💡Okay, so I have this problem where I need to convert the binary number 11 into a decimal number. Hmm, binary numbers are base 2, right? So each digit represents a power of 2. Let me think about how this works.First, I remember that in binary, the rightmost digit is the least significant bit, which is 2^0, and each digit to the left increases the power by one. So for the number 11, there are two digits. The rightmost digit is the first position, and the left digit is the second position.Let me write this out. The rightmost digit is 1, and since it's in the 2^0 position, that's 1 times 2^0, which is 1 times 1, so that's 1. The next digit to the left is also 1, and that's in the 2^1 position. So that's 1 times 2^1, which is 1 times 2, so that's 2.Now, I need to add these two values together to get the decimal equivalent. So 2 plus 1 equals 3. Therefore, the binary number 11 is equal to 3 in decimal.Wait, let me double-check that. If I have 11 in binary, the first 1 is in the 2^1 place, which is 2, and the second 1 is in the 2^0 place, which is 1. Adding them together gives 3. Yeah, that seems right.I guess another way to think about it is to list out the powers of 2. Starting from the right, the positions are 2^0, 2^1, 2^2, and so on. For the number 11, there are only two digits, so we're dealing with 2^0 and 2^1. Each digit tells us whether to include that power of 2 in the total sum. Since both digits are 1, we include both 2^1 and 2^0, which are 2 and 1, adding up to 3.Is there a quicker way to do this without writing it out every time? Maybe memorizing the binary numbers up to a certain point could help, but for now, breaking it down like this seems to work.Just to make sure I'm not making a mistake, I'll think of another example. Let's say I have the binary number 10. The rightmost digit is 0, so that's 0 times 2^0, which is 0. The next digit is 1, which is 1 times 2^1, so that's 2. Adding them together gives 2, which is correct because 10 in binary is 2 in decimal.Another example: 101. The rightmost digit is 1, which is 1 times 2^0 = 1. The middle digit is 0, which is 0 times 2^1 = 0. The leftmost digit is 1, which is 1 times 2^2 = 4. Adding them together: 4 + 0 + 1 = 5. So 101 in binary is 5 in decimal. That checks out.Going back to the original problem, 11 in binary. Using the same method: rightmost digit is 1 times 2^0 = 1, left digit is 1 times 2^1 = 2. Adding them together: 1 + 2 = 3. So yes, 11 in binary is 3 in decimal.I think I've got it. It's all about understanding the place value in binary, which is powers of 2, and then summing those values based on the digits. It might take some practice, but once you get the hang of it, it's pretty straightforward.Maybe to solidify this, I'll try converting a few more binary numbers to decimal. Let's see, how about 100? Rightmost digit is 0, so 0 times 2^0 = 0. Middle digit is 0, so 0 times 2^1 = 0. Leftmost digit is 1, so 1 times 2^2 = 4. Adding them up: 4 + 0 + 0 = 4. So 100 in binary is 4 in decimal. That makes sense.Another one: 111. Rightmost digit is 1 times 2^0 = 1. Middle digit is 1 times 2^1 = 2. Leftmost digit is 1 times 2^2 = 4. Adding them together: 1 + 2 + 4 = 7. So 111 in binary is 7 in decimal. That seems right.Wait, what about a longer binary number, like 1010? Let's break it down. From right to left: 0 times 2^0 = 0, 1 times 2^1 = 2, 0 times 2^2 = 0, 1 times 2^3 = 8. Adding them up: 8 + 0 + 2 + 0 = 10. So 1010 in binary is 10 in decimal. That's correct.This practice is helping me understand the process better. It's all about recognizing the place value and then calculating each digit's contribution to the total sum.Going back to the original problem, 11 in binary. I've confirmed it multiple times, and each time I get 3 in decimal. So I'm pretty confident that's the correct answer.I guess the key takeaway here is that each digit in a binary number represents a power of 2, starting from the rightmost digit as 2^0, and each subsequent digit to the left increases the exponent by one. By multiplying each digit by its corresponding power of 2 and then summing those products, you can convert any binary number to its decimal equivalent.It's interesting how different number systems work. Binary is fundamental in computer science because computers use binary to represent data and perform operations. Understanding binary conversion is essential for anyone studying computer science or related fields.I wonder if there's a quicker method or a formula that can be used for binary to decimal conversion without having to write out each step. Maybe using positional notation or something like that. But for now, breaking it down digit by digit seems to be the most straightforward approach, especially for someone who's just starting out.Another thing I'm thinking about is how binary numbers can represent larger numbers as well. For example, a binary number like 10000000 would be 128 in decimal because it's 2^7. So each additional digit to the left doubles the value of the previous digit. That's pretty cool.I also recall that hexadecimal is another number system used in computing, which is base 16. It uses digits from 0 to 9 and then letters A to F for values 10 to 15. Converting between binary and hexadecimal is often easier because each hexadecimal digit corresponds to four binary digits. But that's a topic for another time.In summary, converting binary to decimal involves understanding the place value of each digit, multiplying each digit by 2 raised to the power of its position (starting from 0 on the right), and then summing those values. Applying this method to the binary number 11 gives us 3 in decimal. It's a simple process once you grasp the concept, but it's crucial to be careful with the exponents to avoid mistakes.

💡Okay, so I have this problem where I need to find all (n-1)-tuples of nonnegative integers a₁, a₂, ..., a_{n-1} such that for every integer m, the sum of these floor functions equals m. The equation is:⌊m/(2ⁿ - 1)⌋ + ⌊(2m + a₁)/(2ⁿ - 1)⌋ + ⌊(4m + a₂)/(2ⁿ - 1)⌋ + ... + ⌊(2^{n-1}m + a_{n-1})/(2ⁿ - 1)⌋ = mHmm, okay. So, first, I need to understand what this equation is saying. It's a sum of floor functions, each of which takes a linear combination of m and some a_k, divides by (2ⁿ - 1), and then takes the floor. The sum of all these floors equals m for every integer m.I remember that floor functions can sometimes be tricky because they depend on the fractional part of the number. So, maybe I can rewrite each term in a way that separates the integer and fractional parts.Let me denote N = 2ⁿ - 1 to simplify the notation. Then, each term becomes ⌊(2ᵏm + a_k)/N⌋. So, the equation becomes:⌊m/N⌋ + ⌊(2m + a₁)/N⌋ + ⌊(4m + a₂)/N⌋ + ... + ⌊(2^{n-1}m + a_{n-1})/N⌋ = mI wonder if there's an identity or theorem that relates sums of floor functions to something simpler. I recall that sometimes, when you have multiple floor functions with different coefficients, you can use properties of modular arithmetic or some kind of telescoping series.Wait, maybe I can think about this in terms of binary representations. Since the coefficients are powers of 2, which are related to binary digits. Maybe each term corresponds to a bit in the binary expansion of m?Let me try plugging in some small values of n to see if I can spot a pattern.Suppose n = 2. Then N = 2² - 1 = 3. The equation becomes:⌊m/3⌋ + ⌊(2m + a₁)/3⌋ = mI need to find a₁ such that this holds for all m. Let me test m = 0, 1, 2, 3.For m = 0:⌊0/3⌋ + ⌊(0 + a₁)/3⌋ = 0 + ⌊a₁/3⌋ = 0So, ⌊a₁/3⌋ = 0 ⇒ a₁ < 3.For m = 1:⌊1/3⌋ + ⌊(2 + a₁)/3⌋ = 0 + ⌊(2 + a₁)/3⌋ = 1So, ⌊(2 + a₁)/3⌋ = 1 ⇒ 1 ≤ (2 + a₁)/3 < 2 ⇒ 1 ≤ 2 + a₁ < 6 ⇒ -1 ≤ a₁ < 4. But a₁ is nonnegative, so 0 ≤ a₁ < 4. But from m=0, a₁ < 3. So, a₁ can be 0, 1, or 2.Wait, but let's check m=2:⌊2/3⌋ + ⌊(4 + a₁)/3⌋ = 0 + ⌊(4 + a₁)/3⌋ = 2So, ⌊(4 + a₁)/3⌋ = 2 ⇒ 2 ≤ (4 + a₁)/3 < 3 ⇒ 6 ≤ 4 + a₁ < 9 ⇒ 2 ≤ a₁ < 5. But from m=0, a₁ < 3, so a₁ must be 2.Wait, so a₁=2. Let me verify for m=3:⌊3/3⌋ + ⌊(6 + 2)/3⌋ = 1 + ⌊8/3⌋ = 1 + 2 = 3. That works.So for n=2, a₁=2.Hmm, interesting. So, in this case, a₁=2=2^{2-1} + 2^{1-1} -1=2 +1 -1=2. Wait, that seems to fit the formula from the previous problem.Wait, maybe in general, a_k=2^{n-1} + 2^{k-1} -1. Let me test that.For n=2, k=1: a₁=2^{1} + 2^{0} -1=2+1-1=2. Correct.Let me test n=3.For n=3, N=7. The equation becomes:⌊m/7⌋ + ⌊(2m + a₁)/7⌋ + ⌊(4m + a₂)/7⌋ = mI need to find a₁ and a₂ such that this holds for all m.Let me try m=0:⌊0/7⌋ + ⌊(0 + a₁)/7⌋ + ⌊(0 + a₂)/7⌋ = 0 + ⌊a₁/7⌋ + ⌊a₂/7⌋ = 0So, a₁ <7 and a₂ <7.Now, m=1:⌊1/7⌋ + ⌊(2 + a₁)/7⌋ + ⌊(4 + a₂)/7⌋ = 0 + ⌊(2 + a₁)/7⌋ + ⌊(4 + a₂)/7⌋ =1So, the sum of two floor terms equals 1. Each floor term is at least 0, so one of them is 1 and the other is 0, or both are 0 and one is 1? Wait, no, the sum is 1, so exactly one of them is 1 and the other is 0.So, either:Case 1: ⌊(2 + a₁)/7⌋=1 and ⌊(4 + a₂)/7⌋=0Which implies:1 ≤ (2 + a₁)/7 <2 ⇒ 5 ≤ a₁ <12But a₁ <7, so 5 ≤ a₁ <7 ⇒ a₁=5 or 6.And:0 ≤ (4 + a₂)/7 <1 ⇒ 0 ≤ a₂ <3But a₂ is nonnegative integer, so a₂=0,1,2.Case 2: ⌊(2 + a₁)/7⌋=0 and ⌊(4 + a₂)/7⌋=1Which implies:0 ≤ (2 + a₁)/7 <1 ⇒ 0 ≤ a₁ <5And:1 ≤ (4 + a₂)/7 <2 ⇒ 3 ≤ a₂ <10But a₂ <7, so 3 ≤ a₂ <7 ⇒ a₂=3,4,5,6.So, in total, possible a₁ and a₂:Either a₁=5,6 and a₂=0,1,2Or a₁=0,1,2,3,4 and a₂=3,4,5,6.Now, let's check m=2:⌊2/7⌋ + ⌊(4 + a₁)/7⌋ + ⌊(8 + a₂)/7⌋ =0 + ⌊(4 + a₁)/7⌋ + ⌊(8 + a₂)/7⌋=2So, the sum of two floor terms equals 2. Each floor term is at least 0, so possible cases:Either both are 1, or one is 2 and the other is 0.But let's see:First, ⌊(4 + a₁)/7⌋ can be 0 or 1, depending on a₁.Similarly, ⌊(8 + a₂)/7⌋ = ⌊1 + (1 + a₂)/7⌋ =1 + ⌊(1 + a₂)/7⌋. Since a₂ <7, ⌊(1 + a₂)/7⌋=0. So, ⌊(8 + a₂)/7⌋=1.So, ⌊(8 + a₂)/7⌋=1 regardless of a₂.Therefore, the equation becomes:⌊(4 + a₁)/7⌋ +1=2 ⇒ ⌊(4 + a₁)/7⌋=1Which implies:1 ≤ (4 + a₁)/7 <2 ⇒7 ≤4 + a₁ <14 ⇒3 ≤a₁ <10But a₁ <7, so 3 ≤a₁ <7 ⇒ a₁=3,4,5,6.But from m=1, we have either a₁=5,6 and a₂=0,1,2 or a₁=0,1,2,3,4 and a₂=3,4,5,6.So, combining with m=2, a₁ must be 3,4,5,6.Therefore, in the first case, a₁=5,6 and a₂=0,1,2.In the second case, a₁=3,4 and a₂=3,4,5,6.Wait, but from m=1, in the second case, a₁=0,1,2,3,4 and a₂=3,4,5,6. But from m=2, a₁ must be 3,4,5,6. So, in the second case, a₁ must be 3,4.So, possible a₁=3,4 and a₂=3,4,5,6.Wait, but let's check m=3:⌊3/7⌋ + ⌊(6 + a₁)/7⌋ + ⌊(12 + a₂)/7⌋=0 + ⌊(6 + a₁)/7⌋ + ⌊(12 + a₂)/7⌋=3So, the sum of two floor terms equals 3.Note that ⌊(12 + a₂)/7⌋=⌊1 + (5 + a₂)/7⌋=1 + ⌊(5 + a₂)/7⌋.Since a₂ <7, ⌊(5 + a₂)/7⌋=0 or 1.If a₂ <2, then 5 + a₂ <7, so ⌊(5 + a₂)/7⌋=0.If a₂ ≥2, then 5 + a₂ ≥7, so ⌊(5 + a₂)/7⌋=1.So, ⌊(12 + a₂)/7⌋=1 + 0=1 if a₂ <2, or 1 +1=2 if a₂ ≥2.Similarly, ⌊(6 + a₁)/7⌋:If a₁ <1, then 6 + a₁ <7, so floor=0.If a₁ ≥1, then 6 + a₁ ≥7, so floor=1.So, depending on a₁ and a₂:Case 1: a₁=5,6 and a₂=0,1,2.Then, ⌊(6 + a₁)/7⌋=⌊6+5=11/7⌋=1, or ⌊6+6=12/7⌋=1.And ⌊(12 + a₂)/7⌋=⌊12 +0=12/7⌋=1, ⌊12 +1=13/7⌋=1, ⌊12 +2=14/7⌋=2.So, for a₂=0,1: ⌊12 + a₂)/7⌋=1, so total sum=1 +1=2≠3.For a₂=2: ⌊14/7⌋=2, so sum=1 +2=3.So, in this case, a₁=5,6 and a₂=2.Case 2: a₁=3,4 and a₂=3,4,5,6.Then, ⌊(6 + a₁)/7⌋=⌊6+3=9/7⌋=1, ⌊6+4=10/7⌋=1.And ⌊(12 + a₂)/7⌋=⌊12 +3=15/7⌋=2, ⌊12 +4=16/7⌋=2, ⌊12 +5=17/7⌋=2, ⌊12 +6=18/7⌋=2.So, sum=1 +2=3. So, this works for any a₂=3,4,5,6.So, from m=3, in case 1, a₂ must be 2, and in case 2, a₂ can be 3,4,5,6.Wait, but in case 1, a₂=0,1,2. But for m=3, only a₂=2 works. So, in case 1, a₂ must be 2.So, possible solutions:Either a₁=5,6 and a₂=2Or a₁=3,4 and a₂=3,4,5,6.Now, let's check m=4:⌊4/7⌋ + ⌊(8 + a₁)/7⌋ + ⌊(16 + a₂)/7⌋=0 + ⌊(8 + a₁)/7⌋ + ⌊(16 + a₂)/7⌋=4Compute each term:⌊(8 + a₁)/7⌋=⌊1 + (1 + a₁)/7⌋=1 + ⌊(1 + a₁)/7⌋. Since a₁ <7, ⌊(1 + a₁)/7⌋=0. So, ⌊(8 + a₁)/7⌋=1.Similarly, ⌊(16 + a₂)/7⌋=⌊2 + (2 + a₂)/7⌋=2 + ⌊(2 + a₂)/7⌋. Since a₂ <7, ⌊(2 + a₂)/7⌋=0. So, ⌊(16 + a₂)/7⌋=2.Thus, sum=1 +2=3≠4. Wait, that's a problem.Wait, no, let me recalculate:Wait, ⌊(8 + a₁)/7⌋:If a₁=5: 8 +5=13, ⌊13/7⌋=1If a₁=6: 8 +6=14, ⌊14/7⌋=2Similarly, a₁=3: 8 +3=11, ⌊11/7⌋=1a₁=4: 8 +4=12, ⌊12/7⌋=1So, for a₁=5:1, a₁=6:2, a₁=3,4:1.Similarly, ⌊(16 + a₂)/7⌋:If a₂=2:16 +2=18, ⌊18/7⌋=2If a₂=3:16 +3=19, ⌊19/7⌋=2a₂=4:20/7=2.857, floor=2a₂=5:21/7=3, floor=3a₂=6:22/7≈3.142, floor=3Wait, so:If a₂=2:2a₂=3,4:2a₂=5,6:3So, for m=4:Case 1: a₁=5,6 and a₂=2If a₁=5: sum=1 +2=3≠4If a₁=6: sum=2 +2=4. So, only a₁=6 and a₂=2 works.Case 2: a₁=3,4 and a₂=3,4,5,6If a₂=3,4: sum=1 +2=3≠4If a₂=5,6: sum=1 +3=4So, in case 2, a₂ must be 5 or6.So, from m=4, in case 1, only a₁=6 and a₂=2 works.In case 2, a₂ must be 5 or6.So, possible solutions:Either a₁=6 and a₂=2Or a₁=3,4 and a₂=5,6.Now, let's check m=5:⌊5/7⌋ + ⌊(10 + a₁)/7⌋ + ⌊(20 + a₂)/7⌋=0 + ⌊(10 + a₁)/7⌋ + ⌊(20 + a₂)/7⌋=5Compute each term:⌊(10 + a₁)/7⌋:a₁=6:10 +6=16, ⌊16/7⌋=2a₁=3:10 +3=13, ⌊13/7⌋=1a₁=4:10 +4=14, ⌊14/7⌋=2a₁=5:10 +5=15, ⌊15/7⌋=2Wait, but in case 1, a₁=6 and a₂=2.So, for a₁=6: ⌊16/7⌋=2a₂=2: ⌊20 +2=22/7⌋=3So, sum=2 +3=5. That works.In case 2, a₁=3,4 and a₂=5,6.For a₁=3: ⌊13/7⌋=1a₂=5: ⌊20 +5=25/7⌋=3Sum=1 +3=4≠5For a₁=4: ⌊14/7⌋=2a₂=5: sum=2 +3=5a₂=6: ⌊20 +6=26/7⌋=3Sum=2 +3=5So, in case 2, a₁=4 and a₂=5,6 works.Wait, but a₁=3 and a₂=5,6 gives sum=4≠5, so only a₁=4 and a₂=5,6 works.So, from m=5, possible solutions:Either a₁=6 and a₂=2Or a₁=4 and a₂=5,6.Now, let's check m=6:⌊6/7⌋ + ⌊(12 + a₁)/7⌋ + ⌊(24 + a₂)/7⌋=0 + ⌊(12 + a₁)/7⌋ + ⌊(24 + a₂)/7⌋=6Compute each term:⌊(12 + a₁)/7⌋:a₁=6:12 +6=18, ⌊18/7⌋=2a₁=4:12 +4=16, ⌊16/7⌋=2a₁=3:12 +3=15, ⌊15/7⌋=2Wait, but in case 1, a₁=6 and a₂=2.So, ⌊18/7⌋=2a₂=2: ⌊24 +2=26/7⌋=3Sum=2 +3=5≠6Wait, that's a problem. So, case 1 doesn't work for m=6.In case 2, a₁=4 and a₂=5,6.a₁=4: ⌊16/7⌋=2a₂=5: ⌊24 +5=29/7⌋=4Sum=2 +4=6a₂=6: ⌊24 +6=30/7⌋=4Sum=2 +4=6So, only case 2 works for m=6.Thus, case 1 is invalid because it fails at m=6.Therefore, the only possible solution is a₁=4 and a₂=5,6.Wait, but let's check m=7:⌊7/7⌋ + ⌊(14 + a₁)/7⌋ + ⌊(28 + a₂)/7⌋=1 + ⌊(14 + a₁)/7⌋ + ⌊(28 + a₂)/7⌋=7Compute each term:⌊(14 + a₁)/7⌋=2 + ⌊a₁/7⌋=2 since a₁ <7.⌊(28 + a₂)/7⌋=4 + ⌊a₂/7⌋=4 since a₂ <7.So, sum=1 +2 +4=7. That works regardless of a₁ and a₂.So, m=7 is satisfied.Wait, but in our previous step, we saw that only a₁=4 and a₂=5,6 works up to m=6.But let's check m=5 again with a₁=4 and a₂=5:⌊5/7⌋=0⌊(10 +4)/7⌋=⌊14/7⌋=2⌊(20 +5)/7⌋=⌊25/7⌋=3Sum=0 +2 +3=5. Correct.Similarly, m=6:⌊6/7⌋=0⌊(12 +4)/7⌋=⌊16/7⌋=2⌊(24 +5)/7⌋=⌊29/7⌋=4Sum=0 +2 +4=6. Correct.And m=5 with a₂=6:⌊20 +6=26/7⌋=3Sum=0 +2 +3=5. Correct.So, seems consistent.But wait, earlier for m=4, in case 2, a₁=4 and a₂=5,6 works.But what about a₁=3? For m=4, a₁=3 and a₂=5:⌊(8 +3)/7⌋=⌊11/7⌋=1⌊(16 +5)/7⌋=⌊21/7⌋=3Sum=1 +3=4. Correct.But earlier, for m=5, a₁=3 and a₂=5:⌊(10 +3)/7⌋=⌊13/7⌋=1⌊(20 +5)/7⌋=⌊25/7⌋=3Sum=1 +3=4≠5. So, it fails.Thus, a₁=3 is invalid.So, only a₁=4 and a₂=5,6 works.Wait, but let's check m=1 with a₁=4 and a₂=5:⌊1/7⌋=0⌊(2 +4)/7⌋=⌊6/7⌋=0⌊(4 +5)/7⌋=⌊9/7⌋=1Sum=0 +0 +1=1. Correct.Similarly, m=2:⌊2/7⌋=0⌊(4 +4)/7⌋=⌊8/7⌋=1⌊(8 +5)/7⌋=⌊13/7⌋=1Sum=0 +1 +1=2. Correct.m=3:⌊3/7⌋=0⌊(6 +4)/7⌋=⌊10/7⌋=1⌊(12 +5)/7⌋=⌊17/7⌋=2Sum=0 +1 +2=3. Correct.m=4:⌊4/7⌋=0⌊(8 +4)/7⌋=⌊12/7⌋=1⌊(16 +5)/7⌋=⌊21/7⌋=3Sum=0 +1 +3=4. Correct.m=5:⌊5/7⌋=0⌊(10 +4)/7⌋=⌊14/7⌋=2⌊(20 +5)/7⌋=⌊25/7⌋=3Sum=0 +2 +3=5. Correct.m=6:⌊6/7⌋=0⌊(12 +4)/7⌋=⌊16/7⌋=2⌊(24 +5)/7⌋=⌊29/7⌋=4Sum=0 +2 +4=6. Correct.m=7:⌊7/7⌋=1⌊(14 +4)/7⌋=⌊18/7⌋=2⌊(28 +5)/7⌋=⌊33/7⌋=4Sum=1 +2 +4=7. Correct.So, it seems that a₁=4 and a₂=5,6 works.Wait, but a₂=6 also works:For m=5, a₂=6:⌊20 +6=26/7⌋=3Sum=0 +2 +3=5.For m=6:⌊24 +6=30/7⌋=4Sum=0 +2 +4=6.So, both a₂=5 and 6 work.But wait, earlier for m=3, a₂=5,6 gives sum=3.Wait, but in the case where a₁=4 and a₂=5,6, it works for all m.But in the case where a₁=6 and a₂=2, it failed at m=6.So, the only valid solution is a₁=4 and a₂=5,6.But wait, let me check a₂=6:For m=1:⌊(4 +6)/7⌋=⌊10/7⌋=1Sum=0 +0 +1=1. Correct.For m=2:⌊(8 +6)/7⌋=⌊14/7⌋=2Sum=0 +2 +1=3≠2. Wait, no, hold on.Wait, when a₁=4 and a₂=6:For m=2:⌊2/7⌋=0⌊(4 +4)/7⌋=⌊8/7⌋=1⌊(8 +6)/7⌋=⌊14/7⌋=2Sum=0 +1 +2=3≠2.Wait, that's a problem. So, a₂=6 doesn't work for m=2.Wait, but earlier, when a₂=6, for m=2:⌊(4 +4)/7⌋=1⌊(8 +6)/7⌋=2Sum=1 +2=3, but m=2 requires sum=2.Wait, so that's a contradiction.Wait, but earlier, when a₂=6, for m=2:⌊(4 +4)/7⌋=1⌊(8 +6)/7⌋=2Sum=1 +2=3≠2.So, a₂=6 doesn't work for m=2.Wait, but earlier, when a₂=5:For m=2:⌊(4 +4)/7⌋=1⌊(8 +5)/7⌋=1Sum=1 +1=2. Correct.So, a₂=5 works, but a₂=6 doesn't.So, only a₂=5 works.Wait, but earlier, for m=4, a₂=5 and 6 both worked.Wait, let me re-examine.When a₂=6:For m=2:⌊(8 +6)/7⌋=2So, sum=1 +2=3≠2.Thus, a₂=6 is invalid.So, only a₂=5 works.Thus, the only solution is a₁=4 and a₂=5.Wait, but let's check m=6 with a₂=5:⌊(24 +5)/7⌋=⌊29/7⌋=4Sum=0 +2 +4=6. Correct.And for m=5:⌊(20 +5)/7⌋=3Sum=0 +2 +3=5. Correct.So, seems consistent.Thus, for n=3, the solution is a₁=4 and a₂=5.Wait, but according to the formula I thought earlier, a_k=2^{n-1} + 2^{k-1} -1.For n=3, k=1: a₁=2^{2} + 2^{0} -1=4 +1 -1=4. Correct.k=2: a₂=2^{2} + 2^{1} -1=4 +2 -1=5. Correct.So, the formula holds.Thus, for n=3, a₁=4 and a₂=5.Similarly, for n=2, a₁=2=2^{1} +2^{0} -1=2+1-1=2.So, the formula seems to hold.Thus, in general, the solution is a_k=2^{n-1} +2^{k-1} -1.Therefore, the answer is a_k=2^{n-1} +2^{k-1} -1 for each k=1,2,...,n-1.I think that's the solution.

💡To find the intersection point of the lines ( y = x ) and ( y = -x + 2 ), we need to solve the system of equations:[left{begin{array}{l}y = x y = -x + 2end{array}right.]Setting the two expressions for ( y ) equal to each other:[x = -x + 2]Solving for ( x ):[x + x = 2 implies 2x = 2 implies x = 1]Substituting ( x = 1 ) into the first equation ( y = x ):[y = 1]Therefore, the coordinates of the intersection point are ( (1, 1) ).

💡Okay, so I have this quadratic function y = x² - (m + 2)x + 3, and there are two parts to this problem. Let me start with part (1).**Problem (1):** The function y = x² - (m + 2)x + 3 has a maximum value M and a minimum value N in the range 1 ≤ x ≤ 2. It's given that M - N ≤ 2. I need to find the range of real numbers m.Alright, so first, I remember that quadratic functions have either a maximum or a minimum depending on the coefficient of x². Since the coefficient here is positive (1), the parabola opens upwards, meaning it has a minimum point. But wait, the problem mentions both a maximum and a minimum in the interval [1, 2]. Hmm, that makes sense because on a closed interval, a continuous function will attain both a maximum and a minimum. So, even though the parabola opens upwards, over a specific interval, it can have both a max and min.To find the maximum and minimum, I should consider the critical point (vertex) of the parabola and evaluate the function at the endpoints of the interval.The vertex of a parabola y = ax² + bx + c is at x = -b/(2a). In this case, a = 1, b = -(m + 2). So, the x-coordinate of the vertex is x = (m + 2)/2.Now, depending on where this vertex is located relative to the interval [1, 2], the maximum and minimum values will occur either at the vertex or at the endpoints.So, let's consider different cases based on the position of the vertex.**Case 1:** The vertex is to the left of the interval, i.e., (m + 2)/2 ≤ 1. That simplifies to m + 2 ≤ 2, so m ≤ 0.In this case, the function is increasing on [1, 2] because the vertex is to the left. So, the minimum value N occurs at x = 1, and the maximum value M occurs at x = 2.Let me compute N and M.At x = 1:y = 1² - (m + 2)(1) + 3 = 1 - m - 2 + 3 = (1 - 2 + 3) - m = 2 - m.At x = 2:y = 2² - (m + 2)(2) + 3 = 4 - 2m - 4 + 3 = (4 - 4 + 3) - 2m = 3 - 2m.So, N = 2 - m and M = 3 - 2m.Given that M - N ≤ 2, let's compute M - N:(3 - 2m) - (2 - m) = 3 - 2m - 2 + m = (3 - 2) + (-2m + m) = 1 - m.So, 1 - m ≤ 2.Solving for m:1 - m ≤ 2 ⇒ -m ≤ 1 ⇒ m ≥ -1.But in this case, we have m ≤ 0. So, combining these, m must satisfy -1 ≤ m ≤ 0.**Case 2:** The vertex is to the right of the interval, i.e., (m + 2)/2 ≥ 2. That simplifies to m + 2 ≥ 4, so m ≥ 2.In this case, the function is decreasing on [1, 2] because the vertex is to the right. So, the maximum value M occurs at x = 1, and the minimum value N occurs at x = 2.Wait, hold on. If the function is decreasing on [1, 2], then as x increases, y decreases. So, the maximum would be at x = 1, and the minimum at x = 2.Wait, but earlier in Case 1, when the vertex was to the left, the function was increasing, so max at x=2 and min at x=1. So, in this case, vertex to the right, function decreasing, so max at x=1, min at x=2.So, let's compute N and M.At x = 1:y = 2 - m (same as before).At x = 2:y = 3 - 2m (same as before).But here, since the function is decreasing, M is at x=1, which is 2 - m, and N is at x=2, which is 3 - 2m.So, M = 2 - m, N = 3 - 2m.Compute M - N:(2 - m) - (3 - 2m) = 2 - m - 3 + 2m = (2 - 3) + (-m + 2m) = -1 + m.Given that M - N ≤ 2:-1 + m ≤ 2 ⇒ m ≤ 3.But in this case, m ≥ 2. So, combining, 2 ≤ m ≤ 3.**Case 3:** The vertex is inside the interval [1, 2]. So, 1 < (m + 2)/2 < 2.Multiplying all parts by 2: 2 < m + 2 < 4 ⇒ 0 < m < 2.So, in this case, the vertex is within the interval, so the minimum occurs at the vertex, and the maximum occurs at one of the endpoints.Wait, but since the parabola opens upwards, the vertex is the minimum point. So, the minimum N is at the vertex, and the maximum M is at one of the endpoints, either x=1 or x=2.So, we need to determine which endpoint gives the larger value.Let me compute y at x=1 and x=2, and compare them.At x=1: y = 2 - m.At x=2: y = 3 - 2m.So, which is larger? Let's subtract:(3 - 2m) - (2 - m) = 1 - m.So, if 1 - m > 0 ⇒ m < 1, then y at x=2 is larger.If m = 1, both are equal.If m > 1, y at x=1 is larger.So, within 0 < m < 2, we have two subcases:**Subcase 3a:** 0 < m < 1.Here, y at x=2 is larger, so M = 3 - 2m, and N is at the vertex.**Subcase 3b:** 1 ≤ m < 2.Here, y at x=1 is larger, so M = 2 - m, and N is at the vertex.So, let's compute N in both subcases.The vertex is at x = (m + 2)/2. Let's compute y at the vertex.y = [(m + 2)/2]^2 - (m + 2)*[(m + 2)/2] + 3.Let me compute this step by step.First, expand [(m + 2)/2]^2:= (m² + 4m + 4)/4.Then, expand (m + 2)*[(m + 2)/2]:= (m + 2)^2 / 2 = (m² + 4m + 4)/2.So, y = (m² + 4m + 4)/4 - (m² + 4m + 4)/2 + 3.Let me combine these terms:First, write all terms with denominator 4:= (m² + 4m + 4)/4 - 2(m² + 4m + 4)/4 + 3.= [ (m² + 4m + 4) - 2(m² + 4m + 4) ] / 4 + 3.= [ -m² - 4m - 4 ] / 4 + 3.= (-m² - 4m - 4)/4 + 3.= (-m² - 4m - 4)/4 + 12/4.= (-m² - 4m - 4 + 12)/4.= (-m² - 4m + 8)/4.= (-m² - 4m + 8)/4.We can factor out a negative sign:= -(m² + 4m - 8)/4.Alternatively, let me compute it as:y = (m + 2)^2 / 4 - (m + 2)^2 / 2 + 3.= (m + 2)^2 (1/4 - 1/2) + 3.= (m + 2)^2 (-1/4) + 3.= - (m + 2)^2 / 4 + 3.So, N = 3 - (m + 2)^2 / 4.That's a simpler expression.So, N = 3 - (m + 2)² / 4.Alright, so now, for Subcase 3a: 0 < m < 1.M = 3 - 2m, N = 3 - (m + 2)² / 4.Compute M - N:(3 - 2m) - [3 - (m + 2)² / 4] = 3 - 2m - 3 + (m + 2)² / 4 = -2m + (m + 2)² / 4.We need M - N ≤ 2:-2m + (m + 2)² / 4 ≤ 2.Let me compute (m + 2)²:= m² + 4m + 4.So, (m + 2)² / 4 = (m² + 4m + 4)/4.So, the inequality becomes:-2m + (m² + 4m + 4)/4 ≤ 2.Multiply both sides by 4 to eliminate the denominator:-8m + m² + 4m + 4 ≤ 8.Combine like terms:m² - 4m + 4 ≤ 8.Subtract 8:m² - 4m - 4 ≤ 0.Solve the quadratic inequality m² - 4m - 4 ≤ 0.First, find the roots:m = [4 ± sqrt(16 + 16)] / 2 = [4 ± sqrt(32)] / 2 = [4 ± 4√2] / 2 = 2 ± 2√2.So, the quadratic is ≤ 0 between the roots: 2 - 2√2 ≤ m ≤ 2 + 2√2.But in Subcase 3a, we have 0 < m < 1.So, intersecting these intervals:2 - 2√2 ≈ 2 - 2.828 ≈ -0.828.So, 2 - 2√2 ≈ -0.828 ≤ m ≤ 2 + 2√2 ≈ 4.828.But in Subcase 3a, m is between 0 and 1. So, the intersection is 0 < m < 1.Therefore, in Subcase 3a, all m in (0,1) satisfy M - N ≤ 2.**Subcase 3b:** 1 ≤ m < 2.Here, M = 2 - m, N = 3 - (m + 2)² / 4.Compute M - N:(2 - m) - [3 - (m + 2)² / 4] = 2 - m - 3 + (m + 2)² / 4 = (-1 - m) + (m + 2)² / 4.We need M - N ≤ 2:(-1 - m) + (m + 2)² / 4 ≤ 2.Again, expand (m + 2)²:= m² + 4m + 4.So, (m + 2)² / 4 = (m² + 4m + 4)/4.Thus, the inequality becomes:-1 - m + (m² + 4m + 4)/4 ≤ 2.Multiply both sides by 4:-4 - 4m + m² + 4m + 4 ≤ 8.Simplify:m² + 0m + 0 ≤ 8.So, m² ≤ 8.Which implies -2√2 ≤ m ≤ 2√2.But in Subcase 3b, m is between 1 and 2.So, intersecting, 1 ≤ m ≤ 2.Thus, in Subcase 3b, all m in [1, 2] satisfy M - N ≤ 2.Wait, but let me double-check the computation:Starting from:(-1 - m) + (m + 2)² / 4 ≤ 2.Compute:-1 - m + (m² + 4m + 4)/4 ≤ 2.Multiply by 4:-4 - 4m + m² + 4m + 4 ≤ 8.Simplify:m² ≤ 8.Yes, that's correct.So, m² ≤ 8 ⇒ |m| ≤ 2√2 ≈ 2.828.But since m is in [1, 2], which is within [-2.828, 2.828], so all m in [1, 2] satisfy this.Therefore, in Subcase 3b, all m in [1, 2] satisfy M - N ≤ 2.So, combining Case 1, 2, and 3:From Case 1: -1 ≤ m ≤ 0.From Case 2: 2 ≤ m ≤ 3.From Case 3: 0 < m < 2.But wait, in Case 3, we have Subcases 3a and 3b, which together cover 0 < m < 2, and all of them satisfy M - N ≤ 2.So, combining all:-1 ≤ m ≤ 3.Wait, but in Case 3, m is between 0 and 2, and in Cases 1 and 2, m is between -1 and 0, and 2 and 3.So, altogether, m ranges from -1 to 3.But let me verify if m = -1 and m = 3 are included.At m = -1:From Case 1: M - N = 1 - (-1) = 2, which satisfies M - N ≤ 2.At m = 3:From Case 2: M - N = -1 + 3 = 2, which also satisfies M - N ≤ 2.So, yes, the endpoints are included.Therefore, the range of m is -1 ≤ m ≤ 3.**Problem (2):** The equation x² - (m + 2)x + 3 = -(2m + 1)x + 2 has a solution in the range 0 ≤ x ≤ 2. Find the range of real numbers m.Alright, let's start by simplifying the equation.Bring all terms to one side:x² - (m + 2)x + 3 + (2m + 1)x - 2 = 0.Combine like terms:x² + [ - (m + 2) + (2m + 1) ]x + (3 - 2) = 0.Simplify the coefficients:For x term:- (m + 2) + (2m + 1) = -m - 2 + 2m + 1 = ( -m + 2m ) + ( -2 + 1 ) = m - 1.Constant term:3 - 2 = 1.So, the equation simplifies to:x² + (m - 1)x + 1 = 0.We need this equation to have at least one solution in the interval [0, 2].Let me denote this quadratic as y = x² + (m - 1)x + 1.We need y = 0 to have a solution in [0, 2].There are a few cases to consider:1. The quadratic has exactly one solution in [0, 2].2. The quadratic has two solutions, at least one of which is in [0, 2].But since it's a quadratic, it can have 0, 1, or 2 real roots.So, let's analyze the conditions.First, let's find the discriminant:Δ = (m - 1)² - 4 * 1 * 1 = (m - 1)² - 4.For real roots, Δ ≥ 0.So, (m - 1)² - 4 ≥ 0 ⇒ (m - 1)² ≥ 4 ⇒ |m - 1| ≥ 2 ⇒ m - 1 ≥ 2 or m - 1 ≤ -2 ⇒ m ≥ 3 or m ≤ -1.So, real roots exist only when m ≤ -1 or m ≥ 3.But we need at least one root in [0, 2]. So, even if the quadratic has two roots, we need at least one in [0, 2].Alternatively, if the quadratic has one root in [0, 2], that's also acceptable.So, let's consider the function y = x² + (m - 1)x + 1.We can use the Intermediate Value Theorem or analyze the sign changes.But perhaps a better approach is to consider the following:For the quadratic to have a root in [0, 2], either:1. y(0) and y(2) have opposite signs, meaning the quadratic crosses the x-axis between 0 and 2.Or2. The quadratic has a repeated root in [0, 2], which would require the discriminant to be zero and the root to be in [0, 2].But since we already have the discriminant condition, let's proceed.First, compute y(0):y(0) = 0 + 0 + 1 = 1.Compute y(2):y(2) = 4 + 2(m - 1) + 1 = 4 + 2m - 2 + 1 = (4 - 2 + 1) + 2m = 3 + 2m.So, y(2) = 3 + 2m.Now, for y(0) = 1 and y(2) = 3 + 2m.If y(2) < 0, then since y(0) = 1 > 0, by Intermediate Value Theorem, there is a root in (0, 2).So, y(2) < 0 ⇒ 3 + 2m < 0 ⇒ 2m < -3 ⇒ m < -3/2.So, if m < -3/2, then y(2) < 0, and since y(0) = 1 > 0, there is a root in (0, 2).But we also need to ensure that the quadratic has real roots, which requires m ≤ -1 or m ≥ 3.But m < -3/2 is more restrictive than m ≤ -1, so in this case, m < -3/2 satisfies both conditions.So, one case is m < -3/2.Now, let's consider when y(2) ≥ 0, i.e., m ≥ -3/2.In this case, y(0) = 1 > 0 and y(2) ≥ 0.So, both endpoints are non-negative. Therefore, for the quadratic to have a root in [0, 2], it must touch the x-axis within [0, 2], meaning the minimum of the quadratic is ≤ 0.Since the quadratic opens upwards (coefficient of x² is positive), the minimum occurs at the vertex.The vertex is at x = -b/(2a) = -(m - 1)/(2*1) = (1 - m)/2.We need the vertex to be within [0, 2] and y at the vertex ≤ 0.So, first, check if the vertex is in [0, 2]:(1 - m)/2 ≥ 0 ⇒ 1 - m ≥ 0 ⇒ m ≤ 1.And,(1 - m)/2 ≤ 2 ⇒ 1 - m ≤ 4 ⇒ -m ≤ 3 ⇒ m ≥ -3.So, the vertex is in [0, 2] when m ≤ 1 and m ≥ -3.But in this case, we are considering m ≥ -3/2.So, combining, m is in [-3/2, 1].Now, compute y at the vertex:y = [(1 - m)/2]^2 + (m - 1)*[(1 - m)/2] + 1.Let me compute this step by step.First, [(1 - m)/2]^2 = (1 - 2m + m²)/4.Second, (m - 1)*[(1 - m)/2] = (m - 1)(1 - m)/2 = -(1 - m)^2 / 2 = -(1 - 2m + m²)/2.So, y = (1 - 2m + m²)/4 - (1 - 2m + m²)/2 + 1.Combine terms:= (1 - 2m + m²)/4 - 2(1 - 2m + m²)/4 + 1.= [1 - 2m + m² - 2 + 4m - 2m²]/4 + 1.= [ (1 - 2) + (-2m + 4m) + (m² - 2m²) ] / 4 + 1.= [ -1 + 2m - m² ] / 4 + 1.= (-m² + 2m - 1)/4 + 1.= (-m² + 2m - 1 + 4)/4.= (-m² + 2m + 3)/4.So, y at the vertex is (-m² + 2m + 3)/4.We need this to be ≤ 0:(-m² + 2m + 3)/4 ≤ 0 ⇒ -m² + 2m + 3 ≤ 0 ⇒ m² - 2m - 3 ≥ 0.Solve m² - 2m - 3 ≥ 0.Factor the quadratic:m² - 2m - 3 = (m - 3)(m + 1).So, (m - 3)(m + 1) ≥ 0.This inequality holds when m ≤ -1 or m ≥ 3.But in our current case, we are considering m ≥ -3/2, and m ≤ 1.So, the intersection is m ≤ -1 or m ≥ 3, but within m ∈ [-3/2, 1].So, m ≤ -1 in [-3/2, 1] is m ∈ [-3/2, -1].And m ≥ 3 is outside of [-3/2, 1], so no overlap there.Therefore, in this case, when m ∈ [-3/2, -1], the quadratic has its vertex in [0, 2] and y at vertex ≤ 0, so the quadratic has a root in [0, 2].So, combining both cases:1. When m < -3/2, y(2) < 0, so root in (0, 2).2. When m ∈ [-3/2, -1], vertex in [0, 2] and y(vertex) ≤ 0, so root in [0, 2].Therefore, overall, m ≤ -1.Wait, but let me check m = -1.At m = -1, the quadratic becomes:x² + (-1 - 1)x + 1 = x² - 2x + 1 = (x - 1)^2.So, it has a double root at x = 1, which is in [0, 2]. So, m = -1 is included.Similarly, at m = -3/2, let's check:y(2) = 3 + 2*(-3/2) = 3 - 3 = 0.So, x=2 is a root when m = -3/2.But wait, in our earlier analysis, when m < -3/2, y(2) < 0, so root in (0, 2). When m = -3/2, y(2) = 0, so x=2 is a root.But in the case when m ∈ [-3/2, -1], we have the vertex in [0, 2] and y(vertex) ≤ 0, so roots in [0, 2].So, m = -3/2 is included as well, since x=2 is a root.But wait, when m = -3/2, the quadratic is:x² + (-3/2 - 1)x + 1 = x² - (5/2)x + 1.Compute y(2):4 - 5 + 1 = 0, so x=2 is a root.But also, the vertex is at x = (1 - m)/2 = (1 - (-3/2))/2 = (1 + 3/2)/2 = (5/2)/2 = 5/4, which is in [0, 2].Compute y at vertex:(-m² + 2m + 3)/4.At m = -3/2:= (- (9/4) + 2*(-3/2) + 3)/4 = (-9/4 - 3 + 3)/4 = (-9/4)/4 = -9/16 < 0.So, the vertex is below the x-axis, so the quadratic crosses the x-axis twice, but since y(0) = 1 > 0 and y(2) = 0, one root is at x=2, and the other is somewhere in (0, 2).Wait, but actually, when m = -3/2, the quadratic has a root at x=2 and another root somewhere else.But since y(0) = 1 > 0, and y(2) = 0, the other root must be less than 0 or greater than 2?Wait, no, because the quadratic is x² - (5/2)x + 1.Let me find the roots:x = [5/2 ± sqrt(25/4 - 4)] / 2 = [5/2 ± sqrt(25/4 - 16/4)] / 2 = [5/2 ± sqrt(9/4)] / 2 = [5/2 ± 3/2] / 2.So, x = (5/2 + 3/2)/2 = (8/2)/2 = 4/2 = 2.And x = (5/2 - 3/2)/2 = (2/2)/2 = 1/2.So, the roots are x=2 and x=1/2. So, both roots are in [0, 2]. So, at m = -3/2, both roots are in [0, 2].Therefore, m = -3/2 is included.So, combining all, the range of m is m ≤ -1.Wait, but earlier, when m ∈ [-3/2, -1], the quadratic has roots in [0, 2], and when m < -3/2, it also has a root in [0, 2]. So, overall, m ≤ -1.But wait, when m = -2, which is less than -3/2, let's check:y(2) = 3 + 2*(-2) = 3 - 4 = -1 < 0.So, y(2) < 0, and y(0) = 1 > 0, so by IVT, there is a root in (0, 2).Similarly, when m = -4, y(2) = 3 + 2*(-4) = 3 - 8 = -5 < 0, so root in (0, 2).So, yes, for all m ≤ -1, the equation has a solution in [0, 2].Wait, but when m is between -1 and 3, the quadratic doesn't have real roots, right? Because discriminant is negative.Wait, no, the discriminant is (m - 1)^2 - 4.So, for m between -1 and 3, discriminant is negative, so no real roots.Wait, but earlier, we considered m ≤ -1 or m ≥ 3 for real roots.So, if m is between -1 and 3, the quadratic has no real roots, so no solution in [0, 2].But when m ≤ -1 or m ≥ 3, the quadratic has real roots.But for m ≥ 3, let's check if any roots are in [0, 2].Compute y(0) = 1, y(2) = 3 + 2m.For m ≥ 3, y(2) = 3 + 2m ≥ 3 + 6 = 9 > 0.So, y(0) = 1 > 0, y(2) > 0.So, both endpoints are positive. So, for the quadratic to have a root in [0, 2], the minimum must be ≤ 0.But the vertex is at x = (1 - m)/2.For m ≥ 3, (1 - m)/2 ≤ (1 - 3)/2 = (-2)/2 = -1 < 0.So, the vertex is at x < 0, which is outside [0, 2].Therefore, in [0, 2], the quadratic is increasing because the vertex is to the left.So, y(0) = 1, y(2) = 3 + 2m > 0, and since it's increasing, it never crosses the x-axis in [0, 2].Therefore, for m ≥ 3, the quadratic has no roots in [0, 2].Thus, the only values of m for which the equation has a solution in [0, 2] are m ≤ -1.Wait, but earlier, when m = -1, the quadratic has a double root at x=1, which is in [0, 2].So, m = -1 is included.Therefore, the range of m is m ≤ -1.So, in conclusion:Problem (1): -1 ≤ m ≤ 3.Problem (2): m ≤ -1.**Final Answer**(1) boxed{[-1, 3]}(2) boxed{(-infty, -1]}

💡Alright, so I have this problem: "How many three-digit numbers are there in which any two adjacent digits differ by 3?" Hmm, okay. Let me try to wrap my head around this.First, let's break down what a three-digit number is. It's a number from 100 to 999, right? So, it has three places: hundreds, tens, and units. Let's denote them as A, B, and C respectively. So, the number is ABC.Now, the condition is that any two adjacent digits differ by 3. That means the difference between A and B should be 3, and the difference between B and C should also be 3. So, mathematically, |A - B| = 3 and |B - C| = 3.Okay, so I need to find all such numbers ABC where these conditions hold. Let's think about how to approach this.Maybe I can start by considering the possible values for A, then for each A, find possible Bs, and for each B, find possible Cs. That sounds systematic.So, A can be from 1 to 9 because it's the hundreds place and can't be zero. For each A, B can be either A + 3 or A - 3, but B has to be a digit between 0 and 9. Similarly, for each B, C can be B + 3 or B - 3, again ensuring it's a valid digit.Let me try to list them out step by step.Starting with A = 1:- B can be 1 + 3 = 4 or 1 - 3 = -2 (but -2 isn't a digit, so only B = 4 is valid).- Now, for B = 4, C can be 4 + 3 = 7 or 4 - 3 = 1.- So, the numbers are 147 and 141. Wait, but 141: is that valid? Let me check: |1 - 4| = 3 and |4 - 1| = 3. Yes, it is. So, two numbers here: 141 and 147.Wait, but earlier I thought only 147 was valid. Hmm, maybe I was too hasty. So, actually, both 141 and 147 are valid. So, that's two numbers for A = 1.Moving on to A = 2:- B can be 2 + 3 = 5 or 2 - 3 = -1 (invalid).- So, B = 5.- For B = 5, C can be 5 + 3 = 8 or 5 - 3 = 2.- So, the numbers are 258 and 252. Both are valid. So, two numbers here.A = 3:- B can be 3 + 3 = 6 or 3 - 3 = 0.- So, B = 6 or B = 0.- For B = 6, C can be 6 + 3 = 9 or 6 - 3 = 3.- So, numbers: 369 and 363.- For B = 0, C can be 0 + 3 = 3 or 0 - 3 = -3 (invalid).- So, only 303 is valid.- So, total three numbers: 303, 363, 369.A = 4:- B can be 4 + 3 = 7 or 4 - 3 = 1.- So, B = 7 or B = 1.- For B = 7, C can be 7 + 3 = 10 (invalid) or 7 - 3 = 4.- So, only 474 is valid.- For B = 1, C can be 1 + 3 = 4 or 1 - 3 = -2 (invalid).- So, only 414 is valid.- Total two numbers: 414, 474.A = 5:- B can be 5 + 3 = 8 or 5 - 3 = 2.- So, B = 8 or B = 2.- For B = 8, C can be 8 + 3 = 11 (invalid) or 8 - 3 = 5.- So, only 585 is valid.- For B = 2, C can be 2 + 3 = 5 or 2 - 3 = -1 (invalid).- So, only 525 is valid.- Total two numbers: 525, 585.A = 6:- B can be 6 + 3 = 9 or 6 - 3 = 3.- So, B = 9 or B = 3.- For B = 9, C can be 9 + 3 = 12 (invalid) or 9 - 3 = 6.- So, only 696 is valid.- For B = 3, C can be 3 + 3 = 6 or 3 - 3 = 0.- So, numbers: 636 and 630.- Total three numbers: 630, 636, 696.A = 7:- B can be 7 + 3 = 10 (invalid) or 7 - 3 = 4.- So, B = 4.- For B = 4, C can be 4 + 3 = 7 or 4 - 3 = 1.- So, numbers: 747 and 741.- Total two numbers: 741, 747.A = 8:- B can be 8 + 3 = 11 (invalid) or 8 - 3 = 5.- So, B = 5.- For B = 5, C can be 5 + 3 = 8 or 5 - 3 = 2.- So, numbers: 858 and 852.- Total two numbers: 852, 858.A = 9:- B can be 9 + 3 = 12 (invalid) or 9 - 3 = 6.- So, B = 6.- For B = 6, C can be 6 + 3 = 9 or 6 - 3 = 3.- So, numbers: 969 and 963.- Total two numbers: 963, 969.Now, let's count all the numbers we've found:- A=1: 141, 147 (2)- A=2: 252, 258 (2)- A=3: 303, 363, 369 (3)- A=4: 414, 474 (2)- A=5: 525, 585 (2)- A=6: 630, 636, 696 (3)- A=7: 741, 747 (2)- A=8: 852, 858 (2)- A=9: 963, 969 (2)Adding them up: 2 + 2 + 3 + 2 + 2 + 3 + 2 + 2 + 2 = 20.Wait, but earlier when I thought about A=1, I initially thought only 147 was valid, but then realized 141 is also valid. So, that's why there are two numbers for A=1. Similarly, for other As, sometimes there are two or three numbers depending on the validity of the digits.So, in total, there are 20 such three-digit numbers where any two adjacent digits differ by 3.I think that's it. I don't see any mistakes in this reasoning. It's a bit tedious to list them all, but I think I covered all possibilities by checking each A from 1 to 9 and then finding valid Bs and Cs accordingly.

💡Okay, so I have this problem here where I need to determine whether the condition ( a = 1 ) is sufficient, necessary, both, or neither for the lines ( l_1: ax + 2y - 1 = 0 ) and ( l_2: x + (a + 1)y + 4 = 0 ) to be parallel. Hmm, let me think about how to approach this.First, I remember that two lines are parallel if their slopes are equal. So, maybe I should find the slopes of these two lines and set them equal to each other. Let me rewrite both equations in slope-intercept form ( y = mx + b ) to find their slopes.Starting with ( l_1: ax + 2y - 1 = 0 ). Let me solve for y:( ax + 2y - 1 = 0 )Subtract ( ax ) and add 1 to both sides:( 2y = -ax + 1 )Divide both sides by 2:( y = (-a/2)x + 1/2 )So, the slope ( m_1 ) of ( l_1 ) is ( -a/2 ).Now, let's do the same for ( l_2: x + (a + 1)y + 4 = 0 ).Solving for y:( x + (a + 1)y + 4 = 0 )Subtract x and 4 from both sides:( (a + 1)y = -x - 4 )Divide both sides by ( a + 1 ):( y = (-1/(a + 1))x - 4/(a + 1) )So, the slope ( m_2 ) of ( l_2 ) is ( -1/(a + 1) ).For the lines to be parallel, their slopes must be equal:( m_1 = m_2 )So,( -a/2 = -1/(a + 1) )Let me simplify this equation. Multiply both sides by 2 to eliminate the denominator on the left:( -a = -2/(a + 1) )Multiply both sides by ( (a + 1) ) to eliminate the denominator on the right:( -a(a + 1) = -2 )Simplify the left side:( -a^2 - a = -2 )Multiply both sides by -1 to make it easier:( a^2 + a = 2 )Subtract 2 from both sides:( a^2 + a - 2 = 0 )Now, I need to solve this quadratic equation. Let me factor it:Looking for two numbers that multiply to -2 and add to 1. Hmm, 2 and -1.So, ( (a + 2)(a - 1) = 0 )Therefore, ( a = -2 ) or ( a = 1 ).Wait, so ( a ) can be either 1 or -2 for the lines to be parallel. That means ( a = 1 ) is one of the solutions, but not the only one. So, if ( a = 1 ), the lines are definitely parallel, but even if ( a = -2 ), they are also parallel.So, does that mean ( a = 1 ) is a sufficient condition? Because if ( a = 1 ), then the lines are parallel. But it's not necessary because there's another value, ( a = -2 ), that also makes them parallel.Let me double-check by plugging in ( a = 1 ) and ( a = -2 ) into the original equations.First, ( a = 1 ):( l_1: 1x + 2y - 1 = 0 ) which simplifies to ( x + 2y = 1 ).( l_2: x + (1 + 1)y + 4 = 0 ) which simplifies to ( x + 2y = -4 ).These are clearly parallel because they have the same coefficients for x and y, just different constants. So, they are parallel.Now, ( a = -2 ):( l_1: -2x + 2y - 1 = 0 ) which simplifies to ( -2x + 2y = 1 ) or dividing by 2: ( -x + y = 0.5 ).( l_2: x + (-2 + 1)y + 4 = 0 ) which simplifies to ( x - y = -4 ).Wait, let me write both in slope-intercept form to check the slopes.For ( l_1: -x + y = 0.5 ) => ( y = x + 0.5 ). So, slope is 1.For ( l_2: x - y = -4 ) => ( -y = -x -4 ) => ( y = x + 4 ). So, slope is also 1.Yes, they have the same slope, so they are parallel. So, both ( a = 1 ) and ( a = -2 ) make the lines parallel.Therefore, ( a = 1 ) is sufficient because it guarantees the lines are parallel, but it's not necessary because there's another value, ( a = -2 ), that also works.So, the condition ( a = 1 ) is sufficient but not necessary. Therefore, the answer should be A.Wait, just to be thorough, what if ( a = 0 )? Let me see.If ( a = 0 ):( l_1: 0x + 2y - 1 = 0 ) => ( 2y = 1 ) => ( y = 0.5 ). That's a horizontal line.( l_2: x + (0 + 1)y + 4 = 0 ) => ( x + y = -4 ). That's a line with slope -1.So, they are not parallel. So, ( a = 0 ) doesn't work, which is fine because we already found the two values that do.Another thing to consider: what if ( a = -1 )? Then, in ( l_2 ), the coefficient of y becomes ( a + 1 = 0 ), making the equation ( x + 0y + 4 = 0 ) => ( x = -4 ), which is a vertical line. Meanwhile, ( l_1 ) becomes ( -x + 2y -1 = 0 ) => ( 2y = x + 1 ) => ( y = 0.5x + 0.5 ), which is a line with slope 0.5. So, they are not parallel either.So, only ( a = 1 ) and ( a = -2 ) make the lines parallel. Therefore, ( a = 1 ) is sufficient but not necessary.I think that's solid. So, the answer is A.**Final Answer**boxed{A}