💡First, determine the total labeled weight of the three bags by multiplying the labeled weight of one bag by the number of bags:3 bags × 200 grams per bag = 600 grams.Next, account for the deviations from the labeled weight for each bag:- The first bag has an excess of +10 grams.- The second bag has a deficiency of -16 grams.- The third bag has a deficiency of -11 grams.Add these deviations to the total labeled weight:600 grams + 10 grams - 16 grams - 11 grams = 583 grams.Therefore, the actual total weight of the three bags of food is 583 grams.

💡First, calculate the total number of toothpicks in a complete grid without any missing blocks. For a grid that is 30 toothpicks long and 20 toothpicks wide, the number of vertical toothpicks is (30 + 1) multiplied by 20, and the number of horizontal toothpicks is (20 + 1) multiplied by 30.Next, account for the missing 2x2 block of toothpicks. This missing block affects both the vertical and horizontal toothpicks. Specifically, two vertical lines and two horizontal lines each lose two toothpicks.Finally, subtract the missing toothpicks from the total count to determine the final number of toothpicks used.

💡嗯，我现在要解决一个三位数的问题，这个数有一些特定的条件。让我仔细读一下题目：“有一个三位数是2的倍数，加1是3的倍数，加2是4的倍数，加3是5的倍数，加4是6的倍数。那么这个数的最小值是多少。”好，首先，我需要找到一个三位数x，满足以下条件：1. x是2的倍数，也就是说x ≡ 0 mod 2。2. x + 1是3的倍数，也就是x ≡ 2 mod 3。3. x + 2是4的倍数，也就是x ≡ 2 mod 4。4. x + 3是5的倍数，也就是x ≡ 2 mod 5。5. x + 4是6的倍数，也就是x ≡ 2 mod 6。看起来所有的条件都指向x ≡ 2 mod某个数。让我把这些条件列出来：- x ≡ 0 mod 2- x ≡ 2 mod 3- x ≡ 2 mod 4- x ≡ 2 mod 5- x ≡ 2 mod 6嗯，这里有一个模式，除了第一个条件，其他都是x ≡ 2 mod某个数。那么，或许我可以把这些条件结合起来，找到一个共同的模数。首先，我注意到x ≡ 2 mod 3，x ≡ 2 mod 4，x ≡ 2 mod 5，x ≡ 2 mod 6。这意味着x - 2同时是3、4、5、6的倍数。也就是说，x - 2是这些数的公倍数。那么，我需要找到x - 2的最小公倍数，然后加上2，得到x的值。因为x是一个三位数，所以我需要找到最小的x满足x >= 100。首先，计算3、4、5、6的最小公倍数（LCM）。让我一步一步来计算：1. 先计算3和4的LCM。3和4互质，所以LCM(3,4)=12。2. 接下来，计算12和5的LCM。12和5互质，所以LCM(12,5)=60。3. 然后，计算60和6的LCM。60已经是6的倍数了，所以LCM(60,6)=60。所以，3、4、5、6的最小公倍数是60。这意味着x - 2必须是60的倍数，即x - 2 = 60k，其中k是一个正整数。因此，x = 60k + 2。现在，我需要找到最小的三位数x，也就是x >= 100。那么，我需要找到最小的k，使得60k + 2 >= 100。解这个不等式：60k + 2 >= 10060k >= 98k >= 98 / 60 ≈ 1.633...因为k必须是整数，所以k >= 2。当k=2时，x = 60*2 + 2 = 122。现在，我需要验证一下122是否满足所有的条件：1. 122是2的倍数吗？是的，因为122 ÷ 2 = 61，没有余数。2. 122 + 1 = 123，是3的倍数吗？123 ÷ 3 = 41，没有余数。3. 122 + 2 = 124，是4的倍数吗？124 ÷ 4 = 31，没有余数。4. 122 + 3 = 125，是5的倍数吗？125 ÷ 5 = 25，没有余数。5. 122 + 4 = 126，是6的倍数吗？126 ÷ 6 = 21，没有余数。看起来122满足所有的条件，而且它是最小的三位数，因为当k=1时，x=62，不是三位数。所以，k=2时，x=122，是符合条件的最小三位数。不过，为了确保我没有遗漏什么，我再检查一下是否还有更小的三位数满足条件。比如，k=2时x=122，k=3时x=182，k=4时x=242，等等。显然，122是最小的。总结一下，通过分析各个条件，我发现x - 2必须是3、4、5、6的公倍数，也就是60的倍数，因此x=60k + 2。找到最小的k使得x是三位数，得到k=2，x=122。验证后，122确实满足所有条件，因此答案是122。

💡Okay, so I have this problem about marbles in a bag. Let me read it again carefully:"In a bag of marbles, 3/7 of the marbles are blue, and the rest are red. If the number of red marbles is tripled and the number of blue marbles is halved, what fraction of the marbles will be red? The options are A) 3/7, B) 5/7, C) 11/14, D) 9/13, E) 8/9."Alright, let's break this down step by step. First, I need to figure out the initial number of blue and red marbles. The problem says that 3/7 of the marbles are blue, so the rest must be red. Since the total fraction must add up to 1, the fraction of red marbles is 1 - 3/7. Let me calculate that:1 - 3/7 = 4/7.So, initially, 3/7 are blue and 4/7 are red.Now, the problem says that the number of red marbles is tripled, and the number of blue marbles is halved. I need to find the new fraction of red marbles after these changes.To make this concrete, maybe I should assume a specific number of marbles to work with. Let's pick a number that's a multiple of 7 to make the fractions easy. Let's say there are 7 marbles in total. Then, according to the fractions:- Blue marbles: 3/7 of 7 = 3 marbles.- Red marbles: 4/7 of 7 = 4 marbles.Okay, so initially, there are 3 blue marbles and 4 red marbles.Now, the number of red marbles is tripled. So, tripling 4 marbles would be:4 * 3 = 12 red marbles.And the number of blue marbles is halved. So, halving 3 marbles would be:3 / 2 = 1.5 marbles.Wait, that's a fraction. Hmm, marbles are whole objects, so having half a marble doesn't make sense. Maybe I should choose a different total number of marbles to avoid fractions. Let me try a multiple of 14, since 7 * 2 = 14, and that might help with the halving.Let's say there are 14 marbles in total. Then:- Blue marbles: 3/7 of 14 = (3 * 14) / 7 = 6 marbles.- Red marbles: 4/7 of 14 = (4 * 14) / 7 = 8 marbles.Okay, that's better. So, 6 blue marbles and 8 red marbles.Now, tripling the red marbles: 8 * 3 = 24 red marbles.Halving the blue marbles: 6 / 2 = 3 blue marbles.So now, the total number of marbles is 3 blue + 24 red = 27 marbles.The fraction of red marbles is 24/27. Let me simplify that:24 ÷ 3 = 827 ÷ 3 = 9So, 24/27 simplifies to 8/9.Looking back at the options, E is 8/9. So, that must be the answer.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Assumed 14 marbles total because it's a multiple of 7 and 2, which would help with halving and tripling without fractions.2. Calculated blue marbles as 6 and red marbles as 8.3. Tripled red marbles: 8 * 3 = 24.4. Halved blue marbles: 6 / 2 = 3.5. Total marbles after change: 3 + 24 = 27.6. Fraction of red marbles: 24/27 = 8/9.Everything seems to check out. So, I think E) 8/9 is the correct answer.**Final Answer**boxed{E}

💡Okay, so I have this problem where I need to find the integers ( n ) such that ( 2^6 + 2^9 + 2^n ) is a perfect square. Hmm, let me break this down step by step.First, I know that ( 2^6 ) is 64 and ( 2^9 ) is 512. So, adding those together, I get ( 64 + 512 = 576 ). So the expression simplifies to ( 576 + 2^n ). Now, I need this sum to be a perfect square.Let me denote ( k^2 = 576 + 2^n ), where ( k ) is some integer. So, ( 2^n = k^2 - 576 ). That means ( k^2 - 2^n = 576 ). Hmm, I can factor this as ( (k - 2^{n/2})(k + 2^{n/2}) = 576 ) if ( n ) is even. But wait, ( 2^n ) is a power of 2, so ( k^2 - 576 ) must also be a power of 2.Let me consider that ( 576 ) is ( 24^2 ). So, ( k^2 - 24^2 = 2^n ). This can be written as ( (k - 24)(k + 24) = 2^n ). Now, since ( 2^n ) is a power of 2, both ( k - 24 ) and ( k + 24 ) must also be powers of 2. Let me denote ( k - 24 = 2^a ) and ( k + 24 = 2^b ), where ( a < b ) because ( k - 24 < k + 24 ).Subtracting these two equations gives ( (k + 24) - (k - 24) = 2^b - 2^a = 48 ). So, ( 2^a (2^{b - a} - 1) = 48 ). Now, 48 factors into ( 16 times 3 ), which is ( 2^4 times 3 ). Since ( 2^{b - a} - 1 ) must be an integer, and it's multiplied by a power of 2, the only way this works is if ( 2^{b - a} - 1 = 3 ) and ( 2^a = 16 ).So, ( 2^{b - a} = 4 ) which means ( b - a = 2 ). And ( 2^a = 16 ) implies ( a = 4 ). Therefore, ( b = a + 2 = 6 ). Now, plugging back into ( k - 24 = 2^4 = 16 ), so ( k = 40 ). Then, ( k + 24 = 64 = 2^6 ), which checks out.So, ( 2^n = (k - 24)(k + 24) = 16 times 64 = 1024 ). Therefore, ( n = 10 ) because ( 2^{10} = 1024 ).Wait, but I should check if there are any other possible values of ( a ) and ( b ) that could satisfy ( 2^a (2^{b - a} - 1) = 48 ). Let me see, 48 can also be factored as ( 8 times 6 ), ( 4 times 12 ), ( 2 times 24 ), etc. But since ( 2^{b - a} - 1 ) must be an odd number (because it's one less than a power of 2), the only valid factorization is where the power of 2 is 16 and the remaining factor is 3. So, I think ( n = 10 ) is the only solution.Let me verify by plugging ( n = 10 ) back into the original expression: ( 2^6 + 2^9 + 2^{10} = 64 + 512 + 1024 = 1600 ). And ( 1600 ) is indeed ( 40^2 ), so it works.I don't think there are any other values of ( n ) that would satisfy this condition because any other factorization would not result in both ( k - 24 ) and ( k + 24 ) being powers of 2. So, I'm pretty confident that ( n = 10 ) is the only solution.

💡Okay, so I'm trying to figure out whether a second round of elections will be necessary in the 1981 presidential elections based on the information given. Let me break down the problem step by step.First, there are several candidates participating in the first round. Each candidate receives exactly half the number of votes as the candidate immediately preceding them. That means if the first candidate gets 'x' votes, the second one gets 'x/2', the third gets 'x/4', and so on. This seems like a geometric sequence where each term is half of the previous one.I need to determine if the first candidate will have enough votes to avoid a second round. In many election systems, a second round is needed if no candidate gets an outright majority. So, I guess I need to find out if the first candidate has more than half of the total votes cast.Let's denote the number of candidates as 'n'. The total number of votes can be calculated by summing up the votes for all candidates. Since each subsequent candidate gets half the votes of the previous one, the total votes would be:Total votes = x + x/2 + x/4 + ... + x/(2^{n-1})This is a geometric series where the first term is 'x' and the common ratio is '1/2'. The sum of a geometric series is given by:Sum = a * (1 - r^n) / (1 - r)Where 'a' is the first term, 'r' is the common ratio, and 'n' is the number of terms.Plugging in the values:Sum = x * (1 - (1/2)^n) / (1 - 1/2) = x * (1 - (1/2)^n) / (1/2) = 2x * (1 - (1/2)^n)So, the total votes are 2x * (1 - (1/2)^n). Now, the first candidate has 'x' votes. To find out if they have a majority, we can compare 'x' to half of the total votes.Half of the total votes = (2x * (1 - (1/2)^n)) / 2 = x * (1 - (1/2)^n)So, we need to check if x > x * (1 - (1/2)^n)Simplifying this inequality:x > x * (1 - (1/2)^n)Divide both sides by x (assuming x > 0):1 > 1 - (1/2)^nSubtract 1 from both sides:0 > - (1/2)^nMultiply both sides by -1 (which reverses the inequality):0 < (1/2)^nSince (1/2)^n is always positive for any positive integer 'n', this inequality holds true. Therefore, the first candidate always has more votes than half of the total votes, meaning they have a majority and no second round is necessary.Wait, but does this make sense? If each candidate gets half the votes of the previous one, the number of votes decreases exponentially. So, the first candidate has significantly more votes than the others combined. Let me test this with a small number of candidates to see if it holds.Let's say there are 3 candidates:- Candidate 1: x votes- Candidate 2: x/2 votes- Candidate 3: x/4 votesTotal votes = x + x/2 + x/4 = (4x + 2x + x)/4 = 7x/4Half of total votes = 7x/8First candidate has x votes, which is more than 7x/8? Wait, x is 8x/8, which is more than 7x/8. So, yes, the first candidate has more than half.What if there are 4 candidates:- Candidate 1: x- Candidate 2: x/2- Candidate 3: x/4- Candidate 4: x/8Total votes = x + x/2 + x/4 + x/8 = (8x + 4x + 2x + x)/8 = 15x/8Half of total votes = 15x/16First candidate has x, which is 16x/16, more than 15x/16. Again, the first candidate has a majority.This seems consistent. So, regardless of the number of candidates, as long as each subsequent candidate gets half the votes of the previous one, the first candidate will always have more than half of the total votes, thus avoiding a second round.But wait, what if there are an infinite number of candidates? Then, the total votes would approach 2x, and half of that is x. So, in that case, the first candidate would have exactly x votes, which is equal to half of the total votes. But since we can't have an infinite number of candidates in reality, this scenario doesn't apply.Therefore, in any practical scenario with a finite number of candidates, the first candidate will always have more than half of the total votes, making a second round unnecessary.