💡First, identify the initial number of candies Tom had, which is 2 pieces.Next, add the candies given to him by his friend: 2 + 7 = 9 pieces.Tom now has a total of 19 candies. To find out how many candies he bought, subtract the candies he already had from the total: 19 - 9 = 10 pieces.Therefore, Tom bought 10 pieces of candy.

💡Okay, so I have this problem here where a random variable ξ follows a normal distribution N(3,4). That means it has a mean of 3 and a variance of 4, right? So the standard deviation would be the square root of 4, which is 2. The problem says that the probability that ξ is less than 2a - 3 is equal to the probability that ξ is greater than a + 2. I need to find the value of a. Hmm, okay. So, mathematically, that's written as P(ξ < 2a - 3) = P(ξ > a + 2). Since ξ is normally distributed, I know that the normal distribution is symmetric around its mean. So, if I can figure out the symmetry point here, maybe I can set up an equation. Let me recall that for a normal distribution N(μ, σ²), the probabilities on either side of the mean are equal. So, if I have two points on the distribution where the probabilities are equal, those points should be equidistant from the mean but on opposite sides. In this case, the mean μ is 3. So, the two points 2a - 3 and a + 2 should be symmetric around 3. That means the distance from 2a - 3 to 3 should be the same as the distance from 3 to a + 2. Wait, actually, since P(ξ < 2a - 3) = P(ξ > a + 2), that implies that 2a - 3 and a + 2 are symmetric around the mean. So, the midpoint between 2a - 3 and a + 2 should be equal to the mean, which is 3. Let me write that down: Midpoint = ( (2a - 3) + (a + 2) ) / 2 = 3So, let's compute that:(2a - 3 + a + 2) / 2 = 3Combine like terms:(3a - 1) / 2 = 3Multiply both sides by 2:3a - 1 = 6Add 1 to both sides:3a = 7Divide both sides by 3:a = 7/3Wait, that seems straightforward. So, a is 7/3. Let me check if that makes sense.If a is 7/3, then 2a - 3 is 2*(7/3) - 3 = 14/3 - 9/3 = 5/3. And a + 2 is 7/3 + 6/3 = 13/3. So, the two points are 5/3 and 13/3. The mean is 3, which is 9/3. The distance from 5/3 to 9/3 is 4/3, and the distance from 9/3 to 13/3 is also 4/3. So, they are symmetric around the mean. That makes sense because the normal distribution is symmetric, so the probabilities on either side of the mean should be equal. Therefore, the value of a is 7/3, which is option A.**Final Answer**boxed{A}

💡Okay, so I have this problem where I need to use the digits 0, 1, 2, 3, and 4 to form integers that satisfy two different conditions. The first part is about forming all four-digit integers, and the second part is about forming five-digit integers without repetition that are greater than 21000. Hmm, let me try to figure this out step by step.Starting with part (I): All four-digit integers. So, I need to create four-digit numbers using the digits 0, 1, 2, 3, and 4. But wait, four-digit numbers can't start with 0, right? Because then it would be a three-digit number. So, the first digit has to be one of the non-zero digits: 1, 2, 3, or 4. That gives me 4 choices for the first digit.Now, for the remaining three digits, I can use any of the remaining digits, including 0, but I can repeat digits unless specified otherwise. Wait, the problem doesn't say anything about repetition, so I think repetition is allowed here. So, for each of the remaining three positions, I can choose any of the five digits. That would mean for each of the second, third, and fourth positions, I have 5 choices each.So, calculating the total number of four-digit integers, it would be 4 (choices for the first digit) multiplied by 5 (choices for the second digit) multiplied by 5 (choices for the third digit) multiplied by 5 (choices for the fourth digit). That would be 4 * 5 * 5 * 5. Let me compute that: 4 * 5 is 20, 20 * 5 is 100, 100 * 5 is 500. So, are there 500 four-digit integers possible?Wait, but hold on. The problem says "use the five digits 0, 1, 2, 3, 4." Does that mean we have to use each digit exactly once? Hmm, it doesn't specify, so I think repetition is allowed unless stated otherwise. So, my initial calculation of 500 should be correct. But let me double-check.If repetition is not allowed, then the number of four-digit integers would be different. For the first digit, 4 choices, then for the second digit, 4 remaining choices (since we can't repeat), third digit, 3 choices, and fourth digit, 2 choices. So, that would be 4 * 4 * 3 * 2 = 96. Hmm, that's a big difference. So, which one is it?Looking back at the problem statement: "Use the five digits 0, 1, 2, 3, 4 to form integers that satisfy the following conditions: (I) All four-digit integers." It doesn't specify whether digits can be repeated or not. Hmm. In combinatorics problems, unless specified, sometimes repetition is allowed, but sometimes it's implied that digits are used without repetition. I'm a bit confused here.Wait, in the second part, it specifically says "without repetition," so maybe in the first part, repetition is allowed. That would make sense because otherwise, the first part would have been specified as well. So, I think for part (I), repetition is allowed, so 500 four-digit integers. But just to be safe, maybe I should consider both cases.But the initial answer I saw had 96, which is the number without repetition. Hmm, maybe I misread the problem. Let me check again. It says "use the five digits 0, 1, 2, 3, 4 to form integers." So, does that mean each digit can be used multiple times? Or does it mean that each digit is used exactly once? Hmm, it's a bit ambiguous.Wait, in the second part, it's specified as "without repetition," so maybe in the first part, repetition is allowed. So, I think I should go with 500 for part (I). But I'm not entirely sure. Maybe I should look for similar problems or think about the context. In many problems, unless specified, digits can be repeated unless it's specified otherwise. So, yeah, I think 500 is correct.Moving on to part (II): Five-digit integers without repetition that are greater than 21000. Okay, so now we're dealing with five-digit numbers, using each digit only once, and the number has to be greater than 21000.First, let's think about the range of five-digit numbers. The smallest five-digit number is 10000, and the largest is 99999. But in our case, we're using only the digits 0, 1, 2, 3, 4, so the largest five-digit number we can form is 44444, but since we can't repeat digits, the largest would be 43210 or something like that.But we need numbers greater than 21000. So, numbers from 21001 up to 43210, but only using the digits 0,1,2,3,4 without repetition.To approach this, let's consider the first digit. Since it's a five-digit number, the first digit can't be 0. So, possible first digits are 1, 2, 3, 4.But since the number has to be greater than 21000, let's think about the possible first digits.If the first digit is 1, then the number would be between 10000 and 19999, which is less than 21000. So, numbers starting with 1 are too small. So, we can exclude first digits of 1.If the first digit is 2, then the number is between 20000 and 29999. But we need numbers greater than 21000, so numbers starting with 21, 22, 23, 24, etc. But since we can't repeat digits, and we're using digits 0,1,2,3,4, let's think about how to count these.If the first digit is 2, then the second digit has to be such that the number is greater than 21000. So, the second digit can't be 0 or 1? Wait, no. Let's think about it.If the first two digits are 21, then the number is 21xxx, which is exactly 21000. But we need numbers greater than 21000. So, 21xxx would be 21000 to 21999, but since our digits are limited, the maximum number starting with 21 would be 21430 or something like that.But wait, actually, if the first two digits are 21, then the number is 21000 plus something, but since we can't have repetition, the digits after 21 have to be unique and not include 2 or 1.Wait, maybe it's better to break it down into cases.Case 1: First digit is 2.Then, the second digit can be 1, 3, or 4. Because if the second digit is 0, then the number would be 20xxx, which is less than 21000. So, to make the number greater than 21000, the second digit must be at least 1. But since the first digit is 2, the second digit can be 1, 3, or 4.Wait, but if the second digit is 1, then the number is 21xxx. But 21xxx is 21000 to 21999, which is exactly the boundary. So, we need numbers greater than 21000, so 21xxx is acceptable as long as the remaining digits make the number greater than 21000.But since we can't have repetition, if the first two digits are 2 and 1, then the remaining digits have to be 0, 3, 4, but arranged in such a way that the number is greater than 21000.Wait, actually, 21000 is the lower bound, so any number starting with 21 and then followed by digits that make it larger than 21000 is acceptable. But since we're using digits without repetition, the number 21000 itself can't be formed because we can't have two 0s. So, the smallest number starting with 21 would be 21034 or something like that, which is greater than 21000.So, in this case, if the first digit is 2, the second digit can be 1, 3, or 4. Let's consider each subcase.Subcase 1: First digit is 2, second digit is 1.Then, the remaining three digits have to be chosen from 0, 3, 4, without repetition. So, how many ways can we arrange these? It's the number of permutations of 3 digits from 3 remaining digits, which is 3! = 6.Subcase 2: First digit is 2, second digit is 3.Then, the remaining three digits are 0, 1, 4. Again, number of permutations is 3! = 6.Subcase 3: First digit is 2, second digit is 4.Then, the remaining three digits are 0, 1, 3. Number of permutations is 3! = 6.So, total numbers when the first digit is 2 is 6 + 6 + 6 = 18.Case 2: First digit is 3 or 4.If the first digit is 3 or 4, then the number is automatically greater than 21000 because 3xxxx and 4xxxx are both greater than 21000.So, for first digit 3 or 4, we have two choices for the first digit. Then, the remaining four digits can be any permutation of the remaining four digits (since we can't repeat digits). So, for each first digit, the number of permutations is 4! = 24.Therefore, total numbers for first digit 3 or 4 is 2 * 24 = 48.Adding up both cases, total numbers greater than 21000 are 18 + 48 = 66.Wait, let me double-check that. For first digit 2, we have 18 numbers, and for first digits 3 and 4, we have 48 numbers. So, 18 + 48 is indeed 66.But just to make sure, let's think about it another way. The total number of five-digit numbers without repetition using digits 0,1,2,3,4 is 4 * 4 * 3 * 2 * 1 = 96. Wait, no, that's not right. Wait, the total number of five-digit numbers without repetition is 4 choices for the first digit (1,2,3,4), then 4 choices for the second digit (including 0 but excluding the first digit), then 3, then 2, then 1. So, 4 * 4 * 3 * 2 * 1 = 96. So, total five-digit numbers without repetition are 96.Out of these 96, how many are greater than 21000? Well, numbers starting with 1 are all less than 21000, so numbers starting with 1 are 1 _ _ _ _, which is 1 followed by permutations of the remaining four digits. So, how many are those? For first digit 1, the remaining four digits can be any permutation of 0,2,3,4, which is 4! = 24. So, 24 numbers start with 1, which are all less than 21000.Therefore, numbers greater than or equal to 20000 are 96 - 24 = 72. But we need numbers greater than 21000, so we have to subtract the numbers between 20000 and 21000.Numbers starting with 20 _ _ _ are between 20000 and 20999. So, how many such numbers are there? First digit is 2, second digit is 0, then the remaining three digits are permutations of 1,3,4. So, that's 3! = 6 numbers.Therefore, numbers greater than 21000 would be total numbers starting with 2 or higher (72) minus numbers starting with 20 (6). So, 72 - 6 = 66. That matches our earlier calculation.So, yeah, the total number of five-digit integers without repetition greater than 21000 is 66.Wait, but in the initial problem statement, it says "use the five digits 0,1,2,3,4." So, does that mean we have to use all five digits? Because in the second part, it's about five-digit integers, so yes, we have to use all five digits without repetition. So, that makes sense.In part (I), it's about four-digit integers, so we can use any of the digits, possibly with repetition, unless specified otherwise. But since the second part specifies without repetition, maybe the first part allows repetition. So, going back to part (I), if repetition is allowed, then the number of four-digit integers is 4 * 5 * 5 * 5 = 500. But if repetition is not allowed, it's 4 * 4 * 3 * 2 = 96.But in the initial answer, it was 96. Hmm, maybe I should consider that in part (I), repetition is not allowed. Let me think again.The problem says "use the five digits 0,1,2,3,4 to form integers." It doesn't specify whether digits can be repeated or not. In part (II), it specifically says "without repetition," so maybe in part (I), repetition is allowed. But sometimes, in problems like this, when they say "use the digits," it might imply using each digit exactly once, but that's not always the case.Wait, in part (I), it's about four-digit integers. So, if we have to use the five digits, but form four-digit numbers, that would imply that we have to use four of the five digits, possibly with repetition. But the problem doesn't specify whether repetition is allowed or not. Hmm.Wait, maybe the problem is that in part (I), it's about forming four-digit integers using the five digits, but without repetition. So, each four-digit number uses four distinct digits from the set {0,1,2,3,4}. That would make sense because in part (II), it's specified as without repetition, so maybe part (I) also implies without repetition.So, if that's the case, then for part (I), the number of four-digit integers without repetition would be calculated as follows:First digit: can't be 0, so 4 choices (1,2,3,4).Then, the remaining three digits can be any of the remaining four digits, including 0, but without repetition.So, for the second digit: 4 choices (since one digit is used already).Third digit: 3 choices.Fourth digit: 2 choices.So, total number is 4 * 4 * 3 * 2 = 96. That's the same as the initial answer.So, maybe in part (I), repetition is not allowed, hence 96 four-digit integers.But then, in part (II), it's specified as without repetition, so that's consistent.So, perhaps the problem implies that in both parts, repetition is not allowed. So, in part (I), it's four-digit numbers without repetition, and in part (II), it's five-digit numbers without repetition.Therefore, the initial answer of 96 for part (I) and 66 for part (II) is correct.But just to make sure, let's think about part (I) again. If repetition is allowed, it's 500, but if not, it's 96. Since the problem didn't specify, but part (II) specifies without repetition, maybe part (I) also implies without repetition. So, 96 is the answer for part (I).Alright, so to summarize:(I) Four-digit integers without repetition: 96.(II) Five-digit integers without repetition greater than 21000: 66.Yeah, that seems consistent.

💡Alright, so I've got these two math problems to solve. Let's take them one at a time. The first one is: -24 - (-15) + (-1) + (-15)Okay, let's break this down. I remember that subtracting a negative is like adding a positive. So, -24 - (-15) should be the same as -24 + 15. Let me write that down:-24 + 15Hmm, what's -24 + 15? Well, if I have a debt of 24 and then I get 15, I still owe 9. So that's -9. Now, the next part is + (-1). Adding a negative is like subtracting. So, -9 - 1 is -10. Then, we have + (-15) again. So, -10 - 15 is -25. Wait, let me double-check that. Starting from -24, adding 15 gets me to -9. Then subtracting 1 gets me to -10, and subtracting another 15 gets me to -25. Yeah, that seems right.Okay, so the first problem is -25.Now, the second problem is:(-27) ÷ frac{3}{2} × frac{2}{3}Hmm, division and multiplication with fractions. I think I need to remember the order of operations, which is division and multiplication from left to right. So, first, I'll do -27 ÷ frac{3}{2}.Dividing by a fraction is the same as multiplying by its reciprocal. So, -27 ÷ frac{3}{2} is the same as -27 × frac{2}{3}. Let me calculate that:-27 × frac{2}{3}Well, 27 divided by 3 is 9, so -9 × 2 is -18.Now, I need to multiply that result by frac{2}{3}. So, -18 × frac{2}{3}.Again, 18 divided by 3 is 6, so -6 × 2 is -12.Let me see if that makes sense. Starting with -27, dividing by frac{3}{2} is like multiplying by frac{2}{3}, which gives -18. Then multiplying by frac{2}{3} gives -12. Yeah, that seems consistent.So, the second problem is -12.Wait, but let me make sure I didn't mix up the order. If I had done the multiplication first, would it change anything? Let's see:(-27) ÷ frac{3}{2} × frac{2}{3}If I first multiply frac{3}{2} × frac{2}{3}, that's 1. So, -27 ÷ 1 is -27. But that's different from what I got before. Hmm, so which one is correct?Oh, I think I remember that division and multiplication are at the same level in order of operations, so you go from left to right. So, I should have done -27 ÷ frac{3}{2} first, then multiplied by frac{2}{3}, which gives -12. If I had done it the other way, I would have gotten -27, which is different. So, I think my first answer of -12 is correct.Yeah, I think that's right. So, the second problem is -12.

💡Okay, so I have this geometry problem here: A circle intersects the sides of a quadrilateral, creating equal chords. I need to prove that a circle can be inscribed in this quadrilateral. Hmm, interesting. Let me try to visualize this.First, let me recall what a quadrilateral is. It's a four-sided polygon, right? So, it has four sides and four vertices. Now, a circle intersects each side of this quadrilateral, and each intersection creates a chord of equal length. So, if I imagine a circle cutting through each side of the quadrilateral, the segments (chords) created on each side are all the same length.I need to show that such a quadrilateral can have an inscribed circle. An inscribed circle, or incircle, is a circle that touches all four sides of the quadrilateral from the inside. For a circle to be inscribed, the quadrilateral must be tangential, meaning it has an incircle tangent to all four sides. I remember that a necessary and sufficient condition for a quadrilateral to be tangential is that the sums of the lengths of its opposite sides are equal. So, if I can show that the sums of the lengths of the opposite sides of this quadrilateral are equal, then it must be tangential, and thus, a circle can be inscribed in it.Alright, so maybe I can use the fact that the chords created by the circle are equal in length to show that the sums of the opposite sides are equal. Let me think about how the chords relate to the sides of the quadrilateral.Each chord is a segment on a side of the quadrilateral, created by the intersection with the circle. Since all chords are equal, let's denote their length as 2a. So, each chord has a length of 2a. Now, if I consider the distance from the center of the circle to each side of the quadrilateral, this distance should be related to the length of the chord.I remember that in a circle, the length of a chord is related to the radius and the distance from the center to the chord. Specifically, the length of a chord is given by 2√(r² - d²), where r is the radius of the circle and d is the distance from the center to the chord. Since all chords are equal, their lengths are the same, which implies that the distance from the center to each side of the quadrilateral is the same. Let's denote this common distance as d.So, if the distance from the center of the circle to each side of the quadrilateral is the same, that means the center is equidistant from all four sides. In other words, the center is the incenter of the quadrilateral. The incenter is the point equidistant from all sides, and it's the center of the incircle.Wait, so if the center of the given circle is equidistant from all four sides of the quadrilateral, that means the quadrilateral has an incircle centered at that point. Therefore, the quadrilateral is tangential, and the given circle is its incircle. So, this would mean that a circle can indeed be inscribed in the quadrilateral.But let me make sure I'm not skipping any steps here. I need to formally show that the sums of the opposite sides are equal. Maybe I can express the lengths of the sides in terms of the chords and the distances.Let's denote the quadrilateral as ABCD, with sides AB, BC, CD, and DA. The circle intersects each side, creating chords of length 2a. Let's denote the points where the circle intersects side AB as P and Q, side BC as R and S, side CD as T and U, and side DA as V and W.Since each chord is of length 2a, the segments AP, PQ, QB, BR, RS, SC, CT, TU, UD, DV, VW, and WA are all related to the chord length. But wait, actually, each side of the quadrilateral is divided into two segments by the circle. For example, side AB is divided into segments AP and QB, each of length a. Similarly, side BC is divided into BR and SC, each of length a, and so on.Wait, no, that might not be accurate. The chord length is 2a, so the segment from one intersection point to the other on a side is 2a. So, for side AB, the chord PQ has length 2a, meaning that AP and QB are each of length a. Similarly, for side BC, the chord RS has length 2a, so BR and SC are each of length a. The same applies to sides CD and DA.So, if I denote the lengths of the sides as follows:- AB = AP + PQ + QB = a + 2a + a = 4a? Wait, no, that can't be right. Because PQ is the chord, which is 2a, but AP and QB are the segments from the vertices to the points of intersection. So, actually, AB is divided into AP and QB, each of length a, with the chord PQ in between. So, AB = AP + PQ + QB = a + 2a + a = 4a. Similarly, BC = BR + RS + SC = a + 2a + a = 4a. Wait, but that would mean all sides are equal, which would make the quadrilateral a rhombus. But a rhombus is a special case of a tangential quadrilateral, but not all tangential quadrilaterals are rhombuses.Hmm, maybe I'm misunderstanding the problem. It says the circle intersects the sides, creating equal chords. So, perhaps each side is intersected by the circle at two points, creating a chord of length 2a on each side. So, each side has a chord of length 2a, but the sides themselves can be longer than 2a.Wait, let me clarify. If a circle intersects a side of a quadrilateral, creating a chord, that chord is a segment on the side. So, if the chord length is 2a, then the side is divided into two segments by the chord. So, for example, on side AB, the chord PQ has length 2a, so AP and QB are the segments from the vertices to the points of intersection. So, AB = AP + PQ + QB. But PQ is the chord, which is 2a, so AB = AP + 2a + QB.But the problem states that all chords are equal, so PQ = RS = TU = VW = 2a. However, the lengths AP, QB, BR, SC, etc., can vary. So, the sides AB, BC, CD, DA can have different lengths, but each has a chord of length 2a.But how does this help me show that the sums of the opposite sides are equal?Maybe I need to consider the distances from the center of the circle to each side. Since all chords are equal, the distance from the center to each side must be equal, as I thought earlier. Let me denote the center of the circle as O. Then, the distance from O to each side is d, and since the chord length is 2a, we have the relationship:Chord length = 2√(r² - d²) = 2aSo, √(r² - d²) = aTherefore, r² - d² = a²So, d² = r² - a²This tells me that the distance from the center O to each side is the same, which is √(r² - a²). Therefore, O is equidistant from all four sides of the quadrilateral, which means it is the incenter of the quadrilateral.Since the incenter exists, the quadrilateral must be tangential, meaning it has an incircle. Therefore, a circle can be inscribed in the quadrilateral.Wait, but I need to make sure that this reasoning is rigorous. Let me try to formalize it.Given a quadrilateral ABCD, and a circle with center O intersecting each side, creating chords of equal length 2a. For each side, the distance from O to the side is d, and from the chord length formula, we have:Chord length = 2√(r² - d²) = 2aSo, √(r² - d²) = aSquaring both sides, r² - d² = a²Therefore, d² = r² - a²Since this is true for each side, the distance d from O to each side is the same. Therefore, O is equidistant from all four sides of the quadrilateral, which is the definition of the incenter. Hence, the quadrilateral is tangential, and a circle can be inscribed in it.Yes, that seems to make sense. So, the key idea is that equal chords imply equal distances from the center to each side, which in turn implies the existence of an incircle.I think I've got it. So, to summarize:1. The circle intersects each side of the quadrilateral, creating chords of equal length 2a.2. For each side, the distance from the center O to the side is d, and from the chord length formula, we have d = √(r² - a²).3. Since all chords are equal, d is the same for all sides.4. Therefore, O is equidistant from all four sides, making it the incenter.5. Hence, the quadrilateral is tangential, and a circle can be inscribed in it.I think that covers it. I should probably write this up more formally for the final answer.

💡Alright, so I have this problem where Dani wrote the integers from 1 to N, and she used the digit 1 fifteen times and the digit 2 fourteen times. I need to find out what N is. Hmm, okay, let's break this down step by step.First, I think I need to figure out how often the digits 1 and 2 appear as we count up from 1 to some number N. So, maybe I should start by looking at how these digits appear in different ranges of numbers.Let me start with the single-digit numbers, from 1 to 9. In this range, each digit from 1 to 9 appears exactly once. So, the digit 1 appears once, and the digit 2 also appears once. That's straightforward.Now, moving on to two-digit numbers. Let's consider the numbers from 10 to 19. In this range, every number has a '1' in the tens place. So, that's ten times the digit 1 right there. Additionally, in the number 12, there's also a '2' in the ones place. So, in this range, the digit 1 appears 11 times (10 in the tens place and 1 in the ones place), and the digit 2 appears once.Next, let's look at the numbers from 20 to 29. Here, every number has a '2' in the tens place, so that's ten times the digit 2. In the number 21, there's also a '1' in the ones place. So, in this range, the digit 1 appears once, and the digit 2 appears 11 times (10 in the tens place and 1 in the ones place).Moving on to the numbers from 30 to 39. In this range, the tens digit is '3', so neither 1 nor 2 appears in the tens place. However, in the number 31, there's a '1' in the ones place, and in the number 32, there's a '2' in the ones place. So, in this range, both digits 1 and 2 appear once each.So, let's summarize what we have so far:- From 1 to 9: - Digit 1: 1 time - Digit 2: 1 time- From 10 to 19: - Digit 1: 11 times - Digit 2: 1 time- From 20 to 29: - Digit 1: 1 time - Digit 2: 11 times- From 30 to 39: - Digit 1: 1 time - Digit 2: 1 timeAdding these up:- Total digit 1: 1 + 11 + 1 + 1 = 14- Total digit 2: 1 + 1 + 11 + 1 = 14Hmm, so up to 39, both digits 1 and 2 have been used 14 times each. But the problem states that digit 1 was used fifteen times and digit 2 was used fourteen times. So, we need to go a bit further.Let's check the next few numbers:- 40: No 1s or 2s.- 41: There's a '1' in the ones place.So, writing the number 41 adds one more '1' to our count. Now, the total count for digit 1 becomes 15, and digit 2 remains at 14.Therefore, N must be 41 because that's when the digit 1 reaches fifteen times and the digit 2 remains at fourteen times.Wait, let me double-check to make sure I didn't miss anything. From 1 to 39, we have 14 of each digit. Then, 40 doesn't add anything, and 41 adds one more '1'. So, yes, that should be correct.I think that's it. N is 41.