💡To determine how many visible factor numbers exist from 200 through 250, inclusive, we need to check each number in this range to see if it is divisible by each of its non-zero digits. First, we note that all numbers in this range start with the digit 2, so we only need to check divisibility by the other digits in the number. We can break down the range into six cases based on the second digit (T) in the number, which can be 0, 1, 2, 3, 4, or 5.**Case 1: T = 0**- Numbers of the form 20U (where U is the units digit).- The number must be divisible by U.- Possible values for U are 0, 1, 2, 4, and 5.- Valid numbers: 200, 201, 202, 204, 205.**Case 2: T = 1**- Numbers of the form 21U.- The number must be divisible by U.- Possible values for U are 0, 1, 2, 3, 5, 6, and 7.- Valid numbers: 210, 211, 212, 213, 215, 216, 217.**Case 3: T = 2**- Numbers of the form 22U.- The number must be divisible by U.- Possible values for U are 0, 1, 2, 4, 5, and 8.- Valid numbers: 220, 221, 222, 224, 225, 228.**Case 4: T = 3**- Numbers of the form 23U.- The number must be divisible by U.- Possible values for U are 0, 1, 2, and 5.- Valid numbers: 230, 231, 232, 235.**Case 5: T = 4**- Numbers of the form 24U.- The number must be divisible by U.- Possible values for U are 0, 1, 2, 3, 4, 5, 6, and 8.- Valid numbers: 240, 241, 242, 243, 244, 245, 246, 248.**Case 6: T = 5**- Only the number 250 is considered.- 250 is divisible by 2, 5, and 0 (though division by 0 is undefined, it is included as per the problem statement).- Valid number: 250.Summarizing the counts from each case:- Case 1: 5 numbers- Case 2: 7 numbers- Case 3: 6 numbers- Case 4: 4 numbers- Case 5: 8 numbers- Case 6: 1 numberAdding these together, we get a total of 31 visible factor numbers between 200 and 250.

💡First, I recognize that sin 70^circ can be expressed as cos 20^circ using the co-function identity.Next, I use the double-angle identity for cosine, which states that cos 2theta = 1 - 2sin^2 theta. By substituting theta = 10^circ, the expression becomes cos 20^circ = 1 - 2sin^2 10^circ.Since sin 10^circ = k, I substitute k into the equation to get cos 20^circ = 1 - 2k^2.Therefore, sin 70^circ = 1 - 2k^2, which corresponds to option D.

💡Okay, so I have this probability problem here, and I need to figure out the probability that the product of three randomly selected distinct integers is odd. The integers are selected between 1 and 2020, inclusive. Each selected integer must be odd, and the sum of its digits must also be odd. The answer choices are given, so I need to determine which one is correct.First, let me understand the problem step by step. We are selecting three distinct integers from 1 to 2020. Each integer must satisfy two conditions: it must be odd, and the sum of its digits must be odd. Then, we need to find the probability that the product of these three integers is odd.Wait, hold on. The product of three integers is odd if and only if all three integers are odd. So, actually, the condition that each integer is odd is already given, right? Because the problem states that each selected integer must be odd. So, if all three integers are odd, their product will automatically be odd. So, does that mean the probability is 1? That can't be right because the answer choices don't include 1.Wait, maybe I misread the problem. Let me check again. It says, "Three distinct integers are selected at random between 1 and 2020, inclusive. Each selected integer must be odd, and the sum of its digits must also be odd. What is the probability p that the product of these three integers is odd?"Hmm, so actually, the selection is done such that each integer is odd and the sum of its digits is odd. So, we're not just selecting any three integers; we're selecting three integers that satisfy both being odd and having an odd digit sum. Then, we need the probability that their product is odd. But since all three integers are already odd, their product is necessarily odd. So, the probability should be 1.But that seems too straightforward, and the answer choices don't include 1. So, maybe I'm misunderstanding the problem. Perhaps the selection is done without considering the digit sum condition, and then we have to check if the product is odd, considering that each integer must be odd and have an odd digit sum.Wait, let me read the problem again carefully: "Three distinct integers are selected at random between 1 and 2020, inclusive. Each selected integer must be odd, and the sum of its digits must also be odd. What is the probability p that the product of these three integers is odd?"Hmm, so it's saying that each selected integer must be odd, and the sum of its digits must also be odd. So, the selection is restricted to integers that are odd and have an odd digit sum. Then, we need the probability that their product is odd. But if all three integers are odd, their product is automatically odd. So, the probability should be 1. But again, the answer choices don't include 1. So, maybe I'm misinterpreting the problem.Alternatively, perhaps the problem is saying that three integers are selected at random, and each must be odd, and the sum of its digits must be odd. So, the selection is done with the condition that each integer is odd and has an odd digit sum. Then, given that, we need the probability that the product is odd. But since all three integers are odd, the product is odd, so the probability is 1. But again, that's not an answer choice.Wait, perhaps the problem is not conditioning on the integers being odd and having an odd digit sum, but rather, it's saying that each integer must be odd, and the sum of its digits must also be odd. So, the integers are selected such that they are odd and their digit sums are odd. Then, given that, we need the probability that the product is odd. But again, since all three are odd, the product is odd, so probability is 1.But since the answer choices don't include 1, I must be misinterpreting the problem. Maybe the problem is asking for the probability that the product is odd, given that each integer is odd and has an odd digit sum. But that still seems like the product is odd with probability 1.Wait, perhaps the problem is not conditioning on the integers being odd and having an odd digit sum, but rather, it's saying that each integer must be odd, and the sum of its digits must also be odd. So, the integers are selected from the set of integers that are odd and have an odd digit sum. Then, the product will be odd if all three are odd, which they are, so probability is 1.But again, the answer choices don't include 1. So, maybe I'm missing something. Let me think differently.Perhaps the problem is not that the integers are selected such that they are odd and have an odd digit sum, but rather, that each integer is selected at random, and we have conditions: each must be odd, and the sum of its digits must be odd. So, it's like two separate conditions: being odd and having an odd digit sum.Wait, but if the integers are selected at random, then the probability that each is odd is 1/2, and the probability that the sum of its digits is odd is some other probability. But the problem says "each selected integer must be odd, and the sum of its digits must also be odd." So, it's not about the probability of selecting such integers, but rather, the integers are selected such that they satisfy both conditions.Wait, maybe the problem is asking for the probability that, when selecting three integers at random, each is odd and has an odd digit sum, and then the product is odd. But since the integers are already selected to be odd, the product is necessarily odd, so the probability is 1. But again, that's not an answer choice.Alternatively, perhaps the problem is asking for the probability that, when selecting three integers at random, each is odd and has an odd digit sum, and then the product is odd. But again, if the integers are selected to be odd, the product is odd, so probability is 1.Wait, maybe the problem is not conditioning on the integers being odd and having an odd digit sum, but rather, it's saying that each integer must be odd, and the sum of its digits must also be odd. So, the integers are selected such that they are odd, and their digit sums are odd. Then, the product is odd if all three are odd, which they are, so probability is 1.But since the answer choices don't include 1, I must be misinterpreting the problem. Maybe the problem is not about selecting integers that are odd and have an odd digit sum, but rather, it's about selecting integers such that each is odd, and the sum of its digits is odd. So, the integers are selected from the set of integers that are odd and have an odd digit sum. Then, the product is odd if all three are odd, which they are, so probability is 1.But again, the answer choices don't include 1. So, perhaps the problem is not about the product being odd, but about something else. Wait, no, the problem clearly states: "What is the probability p that the product of these three integers is odd?"Wait, maybe the problem is not that the integers are selected such that they are odd and have an odd digit sum, but rather, that each integer is selected at random, and we have to consider the probability that each is odd and has an odd digit sum, and then the product is odd. But that would be the same as the probability that each integer is odd and has an odd digit sum, and then the product is odd, which is 1 if all three are odd.Wait, I'm getting confused. Let me try to break it down.First, the total number of integers from 1 to 2020 is 2020.We need to select three distinct integers. So, the total number of ways to select three distinct integers is C(2020, 3).Now, the problem says: "Each selected integer must be odd, and the sum of its digits must also be odd." So, the selection is restricted to integers that are odd and have an odd digit sum.So, first, let's find how many integers between 1 and 2020 are odd. Since every other number is odd, there are 1010 odd numbers.Now, among these 1010 odd numbers, how many have an odd sum of digits?Hmm, that's a bit more involved. The sum of digits being odd depends on the number itself. For an odd number, the sum of its digits can be either odd or even. So, we need to find how many odd numbers between 1 and 2020 have an odd digit sum.This might require some calculation. Let's think about it.First, note that the sum of the digits of a number is odd if and only if the number of odd digits is odd. So, for an odd number, which ends with an odd digit, the sum of its digits is odd if the number of odd digits (including the last digit) is odd.Wait, but actually, the sum of digits is odd if the total number of odd digits is odd. So, for example, if a number has 1, 3, 5, etc., odd digits, then the sum is odd.But considering that the number itself is odd, it must end with an odd digit. So, the last digit is odd, which is one odd digit. Then, the rest of the digits can be either even or odd.So, for the sum of digits to be odd, the total number of odd digits must be odd. Since the last digit is already odd, the remaining digits must have an even number of odd digits to make the total number of odd digits odd.Wait, no. Let me clarify.If the last digit is odd, that's one odd digit. To have the total number of odd digits be odd, the remaining digits must have an even number of odd digits. Because 1 (from the last digit) plus an even number is odd.So, for a number to be odd and have an odd digit sum, it must have an odd number of odd digits, which, given that the last digit is odd, means that the remaining digits must have an even number of odd digits.So, how can we count the number of such numbers?This might be complicated, but perhaps we can approximate or find a pattern.Alternatively, maybe we can note that for numbers, the sum of digits being odd is roughly half the numbers, regardless of the number being odd or even.But wait, for odd numbers, is the sum of digits equally likely to be odd or even?Let me test with small numbers.From 1 to 10:Odd numbers: 1, 3, 5, 7, 9.Sum of digits:1: 1 (odd)3: 3 (odd)5: 5 (odd)7: 7 (odd)9: 9 (odd)So, all odd numbers from 1 to 10 have odd digit sums. Wait, that's interesting.Wait, but 11 is odd, sum of digits is 1+1=2 (even). So, 11 is odd, but sum of digits is even.Similarly, 13: 1+3=4 (even). 15: 1+5=6 (even). 17: 1+7=8 (even). 19: 1+9=10 (even).So, from 11 to 20, the odd numbers have even digit sums.Similarly, 21: 2+1=3 (odd). 23: 2+3=5 (odd). 25: 2+5=7 (odd). 27: 2+7=9 (odd). 29: 2+9=11 (odd).So, from 21 to 30, the odd numbers have odd digit sums.So, it alternates. From 1-10: odd numbers have odd digit sums. 11-20: odd numbers have even digit sums. 21-30: odd numbers have odd digit sums. And so on.So, in each block of 10 numbers, the odd numbers alternate between having odd and even digit sums.Therefore, in each block of 10, half of the odd numbers have odd digit sums, and half have even digit sums.Wait, but in 1-10, all 5 odd numbers have odd digit sums. In 11-20, all 5 odd numbers have even digit sums. So, in each block of 10, the odd numbers alternate between all having odd digit sums and all having even digit sums.Wait, that's not exactly half. It's 5 in one block, 0 in the next, then 5, then 0, etc.Wait, no, actually, in 1-10: 5 odd numbers, all with odd digit sums.In 11-20: 5 odd numbers, all with even digit sums.In 21-30: 5 odd numbers, all with odd digit sums.In 31-40: 5 odd numbers, all with even digit sums.And so on.So, in each block of 10, the odd numbers have either all odd or all even digit sums, alternating each block.Therefore, in the range 1-2020, which is 202 blocks of 10, each block contributing 5 odd numbers.In each pair of blocks (10-20), the first block has odd digit sums, the next has even, and so on.So, in 2020, how many blocks do we have? 2020 / 10 = 202 blocks.Each block of 10 has 5 odd numbers.In each pair of blocks, the first has 5 odd numbers with odd digit sums, the second has 5 odd numbers with even digit sums.Therefore, in 202 blocks, which is 101 pairs of blocks, each pair contributing 5 odd numbers with odd digit sums and 5 with even.Therefore, total number of odd numbers with odd digit sums is 101 * 5 = 505.Similarly, total number of odd numbers with even digit sums is also 505.Wait, but 2020 is exactly 202 blocks of 10, so 202 * 5 = 1010 odd numbers, which matches our earlier count.So, in total, there are 505 odd numbers with odd digit sums and 505 with even digit sums.Therefore, the number of integers between 1 and 2020 that are odd and have an odd digit sum is 505.So, now, the problem is selecting three distinct integers from these 505 integers.Wait, no. Wait, the problem says: "Three distinct integers are selected at random between 1 and 2020, inclusive. Each selected integer must be odd, and the sum of its digits must also be odd."So, the selection is done such that each integer is odd and has an odd digit sum. So, the total number of possible integers to select from is 505.Therefore, the number of ways to select three distinct integers is C(505, 3).But wait, the problem says "selected at random between 1 and 2020", but with the condition that each selected integer must be odd and have an odd digit sum. So, it's equivalent to selecting three integers from the 505 that satisfy both conditions.But then, the product of these three integers is odd if and only if all three are odd, which they are, since we selected them to be odd. So, the product is necessarily odd. Therefore, the probability is 1.But again, the answer choices don't include 1. So, I must be misinterpreting the problem.Wait, perhaps the problem is not conditioning on the integers being odd and having an odd digit sum, but rather, it's saying that each integer must be odd, and the sum of its digits must also be odd. So, the selection is done without any restriction, but we have to compute the probability that each integer is odd and has an odd digit sum, and then the product is odd.But that would be the same as the probability that each integer is odd and has an odd digit sum, and then the product is odd. But since the product is odd if and only if all three are odd, which is already part of the condition, so the probability would be the same as the probability that all three integers are odd and have odd digit sums.But then, the answer choices are fractions like 1/64, 1/8, etc., so maybe we need to compute the probability that all three integers are odd and have odd digit sums, and then the product is odd, which is 1. So, the probability is the same as the probability that all three integers are odd and have odd digit sums.Wait, but the problem says: "Three distinct integers are selected at random between 1 and 2020, inclusive. Each selected integer must be odd, and the sum of its digits must also be odd. What is the probability p that the product of these three integers is odd?"So, the way it's phrased is a bit confusing. It could be interpreted in two ways:1. We select three integers at random, and each must be odd and have an odd digit sum. Then, given that, what's the probability that their product is odd? Since all three are odd, the product is odd, so p=1.2. We select three integers at random, and we want the probability that each is odd and has an odd digit sum, and their product is odd. Since the product is odd if and only if all three are odd, which is already part of the condition, so p is equal to the probability that all three integers are odd and have odd digit sums.But the answer choices don't include 1, so it must be the second interpretation.Therefore, we need to compute the probability that when selecting three distinct integers at random from 1 to 2020, each is odd and has an odd digit sum. Then, since the product is odd if and only if all three are odd, which is already given, so the probability p is equal to the probability that all three integers are odd and have odd digit sums.So, let's compute that.First, the total number of ways to select three distinct integers from 1 to 2020 is C(2020, 3).The number of favorable outcomes is the number of ways to select three distinct integers that are odd and have an odd digit sum. As we calculated earlier, there are 505 such integers.Therefore, the number of favorable outcomes is C(505, 3).So, the probability p is C(505, 3) / C(2020, 3).But let's compute this.First, C(n, 3) = n(n-1)(n-2)/6.So, C(505, 3) = 505 * 504 * 503 / 6.C(2020, 3) = 2020 * 2019 * 2018 / 6.Therefore, p = [505 * 504 * 503 / 6] / [2020 * 2019 * 2018 / 6] = (505 * 504 * 503) / (2020 * 2019 * 2018).Simplify this fraction.First, note that 505 is exactly 2020 / 4, since 2020 / 4 = 505.Similarly, 504 is 2016 / 4, but 2016 is not directly related to 2019 or 2018.Wait, let's see:505 = 2020 / 4504 = 2016 / 4503 is prime, I think.2020 = 4 * 5052019 = 3 * 6732018 = 2 * 1009So, let's write the numerator and denominator:Numerator: 505 * 504 * 503Denominator: 2020 * 2019 * 2018 = (4 * 505) * (3 * 673) * (2 * 1009)So, let's substitute:Numerator: 505 * 504 * 503Denominator: 4 * 505 * 3 * 673 * 2 * 1009We can cancel out the 505 in numerator and denominator.So, we have:Numerator: 504 * 503Denominator: 4 * 3 * 673 * 2 * 1009Simplify denominator:4 * 3 * 2 = 24So, denominator: 24 * 673 * 1009Numerator: 504 * 503Now, let's compute 504 / 24:504 / 24 = 21So, numerator becomes 21 * 503Denominator becomes 673 * 1009So, p = (21 * 503) / (673 * 1009)Now, let's compute this fraction.First, compute numerator: 21 * 50321 * 500 = 1050021 * 3 = 63So, total numerator: 10500 + 63 = 10563Denominator: 673 * 1009Compute 673 * 1000 = 673,000673 * 9 = 6,057So, total denominator: 673,000 + 6,057 = 679,057So, p = 10,563 / 679,057Now, let's compute this fraction.First, let's see how many times 10,563 goes into 679,057.Compute 679,057 / 10,563 ≈ 64.28So, p ≈ 1 / 64.28 ≈ 0.01555Which is approximately 1/64.But let's check:1/64 = 0.015625Our computed p ≈ 0.01555, which is slightly less than 1/64.Wait, but let's see:10,563 * 64 = ?10,563 * 60 = 633,78010,563 * 4 = 42,252Total: 633,780 + 42,252 = 676,032But our denominator is 679,057, which is larger.So, 10,563 * 64 = 676,032Difference: 679,057 - 676,032 = 3,025So, 10,563 * 64 + 3,025 = 679,057Therefore, 10,563 / 679,057 = (10,563 * 64 + 3,025) / (10,563 * 64 + 3,025) = 1 / (64 + 3,025 / 10,563)But 3,025 / 10,563 ≈ 0.286So, p ≈ 1 / (64 + 0.286) ≈ 1 / 64.286 ≈ 0.01555Which is approximately 1/64, but slightly less.Therefore, p ≈ 1/64, but slightly less.Looking at the answer choices:(A) p = 1/64(B) p < 1/64(C) 1/64 < p < 1/8(D) p = 1/8(E) p > 1/8So, since p is approximately 0.01555, which is slightly less than 1/64 ≈ 0.015625, the correct answer is (B) p < 1/64.Wait, but let me double-check my calculations to make sure I didn't make a mistake.First, total number of integers: 2020Number of odd integers: 1010Number of odd integers with odd digit sums: 505So, the number of favorable integers is 505.Therefore, the number of ways to choose 3 is C(505, 3) = 505 * 504 * 503 / 6Total number of ways to choose 3 integers: C(2020, 3) = 2020 * 2019 * 2018 / 6So, p = [505 * 504 * 503] / [2020 * 2019 * 2018]As above, 505 = 2020 / 4, so:p = [ (2020 / 4) * 504 * 503 ] / [2020 * 2019 * 2018]Cancel 2020:p = [ (1/4) * 504 * 503 ] / [2019 * 2018]Now, 504 / 4 = 126So, p = [126 * 503] / [2019 * 2018]Compute 126 * 503:126 * 500 = 63,000126 * 3 = 378Total: 63,000 + 378 = 63,378Denominator: 2019 * 2018Compute 2019 * 2000 = 4,038,0002019 * 18 = 36,342Total: 4,038,000 + 36,342 = 4,074,342So, p = 63,378 / 4,074,342Simplify this fraction.Divide numerator and denominator by 6:63,378 / 6 = 10,5634,074,342 / 6 = 679,057So, p = 10,563 / 679,057 ≈ 0.01555Which is approximately 1/64.28, so slightly less than 1/64.Therefore, p < 1/64.So, the correct answer is (B) p < 1/64.

💡First, I need to understand what 4 tenths and 5 thousandths represent in decimal form. 4 tenths is equivalent to 0.4, and 5 thousandths is equivalent to 0.005.Next, I'll add these two values together to find their sum:0.4 + 0.005 = 0.405To find the average, I need to divide the sum by 2:0.405 ÷ 2 = 0.2025Therefore, the average of 4 tenths and 5 thousandths is 0.2025.

💡First, I need to determine the greatest common divisor (GCD) of 9240 and 6300. I'll use the Euclidean algorithm for this purpose.Starting with the two numbers:- GCD(9240, 6300)I'll subtract 6300 from 9240 to get 2940:- GCD(6300, 2940)Next, subtract 2940 from 6300 to get 3360:- GCD(2940, 3360)Then, subtract 2940 from 3360 to get 420:- GCD(2940, 420)Now, subtract 6 times 420 from 2940 to get 180:- GCD(420, 180)Subtract 2 times 180 from 420 to get 60:- GCD(180, 60)Finally, subtract 3 times 60 from 180 to get 0, so the GCD is 60.Next, I'll factorize 60 into its prime factors:- 60 = 2² × 3¹ × 5¹To find the number of positive divisors, I'll use the formula for the number of divisors based on prime factorization:- (2 + 1) × (1 + 1) × (1 + 1) = 3 × 2 × 2 = 12Therefore, 9240 and 6300 have 12 positive divisors in common.

💡Okay, so I have this problem where I need to place either a 0 or a 1 in each cell of a 5x5 grid. The condition is that the sum of every 2x2 subgrid should be divisible by 3. Also, I need to make sure that both 0s and 1s are present in the grid. Hmm, interesting. Let me try to figure this out step by step.First, I need to understand what it means for the sum of a 2x2 subgrid to be divisible by 3. Since each cell can only be 0 or 1, the sum of four cells can range from 0 to 4. But for the sum to be divisible by 3, the possible sums are 0, 3. Because 1, 2, and 4 are not divisible by 3. So, each 2x2 subgrid must sum to either 0 or 3.Wait, but if the sum is 0, that means all four cells in the subgrid are 0. If the sum is 3, then three of the cells are 1 and one is 0. So, every 2x2 subgrid must either be all 0s or have exactly three 1s and one 0.But the problem also says that both 0s and 1s must appear in the grid. So, I can't have an all-0 grid or an all-1 grid. There must be a mix of 0s and 1s.Now, let's think about how to arrange these 0s and 1s in the 5x5 grid. Maybe I can start by trying to create a pattern that satisfies the condition for every 2x2 subgrid.Let me consider the simplest case first. If I have a 2x2 grid, to satisfy the condition, it must either be all 0s or have three 1s and one 0. But since I need both 0s and 1s in the entire grid, I can't have all 0s. So, in a 2x2 grid, I need three 1s and one 0.But in a 5x5 grid, every 2x2 subgrid must satisfy this condition. That means overlapping subgrids will share some cells, so the placement of 0s and 1s must be consistent across the entire grid.Maybe I can try to create a repeating pattern. For example, if I place a 0 in certain positions and 1s elsewhere, ensuring that every 2x2 block has exactly one 0.Let me try to sketch a possible arrangement. Suppose I place 0s in a checkerboard pattern. That is, alternate 0s and 1s in each row and column. But wait, in a checkerboard pattern, each 2x2 subgrid would have two 0s and two 1s, which sums to 2. That's not divisible by 3. So, that won't work.Hmm, okay, so the checkerboard pattern doesn't satisfy the condition. Maybe I need a different pattern.What if I place 0s in every third cell? Let's see. If I have a row like 1,1,0,1,1, then the next row could be 1,1,0,1,1, and so on. But then, looking at a 2x2 subgrid, it would have two 1s and two 1s, which sums to 4, not divisible by 3. That doesn't work either.Wait, maybe I need to place 0s in a way that every 2x2 subgrid has exactly one 0. So, if I can arrange the 0s such that no two 0s are adjacent in any direction, that might work. Let me try that.Imagine placing 0s in every other cell, but not in a checkerboard pattern. Maybe staggered in such a way that each 2x2 subgrid contains exactly one 0. For example:Row 1: 1,1,0,1,1Row 2: 1,1,0,1,1Row 3: 0,0,1,0,0Row 4: 1,1,0,1,1Row 5: 1,1,0,1,1Wait, let's check the sum of a 2x2 subgrid here. Take the top-left 2x2 subgrid:1,11,1Sum is 4, which is not divisible by 3. Oops, that doesn't work. Maybe I need to adjust the placement.What if I shift the 0s in each row? Like:Row 1: 1,1,0,1,1Row 2: 1,0,1,0,1Row 3: 0,1,1,1,0Row 4: 1,0,1,0,1Row 5: 1,1,0,1,1Now, let's check the top-left 2x2 subgrid:1,11,0Sum is 3, which is good. Next, the subgrid starting at (1,2):1,00,1Sum is 2, which is not divisible by 3. Hmm, that's a problem.Maybe I need a different approach. Perhaps using a periodic pattern where the 0s are spaced out in both rows and columns.Let me try placing 0s in a diagonal pattern. For example:Row 1: 1,1,1,1,1Row 2: 1,0,1,0,1Row 3: 1,1,1,1,1Row 4: 1,0,1,0,1Row 5: 1,1,1,1,1Now, let's check the sums. Take the subgrid starting at (1,1):1,11,0Sum is 3, good.Subgrid starting at (1,2):1,10,1Sum is 3, good.Subgrid starting at (2,1):1,01,1Sum is 3, good.Subgrid starting at (2,2):0,11,1Sum is 3, good.Seems like this pattern might work. Let's check another subgrid, say starting at (3,1):1,11,0Sum is 3, good.And starting at (4,1):1,01,1Sum is 3, good.It looks like this pattern satisfies the condition for all 2x2 subgrids. Plus, both 0s and 1s are present in the grid.Wait, but let me double-check. Is there any 2x2 subgrid that might not sum to 3? Let's see:Take the subgrid starting at (3,2):1,10,1Sum is 3, good.And starting at (4,2):0,11,1Sum is 3, good.Okay, seems consistent. So, this arrangement works.But just to be thorough, let me try another pattern to see if there are multiple solutions or if this is the only way.Suppose I place 0s in a different diagonal:Row 1: 1,1,1,1,1Row 2: 1,1,0,1,1Row 3: 1,0,1,0,1Row 4: 1,1,0,1,1Row 5: 1,1,1,1,1Now, check the sums:Subgrid (1,1):1,11,1Sum is 4, which is not divisible by 3. So, this pattern doesn't work.Hmm, so the initial pattern with 0s in every other cell in rows 2 and 4 seems to work. Let me confirm again.Yes, in that pattern, every 2x2 subgrid has exactly one 0 and three 1s, summing to 3. And since the entire grid has both 0s and 1s, it satisfies all conditions.I think I've found a valid solution. To be completely sure, I'll try to visualize the entire grid and check a few more subgrids.Here's the grid:Row 1: 1,1,1,1,1Row 2: 1,0,1,0,1Row 3: 1,1,1,1,1Row 4: 1,0,1,0,1Row 5: 1,1,1,1,1Now, let's check the subgrid starting at (3,3):1,10,1Sum is 3, good.Subgrid starting at (4,3):0,11,1Sum is 3, good.And the very last 2x2 subgrid starting at (4,4):0,11,1Sum is 3, good.Alright, I think this works. So, the answer is yes, it's possible, and one such arrangement is the grid I've described.