💡Alright, so I've got this problem here: "The sum of three natural numbers, which are perfect squares, is divisible by 9. Prove that two of these numbers can be chosen such that their difference is also divisible by 9." Hmm, okay. Let's try to unpack this step by step.First off, I know that perfect squares modulo 9 can only have certain remainders. Maybe I should start by figuring out what those possible remainders are. Let me list out the squares of numbers from 0 to 8 modulo 9 because any number modulo 9 will be equivalent to one of these.So, calculating each square:- (0^2 = 0), which is 0 mod 9.- (1^2 = 1), which is 1 mod 9.- (2^2 = 4), which is 4 mod 9.- (3^2 = 9), which is 0 mod 9.- (4^2 = 16), which is 7 mod 9 (since 16 - 9 = 7).- (5^2 = 25), which is 7 mod 9 (25 - 18 = 7).- (6^2 = 36), which is 0 mod 9.- (7^2 = 49), which is 4 mod 9 (49 - 45 = 4).- (8^2 = 64), which is 1 mod 9 (64 - 63 = 1).So, the possible remainders when perfect squares are divided by 9 are 0, 1, 4, and 7. That's useful information.Now, the problem states that the sum of three perfect squares is divisible by 9. Let's denote these three perfect squares as (a^2), (b^2), and (c^2). So, (a^2 + b^2 + c^2 equiv 0 pmod{9}).Our goal is to show that among these three squares, at least two of them have a difference divisible by 9. In other words, there exist two squares, say (a^2) and (b^2), such that (a^2 - b^2 equiv 0 pmod{9}), which implies (a^2 equiv b^2 pmod{9}).So, if we can show that at least two of the three squares have the same remainder modulo 9, then their difference will be divisible by 9. That makes sense.Let me think about the possible combinations of remainders for the three squares. Since each square can only be 0, 1, 4, or 7 modulo 9, we need to find all possible triplets of these remainders that add up to 0 modulo 9.Let me list all possible triplets and see which ones sum to 0 modulo 9:1. (0, 0, 0): 0 + 0 + 0 = 0 mod 9. This works.2. (0, 1, 8): Wait, 8 isn't a possible remainder for a square. So, scratch that.3. (0, 4, 5): 5 isn't a possible remainder either. Hmm.4. (0, 7, 2): 2 isn't a possible remainder. Nope.5. (1, 1, 7): 1 + 1 + 7 = 9, which is 0 mod 9. This works.6. (1, 4, 4): 1 + 4 + 4 = 9, which is 0 mod 9. This works.7. (1, 7, 1): Same as (1, 1, 7). Works.8. (4, 4, 1): Same as (1, 4, 4). Works.9. (4, 7, 7): 4 + 7 + 7 = 18, which is 0 mod 9. This works.10. (7, 7, 2): 2 isn't a possible remainder. Doesn't work.Wait, so the valid triplets that sum to 0 mod 9 are:- (0, 0, 0)- (1, 1, 7)- (1, 4, 4)- (4, 7, 7)Looking at these, in each case, at least two of the numbers have the same remainder modulo 9. For example:- In (0, 0, 0), all three are 0.- In (1, 1, 7), two are 1.- In (1, 4, 4), two are 4.- In (4, 7, 7), two are 7.So, in every possible case where the sum of three perfect squares is divisible by 9, at least two of those squares must have the same remainder modulo 9. Therefore, their difference will be divisible by 9.Wait, but what if all three squares have different remainders? For example, could we have (0, 1, 8)? But 8 isn't a possible remainder for a square, so that's not possible. Similarly, other combinations with remainders not in {0, 1, 4, 7} aren't valid. So, actually, there are no valid triplets with all three different remainders that sum to 0 mod 9.Therefore, it's impossible for all three squares to have distinct remainders modulo 9 if their sum is divisible by 9. Hence, at least two must share the same remainder, and their difference will be divisible by 9.I think that covers it. So, the conclusion is that among the three perfect squares, two must have the same remainder modulo 9, making their difference divisible by 9.

💡First, I need to determine the total number of sweets taken by the 48 children. Since each child took 4 sweets, the total sweets taken is:[48 times 4 = 192]Next, the problem states that after the children took their sweets, there was still a third of the original amount left. This means that the 192 sweets represent (2/3) of the original amount of sweets, because (1 - 1/3 = 2/3). So, we have:[2/3 times text{original amount} = 192]To find the original amount, I solve for it by multiplying both sides of the equation by the reciprocal of (2/3), which is (3/2):[text{original amount} = 192 times frac{3}{2}]Calculating the right side of the equation gives me:[192 times frac{3}{2} = 288]Therefore, the original number of sweets in the pack was boxed{288}.

💡To determine the number of different paths from ( A ) to ( C ), we'll analyze the possible routes step by step.1. **Paths through ( B ) and ( D ):** - From ( A ) to ( B ), there are 2 paths. - From ( B ) to ( D ), there are 2 paths. - From ( D ) to ( C ), there are 2 paths. - Total paths through ( B ) and ( D ): ( 2 times 2 times 2 = 8 )2. **Direct paths from ( A ) to ( D ) and then to ( C ):** - From ( A ) to ( D ), there is 1 direct path. - From ( D ) to ( C ), there are 2 paths. - Total paths through ( A ) to ( D ) to ( C ): ( 1 times 2 = 2 )3. **Direct path from ( A ) to ( C ):** - There is 1 direct path from ( A ) to ( C ).4. **Total number of paths:** - Summing up all the paths: ( 8 + 2 + 1 = 11 )

💡First, I need to understand what it means to find the opposite of a number. The opposite of a number is simply the number with its sign changed. If the number is positive, its opposite is negative, and if it's negative, its opposite is positive.Looking at the given number, (-frac{1}{2023}), I see that it is a negative fraction. To find its opposite, I will change the negative sign to a positive sign.So, changing the sign of (-frac{1}{2023}) gives me (frac{1}{2023}).Now, I will compare this result with the provided options:A: (2023) B: (frac{1}{2023}) C: (-2023) D: (-frac{1}{2023})The opposite of (-frac{1}{2023}) is (frac{1}{2023}), which matches option B.

💡Alright, so I have this problem about a Senate committee with 5 Democrats and 5 Republicans, and I need to figure out how many ways they can sit around a circular table without any restrictions. The problem also mentions that two seatings are considered the same if one is a rotation of the other. Hmm, okay, let's break this down step by step.First, I remember that when arranging people around a circular table, the number of arrangements is different from arranging them in a straight line because rotations of the same arrangement are considered identical. For example, if everyone moves one seat to the right, it's still the same arrangement in terms of who is sitting next to whom. So, for a circular arrangement, we usually fix one person's position to eliminate the effect of rotations.In this case, there are 10 people in total (5 Democrats and 5 Republicans). If I were arranging them in a straight line, the number of ways would simply be 10 factorial, which is 10! That's because each of the 10 positions can be filled by any of the remaining people. But since it's a circular table, we have to adjust for the rotational symmetry.To account for rotations, we fix one person's position. This way, we're essentially converting the circular arrangement into a linear one relative to the fixed position. So, if we fix one person, say, the first Democrat, then we only need to arrange the remaining 9 people around the table. That would be 9! ways.Wait, but does the problem specify any restrictions based on party? It says "without restrictions," so I don't need to worry about alternating parties or anything like that. That simplifies things because I don't have to consider separate cases for Democrats and Republicans.So, to recap: for circular arrangements without restrictions, fix one person's position to eliminate rotational duplicates, and then arrange the remaining people linearly. Therefore, the number of unique seating arrangements should be 9!.Let me double-check that. If I have n people around a circular table, the number of distinct arrangements is (n-1)! because fixing one person's position accounts for the n rotations that would otherwise be considered identical. In this case, n is 10, so it's (10-1)! = 9!.Calculating 9! gives me 362,880. That seems right. I don't think I need to consider any additional factors because the problem doesn't mention any restrictions or special conditions beyond the circular arrangement and party distinctions not mattering for the count.Just to be thorough, let's think about it another way. If I didn't fix a position, the total number of arrangements would be 10!, which is 3,628,800. But since rotations are considered the same, each unique arrangement is counted 10 times (once for each possible rotation). So, dividing by 10 gives me 3,628,800 / 10 = 362,880, which matches the earlier result.Okay, I feel confident that 9! is the correct answer here. There's no need to consider permutations within the Democrats or Republicans separately because the problem doesn't impose any restrictions based on party alignment. It's purely about arranging 10 distinguishable individuals around a circular table, accounting for rotational symmetry.So, in conclusion, the number of ways the Senate committee can sit around the circular table without restrictions is 362,880.

💡First, I recognize that the inequality ( x^2 - 2x + 1 leq 0 ) can be rewritten by completing the square. This gives me ( (x - 1)^2 leq 0 ).Next, I recall that a squared term is always non-negative, meaning ( (x - 1)^2 ) is greater than or equal to zero for all real numbers ( x ).Since the inequality requires ( (x - 1)^2 ) to be less than or equal to zero, the only solution is when ( (x - 1)^2 = 0 ).Solving ( (x - 1)^2 = 0 ) gives ( x = 1 ).Therefore, the solution set of the inequality is the set containing only the number 1.