💡Okay, so I have this problem about a cuboid that's painted with stripes on each of its six faces. The cuboid has dimensions where one pair of edges is twice as long as the others. Each face has a stripe that runs from the center of one edge to the center of the opposite edge, and the orientation of each stripe (either horizontal or vertical) is chosen at random. I need to find the probability that there's at least one continuous stripe that encircles the cuboid horizontally.First, let me try to visualize the cuboid. It's like a rectangular box, but one pair of edges is twice as long as the others. So, if I consider the standard cube, which has all edges equal, this cuboid has one dimension doubled. Let's say the length is twice the width and height for simplicity.Now, each face of the cuboid has a stripe running from the center of one edge to the center of the opposite edge. So, on each face, there are two possible orientations for the stripe: either horizontal or vertical. Since the cuboid is not a cube, the orientation might matter differently depending on which face we're talking about.The problem states that each face's stripe orientation is chosen at random. So, for each face, there's a 50% chance of having a horizontal stripe and a 50% chance of having a vertical stripe. Since there are six faces, the total number of possible stripe combinations is 2^6, which is 64. So, the total number of possible outcomes is 64.Now, I need to find the number of favorable outcomes where there's at least one continuous stripe that encircles the cuboid horizontally. Let me think about what it means for a stripe to encircle the cuboid horizontally. It means that the stripe goes around the cuboid in such a way that it forms a closed loop without breaking.Given that the cuboid has one pair of edges twice as long, the continuous stripe would have to go around the longer edges. So, if I imagine looking at the cuboid from the top, the stripe would have to go around the longer sides. But since the cuboid is three-dimensional, the stripe would have to connect on adjacent faces as well.Let me try to break it down. For a continuous stripe to encircle the cuboid horizontally, the stripes on the adjacent faces must align in a way that allows the stripe to continue from one face to the next. So, if I have a stripe on the front face, it needs to connect to a stripe on the right face, which in turn connects to the stripe on the back face, and so on, until it comes back to the starting point.But since each stripe's orientation is chosen randomly, I need to figure out how likely it is that such a continuous path exists. Maybe it's easier to think about the opposite: what's the probability that there is no continuous stripe encircling the cuboid horizontally? Then, subtract that probability from 1 to get the desired probability.However, calculating the probability of no continuous stripe might be complicated because there are multiple ways the stripes could fail to form a loop. Instead, maybe I should directly calculate the number of favorable outcomes where a continuous stripe exists.Let me consider the structure of the cuboid. It has six faces: front, back, left, right, top, and bottom. The front and back faces are the ones with the longer edges, assuming the length is twice the width and height. So, the front and back faces are longer in one dimension, while the left, right, top, and bottom faces are squares or rectangles with the shorter edges.For a stripe to encircle the cuboid horizontally, it needs to go around the longer edges. So, the stripes on the front, right, back, and left faces must align in such a way that they form a continuous loop. The top and bottom faces might not be directly involved in this loop, but their stripe orientations could affect whether the loop is closed or not.Wait, actually, if the loop is horizontal, it would go around the middle of the cuboid, so it might involve the top and bottom faces as well. Hmm, I'm getting a bit confused. Maybe I should draw a diagram or think about the cuboid in terms of its edges and how the stripes connect.Each stripe on a face connects the centers of two opposite edges. So, on the front face, a horizontal stripe would connect the centers of the left and right edges, while a vertical stripe would connect the centers of the top and bottom edges. Similarly, on the right face, a horizontal stripe would connect the centers of the front and back edges, and a vertical stripe would connect the centers of the top and bottom edges.For a continuous stripe to encircle the cuboid horizontally, the stripes on adjacent faces need to connect. So, starting from the front face, if the stripe is horizontal, it connects to the right face's stripe. If the right face's stripe is vertical, it would connect to the top or bottom face, but that might not help in forming a horizontal loop. Alternatively, if the right face's stripe is horizontal, it would connect to the back face, and so on.This seems complicated. Maybe I should think about the possible paths a horizontal stripe can take. Since the cuboid is symmetric, the probability should be the same regardless of which face we start from. So, maybe I can fix a starting face and calculate the probability from there.Let's fix the front face. If the front face has a horizontal stripe, it connects to the right face. For the stripe to continue, the right face must also have a horizontal stripe. Then, the stripe would go to the back face, which also needs a horizontal stripe, and then to the left face, which also needs a horizontal stripe. Finally, the stripe would return to the front face, completing the loop.So, in this case, all four side faces (front, right, back, left) must have horizontal stripes. The top and bottom faces can have any orientation since they're not involved in this loop. So, the number of favorable outcomes for this specific loop is 2^2 = 4 (since the top and bottom can each independently be horizontal or vertical).But wait, is this the only way a horizontal loop can form? Or are there other possible loops that involve different faces?Actually, if the front face has a vertical stripe, it might connect to the top or bottom face, which could then connect to another face, potentially forming a different loop. But since we're specifically looking for a horizontal loop, maybe only the side faces are involved.Hmm, I'm not sure. Maybe I need to consider all possible loops that can encircle the cuboid horizontally. Each loop would involve four faces: front, right, back, and left. For each of these loops, the four side faces must have horizontal stripes, and the top and bottom can be anything.But there are two possible horizontal loops: one going around the front, right, back, and left faces, and another going around the front, left, back, and right faces. Wait, no, that's the same loop in the opposite direction.Alternatively, maybe there are two distinct loops depending on the orientation of the stripes. If the front face has a horizontal stripe going from left to right, and the right face has a horizontal stripe going from front to back, that forms one loop. If the front face has a horizontal stripe going from right to left, and the right face has a horizontal stripe going from back to front, that forms another loop.But actually, these are just the same loop traversed in opposite directions. So, maybe there's only one unique loop that can encircle the cuboid horizontally.Wait, no. Actually, depending on the orientation of the stripes, the loop can be formed in two different ways: clockwise or counterclockwise. But since the cuboid is symmetric, these two cases are essentially the same in terms of probability.So, perhaps the number of favorable outcomes is just 4, as I initially thought: all four side faces have horizontal stripes, and the top and bottom can be anything. So, 2^2 = 4.But wait, is that the only way? What if the top and bottom faces also have horizontal stripes? Would that interfere with the loop? Or would it just create another loop?Actually, if the top and bottom faces have horizontal stripes, they could potentially form their own loops, but those would be vertical loops, not horizontal. So, for a horizontal loop, only the side faces need to have horizontal stripes.Therefore, the number of favorable outcomes is 4, as calculated earlier. So, the probability is 4/64 = 1/16.But wait, let me double-check. If all four side faces have horizontal stripes, that's 1 way for the front, 1 way for the right, 1 way for the back, and 1 way for the left. So, 1*1*1*1 = 1. Then, the top and bottom can each independently be horizontal or vertical, so 2*2 = 4. So, total favorable outcomes are 4.Yes, that seems correct. So, the probability is 4/64 = 1/16.But I'm still a bit unsure because I might be missing other possible loops. Maybe there are other configurations where the stripes on the top and bottom faces also contribute to a horizontal loop. Let me think about that.If the top face has a horizontal stripe, it connects the centers of its left and right edges. Similarly, the bottom face has a horizontal stripe connecting its left and right edges. If the front and back faces have vertical stripes, they connect the centers of their top and bottom edges. So, in this case, the stripes on the top, front, bottom, and back faces could form a loop.Wait, that's a vertical loop, not a horizontal one. So, that's a different loop, but we're only interested in horizontal loops.So, in this case, the horizontal loop is formed by the side faces, and the vertical loop is formed by the top, front, bottom, and back faces. So, these are two different loops, but we're only concerned with the horizontal one.Therefore, the number of favorable outcomes for a horizontal loop is indeed 4, as previously calculated.So, the probability is 4/64 = 1/16.But wait, the answer choices are 1/64, 1/32, 1/16, 1/8, and 1/4. So, 1/16 is one of the options, specifically option C.But I'm still a bit confused because I might be undercounting the number of favorable outcomes. Let me think again.Each face has two possible orientations, so 2^6 = 64 total possibilities.For a horizontal loop to exist, the four side faces (front, right, back, left) must all have horizontal stripes. The top and bottom can be anything.So, the number of favorable outcomes is 1 (for front) * 1 (for right) * 1 (for back) * 1 (for left) * 2 (for top) * 2 (for bottom) = 4.Therefore, the probability is 4/64 = 1/16.Yes, that seems correct. So, the answer should be 1/16, which is option C.

💡First, I need to understand the relationships between the number of men, women, and boys based on their earnings.I'll denote the number of men as M, women as W, and boys as B.Given that all groups earn Rs. 120 in total, I can set up the following equations:1. For men: M * 8 = 1202. For women: W * (120/W) = 1203. For boys: B * (120/B) = 120From the first equation, I can solve for M:M = 120 / 8M = 15Since the number of men is equal to the number of women, W = 15.Finally, since the number of women is equal to the number of boys, B = W = 15.Therefore, there are 15 boys.

💡First, I recognize that Fred currently has 18 baseball cards after Keith bought 22 of them.To find out how many cards Fred had initially, I need to determine the total number of cards before Keith's purchase.Since Keith bought 22 cards, I can add this number to the current number of cards Fred has to find the initial amount.Therefore, adding 22 to 18 gives me the initial number of baseball cards Fred had.

💡Alright, so I have this problem here where I need to multiply two binary numbers, 11011₂ and 1101₂, and then subtract another binary number, 1010₂, from the result. Finally, I have to express the answer in base 2. Hmm, okay, let's break this down step by step.First, I need to remember how to multiply binary numbers. I think it's similar to decimal multiplication, but since it's binary, the digits are only 0 and 1. So, when multiplying, each bit of the second number will multiply the entire first number, shifted appropriately, and then we'll add up the results. Let me try to recall the exact process.So, the two numbers are 11011₂ and 1101₂. Let me write them down:``` 1 1 0 1 1 × 1 1 0 1```I think I need to multiply each bit of the second number (1101) with the first number (11011) starting from the rightmost bit, and then shift left as I move to the next bit. Then, add all the partial products together.Let me start with the rightmost bit of the second number, which is 1. Multiplying 11011 by 1 gives me 11011. Since it's the first bit, I don't need to shift it.Next, the second bit from the right is 0. Multiplying 11011 by 0 gives me 00000. But since this is the second bit, I need to shift it one position to the left. So, it becomes 000000.The third bit from the right is 1. Multiplying 11011 by 1 again gives 11011. This time, since it's the third bit, I need to shift it two positions to the left. So, it becomes 1101100.Finally, the leftmost bit is 1. Multiplying 11011 by 1 gives 11011. This is the fourth bit, so I need to shift it three positions to the left, making it 11011000.Now, I have all the partial products:``` 1 1 0 1 1 + 0 0 0 0 0 0 +1 1 0 1 1 0 0 +1 1 0 1 1 0 0 0```Wait, I think I might have made a mistake here. Let me check. The second partial product was 00000 shifted one position, which should be 000000, but when adding, I need to align them properly. Maybe I should write them one under the other with proper shifting.Let me try writing them aligned:``` 1 1 0 1 1 + 0 0 0 0 0 0 +1 1 0 1 1 0 0 +1 1 0 1 1 0 0 0```But this doesn't look right because the partial products should be shifted left by one each time. Let me adjust that.Actually, the first partial product is 11011 (no shift), the second is 00000 shifted left by one (so 000000), the third is 11011 shifted left by two (1101100), and the fourth is 11011 shifted left by three (11011000). So, when adding, they should look like this:``` 1 1 0 1 1 + 0 0 0 0 0 0 +1 1 0 1 1 0 0 +1 1 0 1 1 0 0 0```Now, adding them up. Let's start from the rightmost bit.Rightmost bit: 1 (from first partial product) + 0 (second) + 0 (third) + 0 (fourth) = 1.Next bit: 0 (first) + 0 (second) + 0 (third) + 0 (fourth) = 0.Next: 1 (first) + 0 (second) + 1 (third) + 0 (fourth) = 1 + 1 = 10 in binary, so write down 0 and carry over 1.Next: 0 (first) + 0 (second) + 1 (third) + 1 (fourth) + carryover 1 = 0 + 0 + 1 + 1 + 1 = 3, which is 11 in binary. So write down 1 and carry over 1.Next: 1 (first) + 0 (second) + 0 (third) + 1 (fourth) + carryover 1 = 1 + 0 + 0 + 1 + 1 = 3, again 11. Write 1, carry over 1.Next: 1 (first) + 0 (second) + 1 (third) + 1 (fourth) + carryover 1 = 1 + 0 + 1 + 1 + 1 = 4, which is 100 in binary. So write down 0, carry over 10.Wait, that seems complicated. Maybe I should write it out step by step.Let me write the partial products with proper alignment:``` 1 1 0 1 1 + 0 0 0 0 0 0 +1 1 0 1 1 0 0 +1 1 0 1 1 0 0 0```Now, let's add them column by column from right to left.Starting from the right:Column 1: 1 (from first) + 0 (second) + 0 (third) + 0 (fourth) = 1.Column 2: 1 (first) + 0 (second) + 0 (third) + 0 (fourth) = 1.Column 3: 0 (first) + 0 (second) + 1 (third) + 0 (fourth) = 1.Column 4: 1 (first) + 0 (second) + 1 (third) + 0 (fourth) = 1 + 1 = 10. So write 0, carry over 1.Column 5: 1 (first) + 0 (second) + 0 (third) + 1 (fourth) + carryover 1 = 1 + 0 + 0 + 1 + 1 = 3, which is 11. So write 1, carry over 1.Column 6: 0 (first) + 0 (second) + 1 (third) + 1 (fourth) + carryover 1 = 0 + 0 + 1 + 1 + 1 = 3, which is 11. So write 1, carry over 1.Column 7: 1 (first) + 0 (second) + 1 (third) + 1 (fourth) + carryover 1 = 1 + 0 + 1 + 1 + 1 = 4, which is 100. So write 0, carry over 10.Wait, that's getting messy. Maybe I should use a different approach. Perhaps convert the binary numbers to decimal, multiply, subtract, and then convert back to binary.Let me try that.First, convert 11011₂ to decimal.11011₂ = 1×2⁴ + 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = 16 + 8 + 0 + 2 + 1 = 27.Next, convert 1101₂ to decimal.1101₂ = 1×2³ + 1×2² + 0×2¹ + 1×2⁰ = 8 + 4 + 0 + 1 = 13.Now, multiply 27 and 13.27 × 13 = 351.Next, convert 1010₂ to decimal.1010₂ = 1×2³ + 0×2² + 1×2¹ + 0×2⁰ = 8 + 0 + 2 + 0 = 10.Now, subtract 10 from 351.351 - 10 = 341.Finally, convert 341 back to binary.To convert 341 to binary, I'll divide by 2 and keep track of the remainders.341 ÷ 2 = 170 remainder 1170 ÷ 2 = 85 remainder 085 ÷ 2 = 42 remainder 142 ÷ 2 = 21 remainder 021 ÷ 2 = 10 remainder 110 ÷ 2 = 5 remainder 05 ÷ 2 = 2 remainder 12 ÷ 2 = 1 remainder 01 ÷ 2 = 0 remainder 1Reading the remainders from bottom to top: 1 0 1 0 1 0 1 0 1.Wait, that's 9 bits: 101010101₂.But let me double-check:128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255, which is more than 341. Wait, no, 256 is 2⁸, so 256 + 64 = 320, which is less than 341.Wait, maybe I made a mistake in the conversion.Let me try again.341 ÷ 2 = 170 r1170 ÷ 2 = 85 r085 ÷ 2 = 42 r142 ÷ 2 = 21 r021 ÷ 2 = 10 r110 ÷ 2 = 5 r05 ÷ 2 = 2 r12 ÷ 2 = 1 r01 ÷ 2 = 0 r1So the remainders are: 1,0,1,0,1,0,1,0,1.So the binary is 101010101₂.But let's verify:128 + 32 + 8 + 1 = 169, which is less than 341. Wait, that can't be right.Wait, perhaps I need to consider higher powers.Let me list the powers of 2:2⁰ = 12¹ = 22² = 42³ = 82⁴ = 162⁵ = 322⁶ = 642⁷ = 1282⁸ = 2562⁹ = 512So, 256 is the highest power less than 341.341 - 256 = 85.Now, 64 is the next highest power less than 85.85 - 64 = 21.Next, 16 is less than 21.21 - 16 = 5.Next, 4 is less than 5.5 - 4 = 1.Finally, 1 is 1.So, the binary representation is:256 (1) + 64 (1) + 16 (1) + 4 (1) + 1 (1) = 101010101₂.Wait, but 256 + 64 = 320, plus 16 is 336, plus 4 is 340, plus 1 is 341. Yes, that's correct.So, 341 in binary is 101010101₂.But earlier, when I tried to add the partial products directly in binary, I got 110011101₂, which is different. Hmm, that means I must have made a mistake in the binary multiplication.Let me try the binary multiplication again more carefully.Multiplying 11011₂ by 1101₂.Let's write it out step by step:``` 1 1 0 1 1 × 1 1 0 1```We'll multiply each bit of the second number (1101) with the first number (11011), shifting appropriately.Starting from the rightmost bit:1st bit (1): 11011 × 1 = 110112nd bit (0): 11011 × 0 = 00000, shifted left by 1: 0000003rd bit (1): 11011 × 1 = 11011, shifted left by 2: 11011004th bit (1): 11011 × 1 = 11011, shifted left by 3: 11011000Now, adding them up:``` 1 1 0 1 1 + 0 0 0 0 0 0 +1 1 0 1 1 0 0 +1 1 0 1 1 0 0 0```Let's add them column by column:Starting from the right:Column 1: 1Column 2: 1Column 3: 0 + 0 + 1 = 1Column 4: 1 + 0 + 1 = 10, write 0, carry 1Column 5: 1 + 0 + 0 + 1 + carry 1 = 3, write 1, carry 1Column 6: 0 + 0 + 1 + 1 + carry 1 = 3, write 1, carry 1Column 7: 1 + 0 + 1 + 1 + carry 1 = 4, write 0, carry 10Wait, that's not right. Let me write it out properly.Let me write the partial products with proper alignment:``` 1 1 0 1 1 + 0 0 0 0 0 0 +1 1 0 1 1 0 0 +1 1 0 1 1 0 0 0```Now, aligning them:``` 0 0 1 1 0 1 1 + 0 0 0 0 0 0 0 +0 1 1 0 1 1 0 0 +1 1 0 1 1 0 0 0```Wait, that's not correct. I think I need to pad the shorter partial products with zeros to match the length.Let me adjust:The first partial product is 11011, which is 5 bits.The second is 000000, which is 6 bits.The third is 1101100, which is 7 bits.The fourth is 11011000, which is 8 bits.So, to add them, I need to align them properly:``` 0 0 1 1 0 1 1 + 0 0 0 0 0 0 0 +0 1 1 0 1 1 0 0 +1 1 0 1 1 0 0 0```But this is getting confusing. Maybe I should write them all with the same number of bits by padding with zeros on the left.Let's make them all 8 bits:First partial product: 0011011 → 0011011 (7 bits, so add one zero: 00011011)Wait, no, 11011 is 5 bits. To make it 8 bits, add three zeros: 00011011.Second partial product: 000000 is 6 bits, add two zeros: 00000000.Third partial product: 1101100 is 7 bits, add one zero: 01101100.Fourth partial product: 11011000 is already 8 bits.Now, adding them:```00011011000000000110110011011000```Now, let's add them column by column from right to left.Column 1: 1 + 0 + 0 + 0 = 1Column 2: 1 + 0 + 0 + 0 = 1Column 3: 0 + 0 + 0 + 0 = 0Column 4: 1 + 0 + 1 + 0 = 1Column 5: 1 + 0 + 1 + 1 = 3, write 1, carry 1Column 6: 0 + 0 + 1 + 1 + carry 1 = 3, write 1, carry 1Column 7: 0 + 0 + 0 + 1 + carry 1 = 2, write 0, carry 1Column 8: 0 + 0 + 1 + 1 + carry 1 = 3, write 1, carry 1Since we're at the leftmost bit, we need to add the carryover 1, so it becomes 1.Putting it all together: 1 1 0 0 1 1 1 0 1.Wait, that's 9 bits: 110011101₂.But earlier, when I converted to decimal, I got 101010101₂, which is different. So there's a discrepancy here. That means I must have made a mistake in either the binary multiplication or the decimal conversion.Let me check the decimal conversion again.11011₂ is 27, 1101₂ is 13, 27 × 13 = 351, 351 - 10 = 341, which is 101010101₂.But the binary multiplication gave me 110011101₂, which is different.Wait, let me convert 110011101₂ to decimal to see what it is.110011101₂ = 1×2⁸ + 1×2⁷ + 0×2⁶ + 0×2⁵ + 1×2⁴ + 1×2³ + 1×2² + 0×2¹ + 1×2⁰= 256 + 128 + 0 + 0 + 16 + 8 + 4 + 0 + 1 = 256 + 128 = 384 + 16 = 400 + 8 = 408 + 4 = 412 + 1 = 413.Wait, that's way higher than 341. So clearly, I made a mistake in the binary multiplication.Let me try multiplying 11011₂ by 1101₂ again, but this time more carefully.First, write down the two numbers:``` 1 1 0 1 1 × 1 1 0 1```Multiply each bit of the second number with the first number, shifting appropriately.Starting from the rightmost bit of the second number:1st bit (1): 11011 × 1 = 110112nd bit (0): 11011 × 0 = 00000, shifted left by 1: 0000003rd bit (1): 11011 × 1 = 11011, shifted left by 2: 11011004th bit (1): 11011 × 1 = 11011, shifted left by 3: 11011000Now, add these partial products:``` 1 1 0 1 1 + 0 0 0 0 0 0 +1 1 0 1 1 0 0 +1 1 0 1 1 0 0 0```Let's add them step by step.First, add 11011 and 000000:``` 0 0 1 1 0 1 1+ 0 0 0 0 0 0 0= 0 0 1 1 0 1 1```Next, add 1101100:``` 0 0 1 1 0 1 1+0 1 1 0 1 1 0 0```Align them:``` 0 0 1 1 0 1 1+0 1 1 0 1 1 0 0```Add column by column:Starting from the right:1 + 0 = 11 + 0 = 10 + 1 = 11 + 1 = 10, write 0, carry 11 + 0 + carry 1 = 2, write 0, carry 10 + 1 + carry 1 = 2, write 0, carry 10 + 1 + carry 1 = 2, write 0, carry 10 + 0 + carry 1 = 1So, the result is 1 0 0 0 0 1 0 1.Wait, that doesn't seem right. Let me write it out properly.Wait, actually, when adding 0011011 (7 bits) and 01101100 (8 bits), we need to align them properly:``` 0 0 1 1 0 1 1+0 1 1 0 1 1 0 0```So, the first number is 7 bits, the second is 8 bits. To add them, we need to make them the same length by padding the shorter one with a zero on the left.So, 0011011 becomes 00011011 (8 bits), and 01101100 is already 8 bits.Now, adding:```0 0 0 1 1 0 1 1+0 1 1 0 1 1 0 0```Let's add column by column from right to left:Column 1: 1 + 0 = 1Column 2: 1 + 0 = 1Column 3: 0 + 1 = 1Column 4: 1 + 1 = 10, write 0, carry 1Column 5: 1 + 1 + carry 1 = 11, write 1, carry 1Column 6: 0 + 0 + carry 1 = 1Column 7: 0 + 1 = 1Column 8: 0 + 1 = 1So, the result is 1 1 0 0 1 1 1 1.Wait, that's 8 bits: 11001111₂.Now, add the last partial product: 11011000.So, we have:``` 1 1 0 0 1 1 1 1+1 1 0 1 1 0 0 0```Align them:```1 1 0 0 1 1 1 1+1 1 0 1 1 0 0 0```Add column by column:Column 1: 1 + 0 = 1Column 2: 1 + 0 = 1Column 3: 1 + 0 = 1Column 4: 1 + 1 = 10, write 0, carry 1Column 5: 1 + 1 + carry 1 = 11, write 1, carry 1Column 6: 0 + 0 + carry 1 = 1Column 7: 0 + 1 = 1Column 8: 1 + 1 = 10, write 0, carry 1Since we're at the leftmost bit, we need to add the carryover 1, so it becomes 1.Putting it all together: 1 0 1 0 0 1 1 1 1.Wait, that's 9 bits: 101001111₂.But earlier, when I converted 341 to binary, I got 101010101₂, which is different. So, clearly, there's a mistake here.I think the confusion arises from how I'm handling the partial products. Maybe I should use a different method, like the lattice method or the standard multiplication algorithm.Alternatively, perhaps I should convert both numbers to decimal, multiply, subtract, and then convert back to binary, as I did earlier, which gave me 101010101₂.But since the binary multiplication is giving me a different result, I must be making a mistake in the process.Let me try multiplying 11011₂ by 1101₂ again, but this time using the standard multiplication algorithm.First, write down the two numbers:``` 1 1 0 1 1 × 1 1 0 1```Multiply each bit of the second number with the first number, starting from the right.1st bit (1): 11011 × 1 = 110112nd bit (0): 11011 × 0 = 00000, shifted left by 1: 0000003rd bit (1): 11011 × 1 = 11011, shifted left by 2: 11011004th bit (1): 11011 × 1 = 11011, shifted left by 3: 11011000Now, add these partial products:``` 1 1 0 1 1 + 0 0 0 0 0 0 +1 1 0 1 1 0 0 +1 1 0 1 1 0 0 0```Let's add them step by step.First, add 11011 and 000000:``` 0 0 1 1 0 1 1+ 0 0 0 0 0 0 0= 0 0 1 1 0 1 1```Next, add 1101100:``` 0 0 1 1 0 1 1+0 1 1 0 1 1 0 0```Align them properly by padding the shorter one with zeros:```0 0 1 1 0 1 1+0 1 1 0 1 1 0 0```Convert to 8 bits:```0 0 1 1 0 1 1 → 0 0 1 1 0 1 1 0+0 1 1 0 1 1 0 0```Now, add:```0 0 1 1 0 1 1 0+0 1 1 0 1 1 0 0```Column 1: 0 + 0 = 0Column 2: 1 + 0 = 1Column 3: 1 + 1 = 10, write 0, carry 1Column 4: 0 + 1 + carry 1 = 2, write 0, carry 1Column 5: 1 + 1 + carry 1 = 3, write 1, carry 1Column 6: 0 + 1 + carry 1 = 2, write 0, carry 1Column 7: 1 + 0 + carry 1 = 2, write 0, carry 1Column 8: 0 + 0 + carry 1 = 1So, the result is 1 0 0 0 1 0 0 1.Wait, that's 8 bits: 10001001₂.Now, add the last partial product: 11011000.So, we have:```1 0 0 0 1 0 0 1+1 1 0 1 1 0 0 0```Align them:```1 0 0 0 1 0 0 1+1 1 0 1 1 0 0 0```Add column by column:Column 1: 1 + 0 = 1Column 2: 0 + 0 = 0Column 3: 0 + 0 = 0Column 4: 0 + 1 = 1Column 5: 1 + 1 = 10, write 0, carry 1Column 6: 0 + 0 + carry 1 = 1Column 7: 0 + 0 = 0Column 8: 1 + 1 = 10, write 0, carry 1Since we're at the leftmost bit, we need to add the carryover 1, so it becomes 1.Putting it all together: 1 0 1 0 0 0 0 1 0 1.Wait, that's 10 bits: 1010000101₂.But earlier, when I converted 341 to binary, I got 101010101₂, which is different. So, clearly, I'm making a mistake in the binary multiplication process.I think the confusion is arising from how I'm handling the partial products and their shifts. Maybe I should use a different approach, like the standard multiplication algorithm used in binary.Alternatively, perhaps I should stick to converting to decimal, perform the operations, and then convert back to binary, as that seems to give me a consistent result.Given that, I'll proceed with the decimal method:11011₂ = 271101₂ = 1327 × 13 = 3511010₂ = 10351 - 10 = 341341 in binary is 101010101₂.Therefore, the final answer is 101010101₂.But wait, earlier, when I tried binary multiplication, I got 110011101₂, which is 413 in decimal, which is incorrect. So, I must have made a mistake in the binary multiplication process.To avoid confusion, I'll stick with the decimal method, which gives me a consistent and correct result.So, the final answer is 101010101₂.

💡First, I note that Joe's dad has 21 marshmallows and roasts a third of them. Calculating this, Joe's dad roasts ( frac{1}{3} times 21 = 7 ) marshmallows.Next, I let ( J ) represent the number of marshmallows Joe has. Joe roasts half of his marshmallows, which is ( frac{1}{2}J ).Together, they roasted a total of 49 marshmallows. This leads to the equation:[7 + frac{1}{2}J = 49]Solving for ( J ), I subtract 7 from both sides:[frac{1}{2}J = 42]Then, I multiply both sides by 2 to find ( J ):[J = 84]Finally, to find the ratio of the number of marshmallows Joe has to the number his dad has, I compare 84 to 21. Simplifying this ratio by dividing both numbers by their greatest common divisor (21), I get:[frac{84}{21} : frac{21}{21} = 4 : 1]

💡Okay, so I've got this problem here about tourists who went on a journey. Half of it, they walked, and the other half, they took a bus. The whole trip took them 5.5 hours. Then, if they had taken the bus for the entire journey, it would have only taken 1 hour. I need to figure out two things: first, how long would it take them to walk the entire journey, and second, how many times faster is the bus compared to walking.Alright, let's break this down. Let's assume the total distance of the journey is D. So, half of that distance is D/2, and the other half is also D/2. They walked one half and took the bus for the other half. The total time they spent was 5.5 hours.Now, if they had taken the bus for the entire journey, it would have taken them 1 hour. That means the bus is pretty fast. So, from this, I can figure out the speed of the bus. Speed is distance divided by time, right? So, the bus's speed would be D divided by 1 hour, which is D per hour. So, the bus speed is D/hour.Now, for the walking part. They walked half the distance, which is D/2, and took some time to do that. Let's call the time they spent walking T_walk. Similarly, the time they spent on the bus for the other half would be T_bus. We know that T_walk plus T_bus equals 5.5 hours.But wait, we also know that if they had taken the bus for the entire journey, it would have taken 1 hour. So, the bus speed is D per hour, as I figured out earlier. Therefore, the time it takes to go D/2 by bus would be (D/2) divided by (D/hour), which simplifies to (D/2) * (hour/D) = hour/2. So, T_bus is 0.5 hours.Now, since the total time is 5.5 hours, and T_bus is 0.5 hours, then T_walk must be 5.5 - 0.5 = 5 hours. So, walking half the distance took them 5 hours.Therefore, walking the entire distance would take twice that time, right? So, 5 hours times 2 is 10 hours. So, walking the entire journey would take them 10 hours.Now, to find out how many times faster the bus is compared to walking. Let's think about speeds. The bus speed is D per hour, and walking speed is (D/2) divided by 5 hours, which is (D/2) / 5 = D/10 per hour.So, the bus speed is D per hour, and walking speed is D/10 per hour. To find out how many times faster the bus is, we divide the bus speed by the walking speed: (D/hour) / (D/10/hour) = (D/hour) * (10/hour/D) = 10. So, the bus is 10 times faster than walking.Wait, let me double-check that. If the bus takes 1 hour for the whole journey, and walking takes 10 hours, then the bus is indeed 10 times faster. Yeah, that makes sense.So, putting it all together, walking the entire journey would take 10 hours, and the bus is 10 times faster than walking.