 225. The last two digits are 25 and 25, same again.Wait, but what if a and b are both odd? Let me try a = 27 and b = 23. Then, a + b = 50, which is divisible by 50. a² = 729 and b² = 529. The last two digits are 29 and 29, same.Another case: a = 43 and b = 43. Then, a - b = 0, which is divisible by 50. a² = 1849 and b² = 1849. The last two digits are 49 and 49, same.Wait, but 0 is divisible by 50, right? So, if a = b, their difference is 0, which is divisible by 50, and obviously, their squares are the same.Another test case: a = 125 and b = 75. Then, a - b = 50, which is divisible by 50. a² = 15625 and b² = 5625. The last two digits are 25 and 25, same.Hmm, all these examples seem to confirm the statement. So, the logic seems solid.But let me think about the converse. If two numbers have squares with the same last two digits, does that mean their sum or difference is divisible by 50? Hmm, not necessarily. For example, take a = 10 and b = 10. Their squares are both 100, same last two digits. But their sum is 20, which is not divisible by 50, and their difference is 0, which is divisible by 50. So, in this case, the difference is divisible by 50, but the sum isn't. So, the converse isn't necessarily true, but the original statement is about if the sum or difference is divisible by 50, then the last two digits of the squares are the same. So, the converse isn't required here.Another thought: What if a and b are such that their sum is divisible by 50, but their difference isn't? Or vice versa? For example, a = 25 and b = 25. Their sum is 50, difference is 0. Both cases, the squares have same last two digits.Another example: a = 75 and b = 25. Their difference is 50, sum is 100, which is also divisible by 50. So, both sum and difference are divisible by 50 in this case.Wait, is it possible for the sum to be divisible by 50 but the difference not to be? Let's see. Let a = 30 and b = 20. Their sum is 50, which is divisible by 50. Their difference is 10, which is not divisible by 50. So, in this case, the sum is divisible by 50, but the difference isn't. But still, their squares have the same last two digits.Wait, a = 30, b = 20. a² = 900, b² = 400. The last two digits are 00 and 00, which are the same. So, even if only the sum is divisible by 50, the squares still have the same last two digits.Similarly, if only the difference is divisible by 50, like a = 75, b = 25, their difference is 50, sum is 100, which is also divisible by 50. But in the previous example, a = 30, b = 20, their sum is 50, difference is 10. So, only the sum is divisible by 50, but the squares still have the same last two digits.So, in both cases, whether the sum or difference is divisible by 50, the squares have the same last two digits.Wait, but in the case where only the sum is divisible by 50, like a = 30, b = 20, their difference is 10, which is not divisible by 50, but their squares still have the same last two digits. So, the key is that either the sum or the difference being divisible by 50 is sufficient to ensure that the squares have the same last two digits.So, going back to the proof, I think the key is that if either a + b or a - b is divisible by 50, then a² - b² is divisible by 100, which implies that a² ≡ b² mod 100, meaning their last two digits are the same.But let me formalize this.Given that either a + b is divisible by 50 or a - b is divisible by 50.Case 1: a + b = 50k for some integer k.Then, a² - b² = (a - b)(a + b) = (a - b)(50k).If a - b is even, then a - b = 2m for some integer m.Thus, a² - b² = 2m * 50k = 100mk, which is divisible by 100.Therefore, a² ≡ b² mod 100.Case 2: a - b = 50m for some integer m.Then, a² - b² = (a - b)(a + b) = (50m)(a + b).If a + b is even, then a + b = 2n for some integer n.Thus, a² - b² = 50m * 2n = 100mn, which is divisible by 100.Therefore, a² ≡ b² mod 100.Now, why is a - b even in Case 1 and a + b even in Case 2?In Case 1: a + b is divisible by 50, which is even. Therefore, a + b is even. Since a + b is even, both a and b must be both even or both odd. Therefore, a - b is also even because:- If both are even, their difference is even.- If both are odd, their difference is even (since odd - odd = even).Similarly, in Case 2: a - b is divisible by 50, which is even. Therefore, a - b is even. Hence, a and b must be both even or both odd. Therefore, a + b is also even.So, in both cases, the other term (a - b or a + b) is even, ensuring that a² - b² is divisible by 100.Therefore, the last two digits of a² and b² are the same.I think this covers all the necessary steps. I've also tested with examples, and they all seem to confirm the statement. So, I feel confident that this proof is correct.