answer:1. **Introduction**: - We are given a set of 43 positive integers. - We aim to prove that there exist two integers (a) and (b) within this set such that (a^2 - b^2) is divisible by 100. 2. **Understanding Squares Modulo 100**: - Consider the expression (x^2 mod 100). For any integer (x), (x) and (100k + x) (where (k) is an integer) have the same remainder when divided by 100. - Thus, (x^2 mod 100) and ((100k + x)^2 mod 100) will result in the same remainder. - This consideration implies that we only need to evaluate the remainders (0, 1, 2, ldots, 99) when squared, i.e., (x^2 mod 100) for (x) taking values from 0 to 99. 3. **Possible Remainders of Squares Modulo 100**: - Any integer squared can yield only certain particular remainders when divided by 100. Specifically, these remainders are between 0 and 99. - However, since (100 - x) also maps back to the same remainder modulo 100, some remainders are repeated. - By detailed analysis or computation, it is known that there are exactly 51 unique possible remainders for (x^2 mod 100). 4. **Pigeonhole Principle Application**: - Given the 43 numbers, there are 43 squares. - Since there are only 51 unique remainders possible for squares modulo 100 and we have 43 numbers, placing 43 numbers into 51 possible remainder slots, necessarily some slots will contain numbers that are multiples of 5 (multiple of 5 remainders). 5. **Consideration of Multiples of 5**: - Suppose we have three such remainders, say (5k), (5ell), and (5m), where (k), (ell), and (m) are integers. - We now examine their squares. If any pair among (k), (ell), and (m) have the same parity (either both even or both odd), we let those be (k) and (ell). - Thus: [ a = 5k, quad b = 5ell ] [ a^2 - b^2 = (5k)^2 - (5ell)^2 = 25(k^2 - ell^2) ] - Factorizing further, [ a^2 - b^2 = 25(k - ell)(k + ell) ] - If (k) and (ell) are of the same parity, then (k - ell) and (k + ell) are even integers, implying that their product is divisible by 4. 6. **Combining Divisibilities**: - Since (25 times 4 = 100), (a^2 - b^2) is divisible by 100. - By the pigeonhole principle, if we have fewer than 51 remainders, we are forced to include a sufficient number of multiples of 5, and among them, pairs will exist which satisfy the divisibility requirements. 7. **Conclusion**: - Among any 43 positive integers, one can always find two integers (a) and (b) such that (a^2 - b^2) is divisible by 100. Therefore, the result: [ boxed{} ]

answer:# Step-by-Step Solution Part 1: Factorize the expression x^{2}-6x+8 1. Start with the given expression: x^{2}-6x+8 2. Add and subtract 9 (the square of half the coefficient of x) to complete the square: = x^{2}-6x+9-1 3. Recognize the perfect square trinomial and rewrite: = (x-3)^{2}-1^{2} 4. Factorize using the difference of squares formula: = (x-3+1)(x-3-1) 5. Simplify the expression: = (x-2)(x-4) So, the factorized form of the expression is boxed{(x-2)(x-4)}. Part 2: Find the minimum value of x^{2}-6x+8 1. Start with the expression in its completed square form: x^{2}-6x+8 = (x-3)^{2}-1 2. Recognize that a square term (x-3)^{2} is always greater than or equal to 0, and thus the minimum value of the expression occurs when (x-3)^{2} = 0. 3. Solve for x when (x-3)^{2} = 0: x = 3 4. Substitute x = 3 into the expression to find the minimum value: x^{2}-6x+8 = (3-3)^{2}-1 = -1 Therefore, when x=3, the original algebraic expression x^{2}-6x+8 has a minimum value of boxed{-1}.

answer:1. **Understanding the weekday progression:** Given that the 250^{text{th}} day of year P is a Sunday. The technique to reduce days modulo 7: 250 mod 7 = 5, so a week earlier, i.e., the 250 - 7 = 243^{text{th}} day, is also a Sunday, and continuing this, we eventually reach the 50^{text{th}} day by subtracting multiples of 7. Specifically, we subtract 28 weeks: 250 - 28 times 7 = 50. Thus, the 50^{text{th}} day of year P is a Sunday. 2. **Checking for the leap year effect on P+1:** Since the 150^{text{th}} day of year P+1 (a leap year) is also a Sunday, we calculate 150 - 7n where n is the number of weeks. The calculation directly aligns at 21 weeks earlier: 150 - 21 times 7 = 3. However, counting from the beginning of the year, we calculate each step modulo 7, so 150 mod 7 = 3, which indicates that the first day of P+1 is 3 days after Sunday, i.e., Wednesday. 3. **Calculating 50^{text{th}} day of year P-1:** Since the first day of P+1 is 2 days (the normal increment without a leap year plus one due to leap year) after the last day of P (which is Sunday), the first day of P is a Sunday. Then the first day of P-1 (not a leap year) would be one day prior i.e., Saturday. Hence, the 50^{text{th}} day of year P-1 is 7 weeks or 49 days after the first day plus one day leads to: Saturday + 1 day = Sunday. # Conclusion: The 50^{text{th}} day of year P-1 occurred on a text{Sunday}. The final answer is boxed{text{C: Sunday}}

answer:If Mary wants to double the original recipe, she needs to double the amount of each ingredient. For flour, the original recipe calls for 2.5 cups, so doubled would be 2.5 cups x 2 = 5 cups. Mary has already put in 4 cups of flour, so she doesn't need to add any more flour because she has put in less than required. For sugar, the original recipe calls for 3.75 cups, so doubled would be 3.75 cups x 2 = 7.5 cups. Mary has already put in 4.5 cups of sugar. To find out how much more sugar she needs to add, we subtract the amount she has already put in from the total amount she needs: 7.5 cups (needed) - 4.5 cups (already added) = 3 cups Mary needs to add boxed{3} more cups of sugar to the recipe.