answer：Okay, so I have this problem where I need to find positive integers ( a ) and ( b ) such that ( a^2 - b^4 = 2009 ). Then, I have to find ( a + b ). Hmm, let me think about how to approach this.First, I know that both ( a ) and ( b ) are positive integers, so I can try plugging in some values for ( b ) and see if ( a ) comes out as an integer. Since ( b ) is raised to the fourth power, it might not be too large because ( b^4 ) grows quickly. Let me list out some fourth powers to get a sense of the possible values.Let's see:- ( 1^4 = 1 )- ( 2^4 = 16 )- ( 3^4 = 81 )- ( 4^4 = 256 )- ( 5^4 = 625 )- ( 6^4 = 1296 )- ( 7^4 = 2401 )Wait, ( 7^4 ) is 2401, which is already larger than 2009. So, ( b ) can't be 7 or higher because ( b^4 ) would make ( a^2 ) negative, which isn't possible since ( a ) is a positive integer. So, ( b ) can only be from 1 to 6.Now, let me compute ( a^2 = 2009 + b^4 ) for each ( b ) from 1 to 6 and check if the result is a perfect square.Starting with ( b = 1 ):( a^2 = 2009 + 1 = 2010 ). Is 2010 a perfect square? Let me see. The square of 44 is 1936, and 45 squared is 2025. So, 2010 is between 44² and 45². Not a perfect square.Next, ( b = 2 ):( a^2 = 2009 + 16 = 2025 ). Hmm, 2025. I remember that 45 squared is 2025. So, ( a = 45 ). That works!Wait, so if ( b = 2 ), then ( a = 45 ). So, ( a + b = 45 + 2 = 47 ). Is that the answer? Let me check the other values just to be thorough.( b = 3 ):( a^2 = 2009 + 81 = 2090 ). Is 2090 a perfect square? 45² is 2025, 46² is 2116. So, 2090 is between them, not a perfect square.( b = 4 ):( a^2 = 2009 + 256 = 2265 ). Let's see. 47² is 2209, 48² is 2304. So, 2265 is in between, not a perfect square.( b = 5 ):( a^2 = 2009 + 625 = 2634 ). 51² is 2601, 52² is 2704. So, 2634 is in between, not a perfect square.( b = 6 ):( a^2 = 2009 + 1296 = 3305 ). 57² is 3249, 58² is 3364. So, 3305 is between them, not a perfect square.So, the only value that works is ( b = 2 ), giving ( a = 45 ). Therefore, ( a + b = 45 + 2 = 47 ).Wait, let me double-check my calculations to make sure I didn't make a mistake.For ( b = 2 ):( 2^4 = 16 ), so ( a^2 = 2009 + 16 = 2025 ). And 45² is indeed 2025 because 45 times 45 is 2025. So that's correct.Just to be thorough, let me check ( b = 1 ) again:( 1^4 = 1 ), so ( a^2 = 2010 ). 44² is 1936, 45² is 2025, so 2010 isn't a square. Correct.( b = 3 ): 3^4 is 81, so 2009 + 81 is 2090. 45² is 2025, 46² is 2116, so 2090 isn't a square. Correct.( b = 4 ): 4^4 is 256, so 2009 + 256 is 2265. 47² is 2209, 48² is 2304, so 2265 isn't a square. Correct.( b = 5 ): 5^4 is 625, so 2009 + 625 is 2634. 51² is 2601, 52² is 2704, so 2634 isn't a square. Correct.( b = 6 ): 6^4 is 1296, so 2009 + 1296 is 3305. 57² is 3249, 58² is 3364, so 3305 isn't a square. Correct.So, yeah, the only solution is ( b = 2 ) and ( a = 45 ), so ( a + b = 47 ).I think that's solid. I don't see any mistakes in my reasoning.**Final Answer**The value of ( a + b ) is boxed{47}.

answer：Okay, so I have this problem about two circles, one larger with center P and radius p, and the smaller one with center Q and radius q. They tell me to draw the line segment PQ, which connects the two centers. Then, I need to figure out which of the given statements is false. The options are about the relationships between p, q, and the length of PQ.Let me recall what I know about circles and their positions relative to each other. When two circles are drawn, the distance between their centers (PQ) can determine their relationship—whether they're separate, touching, intersecting, or one inside the other.First, if the distance PQ is equal to p + q, that means the two circles are externally tangent to each other. They just touch at one point without overlapping. So, statement B says "p + q can be equal to PQ," which is definitely true because that's the case when they're externally tangent.Next, if PQ is equal to p - q (assuming p > q), then the smaller circle is just touching the larger one from the inside. That's internal tangency. So, statement A says "p - q can be equal to PQ," which is also true because that's when the smaller circle is inside the larger one and touching at one point.Now, statement C says "p + q can be less than PQ." Hmm, let me think. If p + q is less than PQ, that would mean the two circles are separate and don't intersect at all. Each circle is entirely outside the other. Is that possible? Yes, definitely. For example, if you have two circles far apart, their centers are much farther apart than the sum of their radii. So, statement C is true as well.Moving on to statement D: "p - q can be less than PQ." Wait, p - q is the difference in radii. If p - q is less than PQ, that means the distance between centers is greater than the difference in radii. What does that imply? If PQ is greater than p - q, then the smaller circle isn't entirely inside the larger one. They might be separate or intersecting. But can p - q be less than PQ? Yes, because if the circles are separate or intersecting, PQ is greater than p - q. So, statement D is also true.Wait, hold on. Let me double-check. If p - q is less than PQ, that means the circles are not one inside the other. So, if PQ is greater than p - q, they could be separate or intersecting. So, yes, p - q can indeed be less than PQ. So, statement D is true.So, all statements A, B, C, D are true? That would mean the answer is E, none of these. But let me make sure I didn't make a mistake.Wait, statement C says "p + q can be less than PQ." So, can p + q be less than PQ? Yes, that's when the circles are separate. So, that's true.Statement D: "p - q can be less than PQ." So, if PQ is greater than p - q, that's when the circles are either separate or intersecting. So, yes, that's possible. So, D is true.Hmm, so all statements A through D are true, meaning none of them are false. So, the answer should be E, none of these.But wait, let me think again. Maybe I misread the question. It says "the larger of two circles" with center P and radius p, and the smaller with center Q and radius q. So, p is larger than q. So, p - q is positive.So, for statement C: p + q can be less than PQ. That's when the circles are separate. So, yes, that's possible.Statement D: p - q can be less than PQ. So, if PQ is greater than p - q, which is the case when the circles are not one inside the other. Since p > q, if PQ is greater than p - q, the smaller circle isn't entirely inside the larger one. So, that's also possible.Wait, so all statements are possible, meaning none are false. So, E is the correct answer.But let me check if any of them can't be true. For example, can p + q be less than PQ? Yes, as in separate circles.Can p + q be equal to PQ? Yes, externally tangent.Can p - q be equal to PQ? Yes, internally tangent.Can p - q be less than PQ? Yes, when the circles are separate or intersecting.So, all four statements are possible. Therefore, none of them are false. So, the answer is E.Wait, but the question is asking which statement is false. So, if all are true, then E is correct because none are false.Alternatively, maybe I made a mistake in interpreting statement C. Let me read it again: "p + q can be less than PQ." So, can the sum of the radii be less than the distance between centers? Yes, that's when they don't intersect and are separate. So, that's possible.Similarly, statement D: "p - q can be less than PQ." So, can the difference of the radii be less than the distance? Yes, that's when the smaller circle isn't entirely inside the larger one, so they can be separate or intersecting.So, yes, all statements are true. So, E is the correct answer.**Final Answer**boxed{E}

answer：First, I need to calculate the distance the driver travels each day. On the first part of the trip, the driver goes at 30 miles per hour for 3 hours. So, the distance for this part is 30 multiplied by 3, which equals 90 miles.Next, on the second part of the trip, the driver travels at 25 miles per hour for 4 hours. The distance for this part is 25 multiplied by 4, resulting in 100 miles.Adding both distances together gives the total daily distance: 90 miles plus 100 miles equals 190 miles per day.Since the driver works from Monday to Saturday, that's a total of 6 days. To find the weekly distance, I multiply the daily distance by 6: 190 miles per day multiplied by 6 days equals 1,140 miles.Therefore, the driver travels a total of 1,140 miles in a week.

answer：To find the smallest fraction greater than 4/5 with both numerator and denominator as two-digit numbers, I'll start by understanding the target value. The fraction 4/5 is equal to 0.8.Next, I'll consider all two-digit numbers for the numerator (a) and denominator (b), ensuring that a < b to maintain the fraction as proper. My goal is to find the smallest possible fraction a/b that is just above 0.8.I'll calculate the minimum value for the numerator a for each possible denominator b using the inequality a > 0.8b. Since a must be an integer, I'll round up the result of 0.8b to the nearest whole number.For each denominator b from 10 to 99, I'll determine the smallest a that satisfies a > 0.8b and a < b. Then, I'll calculate the fraction a/b and keep track of the smallest fraction that is greater than 0.8.After evaluating all possible denominators, the fraction with the smallest value that meets the criteria will be the answer. The numerator of this fraction will be the solution to the problem.

answer：Okay, so I have this problem about Pascal's triangle, and I need to compute the difference between two sums involving the 2005th, 2006th, and 2007th rows. Let me try to unpack this step by step.First, let me make sure I understand the problem correctly. We have three sequences: (a_i), (b_i), and (c_i), which correspond to the elements in the 2005th, 2006th, and 2007th rows of Pascal's triangle, respectively. Each row is read from left to right, starting at i = 0. So, for example, the 2005th row has elements a_0, a_1, ..., a_{2005}, the 2006th row has b_0, b_1, ..., b_{2006}, and the 2007th row has c_0, c_1, ..., c_{2007}.The problem asks me to compute the difference between two sums:1. The first sum is from i = 0 to 2006 of (b_i / c_i).2. The second sum is from i = 0 to 2005 of (a_i / b_i).So, mathematically, I need to find:[sum_{i = 0}^{2006} frac{b_i}{c_i} - sum_{i = 0}^{2005} frac{a_i}{b_i}]Alright, let's think about how to approach this. Since we're dealing with Pascal's triangle, I know that each element in a row can be expressed using combinations. Specifically, the element at position i in the nth row is given by:[binom{n}{i}]So, for our sequences:- a_i = (binom{2005}{i})- b_i = (binom{2006}{i})- c_i = (binom{2007}{i})Therefore, the fractions in the sums can be rewritten as:- (frac{b_i}{c_i} = frac{binom{2006}{i}}{binom{2007}{i}})- (frac{a_i}{b_i} = frac{binom{2005}{i}}{binom{2006}{i}})Hmm, maybe I can simplify these fractions using properties of combinations. Let me recall that:[binom{n}{i} = frac{n!}{i!(n - i)!}]So, let's compute (frac{binom{n-1}{i}}{binom{n}{i}}):[frac{binom{n-1}{i}}{binom{n}{i}} = frac{frac{(n-1)!}{i!(n - 1 - i)!}}{frac{n!}{i!(n - i)!}} = frac{(n - 1)! cdot (n - i)!}{n! cdot (n - 1 - i)!} = frac{(n - i)}{n}]Wait, let me check that again. The (n - 1)! cancels with the n! as n! = n*(n - 1)! So, we have:[frac{(n - 1)!}{n!} = frac{1}{n}]And then, the (n - i)! in the numerator and (n - 1 - i)! in the denominator. Let's see:(n - i)! = (n - i)(n - 1 - i)! So, when we divide (n - i)! by (n - 1 - i)!, we get (n - i).So, putting it all together:[frac{binom{n-1}{i}}{binom{n}{i}} = frac{(n - i)}{n}]Yes, that seems correct.So, applying this to our fractions:1. (frac{b_i}{c_i} = frac{binom{2006}{i}}{binom{2007}{i}} = frac{2007 - i}{2007})Wait, hold on. Let me see. If n = 2007, then (frac{binom{2006}{i}}{binom{2007}{i}} = frac{2007 - i}{2007}). Is that right?Wait, let me substitute n = 2007 into the formula:[frac{binom{n - 1}{i}}{binom{n}{i}} = frac{n - i}{n}]So, yes, that would mean:[frac{binom{2006}{i}}{binom{2007}{i}} = frac{2007 - i}{2007}]Similarly, for the other fraction:[frac{a_i}{b_i} = frac{binom{2005}{i}}{binom{2006}{i}} = frac{2006 - i}{2006}]So, substituting these back into the sums, we have:First sum:[sum_{i = 0}^{2006} frac{2007 - i}{2007}]Second sum:[sum_{i = 0}^{2005} frac{2006 - i}{2006}]So, now, I need to compute these two sums and then subtract them.Let me compute the first sum:[sum_{i = 0}^{2006} frac{2007 - i}{2007}]I can factor out 1/2007:[frac{1}{2007} sum_{i = 0}^{2006} (2007 - i)]Similarly, for the second sum:[sum_{i = 0}^{2005} frac{2006 - i}{2006} = frac{1}{2006} sum_{i = 0}^{2005} (2006 - i)]So, now, let me compute each sum separately.Starting with the first sum:[S_1 = frac{1}{2007} sum_{i = 0}^{2006} (2007 - i)]Let me change the index to make it easier. Let j = 2007 - i. When i = 0, j = 2007. When i = 2006, j = 1. So, the sum becomes:[sum_{j = 1}^{2007} j]But wait, when i goes from 0 to 2006, j goes from 2007 down to 1. So, the sum is the same as summing from 1 to 2007.Therefore:[S_1 = frac{1}{2007} cdot sum_{j = 1}^{2007} j]We know that the sum of the first n integers is (frac{n(n + 1)}{2}), so:[sum_{j = 1}^{2007} j = frac{2007 cdot 2008}{2}]Therefore, plugging back into S1:[S_1 = frac{1}{2007} cdot frac{2007 cdot 2008}{2} = frac{2008}{2} = 1004]Wait, that's nice! So, the first sum simplifies to 1004.Now, moving on to the second sum:[S_2 = frac{1}{2006} sum_{i = 0}^{2005} (2006 - i)]Similarly, let me change the index. Let k = 2006 - i. When i = 0, k = 2006. When i = 2005, k = 1. So, the sum becomes:[sum_{k = 1}^{2006} k]Again, the sum of the first n integers:[sum_{k = 1}^{2006} k = frac{2006 cdot 2007}{2}]Therefore, plugging back into S2:[S_2 = frac{1}{2006} cdot frac{2006 cdot 2007}{2} = frac{2007}{2} = 1003.5]Wait, 2007 divided by 2 is 1003.5. Hmm, so S2 is 1003.5.But wait, let me double-check that. So, S2 is 1/2006 times the sum from i=0 to 2005 of (2006 - i). When we change variables, k = 2006 - i, so when i=0, k=2006; when i=2005, k=1. So, the sum is from k=1 to 2006, which is indeed (2006)(2007)/2. Then, multiplied by 1/2006, gives (2007)/2, which is 1003.5.So, now, putting it all together:The expression we need is S1 - S2 = 1004 - 1003.5 = 0.5.Wait, 1004 minus 1003.5 is 0.5, which is 1/2.So, the difference between the two sums is 1/2.But let me just make sure I didn't make a mistake in my calculations.Starting with S1:Sum from i=0 to 2006 of (2007 - i)/2007.Which is 1/2007 times sum from i=0 to 2006 of (2007 - i).Changing variable j = 2007 - i, so when i=0, j=2007; when i=2006, j=1. So, sum from j=1 to 2007 of j.Sum is (2007)(2008)/2, so S1 = (2007)(2008)/(2*2007) = 2008/2 = 1004. That seems correct.Similarly, S2:Sum from i=0 to 2005 of (2006 - i)/2006.Change variable k = 2006 - i, so when i=0, k=2006; when i=2005, k=1. So, sum from k=1 to 2006 of k.Sum is (2006)(2007)/2, so S2 = (2006)(2007)/(2*2006) = 2007/2 = 1003.5. Correct.Therefore, S1 - S2 = 1004 - 1003.5 = 0.5, which is 1/2.So, the answer is 1/2.But just to make sure, let me think about another approach or see if there's another way to verify this result.Alternatively, perhaps there's a telescoping series or some identity that can be used here.Wait, another thought: the terms in the sums are fractions of consecutive binomial coefficients. Maybe there's a telescoping nature when considering the differences or something.But in this case, since we've already computed the sums directly and found a clean answer, and the steps seem to check out, I think 1/2 is the correct answer.But let me think again about the fractions:We had:[frac{binom{n-1}{i}}{binom{n}{i}} = frac{n - i}{n}]Which is correct because:[frac{binom{n-1}{i}}{binom{n}{i}} = frac{(n-1)! / (i! (n - 1 - i)!)}{n! / (i! (n - i)!)} = frac{(n - 1)! (n - i)!}{n! (n - 1 - i)!} = frac{(n - i)}{n}]Yes, that's correct.So, applying that formula, we correctly converted the binomial coefficients into linear terms, which made the sums manageable.Therefore, I feel confident that the difference between the two sums is indeed 1/2.**Final Answer**boxed{dfrac{1}{2}}

answer：Alright, so I have this problem here where I need to find the number of integer points (points with integer x and y coordinates) that satisfy a system of three inequalities. The inequalities are:1. ( y leq 3x )2. ( y geq frac{1}{3}x )3. ( x + y leq 100 )Hmm, okay. So, I need to figure out all the integer points (x, y) that lie within the region defined by these three inequalities. Let me try to visualize this region first.First, let me sketch the lines corresponding to each inequality.1. The line ( y = 3x ) is a straight line passing through the origin with a slope of 3. Since the inequality is ( y leq 3x ), the region under consideration is below this line.2. The line ( y = frac{1}{3}x ) is another straight line through the origin with a slope of 1/3. The inequality ( y geq frac{1}{3}x ) means we're looking at the region above this line.3. The line ( x + y = 100 ) is a straight line with a slope of -1 and y-intercept at 100. The inequality ( x + y leq 100 ) means we're considering the region below this line.So, putting it all together, the feasible region is the area that is below ( y = 3x ), above ( y = frac{1}{3}x ), and below ( x + y = 100 ). This should form a sort of triangular or polygonal region in the first quadrant, since all these lines pass through the origin or have positive intercepts.Since we're dealing with integer points, I need to count all the integer coordinate pairs (x, y) that lie within this region. Let me think about how to approach this systematically.First, I can consider the boundaries. The region is bounded by three lines, so it's a polygon. To count the integer points, I can iterate over possible x-values and for each x, determine the range of y-values that satisfy all three inequalities. Then, count the integer y's in that range for each x.But before I start iterating, maybe I should find the points of intersection of these lines to understand the limits of the region.Let me find where ( y = 3x ) intersects ( x + y = 100 ). Substituting ( y = 3x ) into ( x + y = 100 ):( x + 3x = 100 )( 4x = 100 )( x = 25 )Then, ( y = 3*25 = 75 )So, the intersection point is (25, 75).Next, where does ( y = frac{1}{3}x ) intersect ( x + y = 100 )? Substitute ( y = frac{1}{3}x ) into ( x + y = 100 ):( x + frac{1}{3}x = 100 )( frac{4}{3}x = 100 )( x = 75 )Then, ( y = frac{1}{3}*75 = 25 )So, the intersection point is (75, 25).Also, both lines ( y = 3x ) and ( y = frac{1}{3}x ) pass through the origin (0,0). So, the feasible region is a triangle with vertices at (0,0), (25,75), and (75,25).Wait, is that correct? Let me verify.Yes, because the three lines intersect each other at these three points: origin, (25,75), and (75,25). So, the feasible region is indeed a triangle with these three vertices.So, now I need to count all integer points within this triangle. Since the triangle is in the first quadrant, both x and y are non-negative integers.To count the integer points, I can use Pick's Theorem, which relates the area of a polygon with integer coordinates to the number of interior and boundary points. The formula is:( A = I + frac{B}{2} - 1 )Where:- ( A ) is the area of the polygon,- ( I ) is the number of interior integer points,- ( B ) is the number of boundary integer points.But in this case, I need the total number of integer points, which is ( I + B ). So, if I can compute ( A ) and ( B ), I can solve for ( I ) and then add ( B ) to get the total.Alternatively, since the region is a triangle, maybe I can compute the area and then use Pick's Theorem to find the number of integer points. Let me try that.First, let's compute the area of the triangle with vertices at (0,0), (25,75), and (75,25).The formula for the area of a triangle given three vertices is:( A = frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| )Plugging in the coordinates:( x_1 = 0, y_1 = 0 )( x_2 = 25, y_2 = 75 )( x_3 = 75, y_3 = 25 )So,( A = frac{1}{2} |0*(75 - 25) + 25*(25 - 0) + 75*(0 - 75)| )Simplify each term:First term: 0*(50) = 0Second term: 25*(25) = 625Third term: 75*(-75) = -5625So,( A = frac{1}{2} |0 + 625 - 5625| = frac{1}{2} | -5000 | = frac{1}{2} * 5000 = 2500 )So, the area is 2500.Now, let's compute the number of boundary points ( B ). The boundary consists of three edges: from (0,0) to (25,75), from (25,75) to (75,25), and from (75,25) back to (0,0).For each edge, we can compute the number of integer points using the greatest common divisor (gcd) of the differences in coordinates plus 1.For the edge from (0,0) to (25,75):The differences are ( Delta x = 25 - 0 = 25 ), ( Delta y = 75 - 0 = 75 ).The gcd of 25 and 75 is 25. So, the number of integer points on this edge is gcd(25,75) + 1 = 25 + 1 = 26.Similarly, for the edge from (25,75) to (75,25):Differences: ( Delta x = 75 - 25 = 50 ), ( Delta y = 25 - 75 = -50 ).The gcd of 50 and 50 is 50. So, the number of integer points is 50 + 1 = 51.Wait, but hold on. The formula is gcd(|Δx|, |Δy|) + 1. So, for the edge from (25,75) to (75,25), the differences are 50 and -50, so gcd(50,50)=50, so number of points is 50 + 1 = 51.But wait, actually, in Pick's theorem, the boundary points are counted, but we have to be careful not to double-count the vertices.Each edge contributes its number of points, but the vertices are shared between edges. So, if I add up the points on each edge, I will have counted each vertex twice. Therefore, to get the total boundary points, I need to sum the points on each edge and subtract 3 (since there are three vertices, each counted twice).So, let's compute:Edge 1: (0,0) to (25,75): 26 pointsEdge 2: (25,75) to (75,25): 51 pointsEdge 3: (75,25) to (0,0): Let's compute this.From (75,25) to (0,0):Δx = 0 - 75 = -75, Δy = 0 - 25 = -25gcd(75,25) = 25, so number of points is 25 + 1 = 26.So, Edge 3: 26 points.Total boundary points before correction: 26 + 51 + 26 = 103But we have overcounted the 3 vertices (each vertex is shared by two edges). So, each vertex is counted twice, so total overcount is 3 points. Therefore, total boundary points ( B = 103 - 3 = 100 ).Wait, let me check that again. Each edge includes its endpoints. So, for three edges, each vertex is included in two edges. So, we have 3 vertices each counted twice, so total overcount is 3. Therefore, total boundary points is 26 + 51 + 26 - 3*1 = 103 - 3 = 100.Yes, that seems correct.So, ( B = 100 ).Now, using Pick's theorem:( A = I + frac{B}{2} - 1 )We have ( A = 2500 ), ( B = 100 ). Let's plug in:( 2500 = I + frac{100}{2} - 1 )Simplify:( 2500 = I + 50 - 1 )( 2500 = I + 49 )Therefore, ( I = 2500 - 49 = 2451 )So, the number of interior points is 2451, and the number of boundary points is 100. Therefore, the total number of integer points is ( I + B = 2451 + 100 = 2551 ).Wait, but before I conclude, let me verify if this makes sense. So, the area is 2500, and the number of integer points is 2551. That seems plausible because Pick's theorem relates the area to the number of points, but let me think if I did everything correctly.Wait, another way to compute the number of integer points is to iterate over x from 0 to 75 and for each x, find the range of y that satisfies all three inequalities, then count the integer y's in that range. Maybe I can do that as a check.So, let's consider x from 0 to 75, since the triangle goes from x=0 to x=75.For each x, y must satisfy:1. ( y leq 3x )2. ( y geq frac{1}{3}x )3. ( y leq 100 - x )So, for each x, the lower bound of y is ( lceil frac{1}{3}x rceil ) and the upper bound is ( min(3x, 100 - x) ). The number of integer y's is then ( text{upper} - text{lower} + 1 ), provided that upper >= lower.But since x and y are integers, I have to be careful with the ceiling function.Alternatively, since x is integer, ( frac{1}{3}x ) may not be integer, so the lower bound is the smallest integer greater than or equal to ( frac{1}{3}x ). Similarly, the upper bound is the largest integer less than or equal to ( min(3x, 100 - x) ).So, let's denote:For each integer x from 0 to 75:- Lower y: ( lceil frac{x}{3} rceil )- Upper y: ( lfloor min(3x, 100 - x) rfloor )But since x is integer, ( frac{x}{3} ) can be integer or fractional. So, ( lceil frac{x}{3} rceil ) is equal to ( lfloor frac{x + 2}{3} rfloor ). Hmm, maybe that's complicating things.Alternatively, for each x, compute ( frac{x}{3} ) and round up to the nearest integer, and compute ( 3x ) and ( 100 - x ), take the minimum, and then count the integers between these two values inclusive.But this might be tedious, but perhaps I can find a pattern or break it into ranges where the upper bound switches from 3x to 100 - x.So, let's find the x where 3x = 100 - x. Solving for x:( 3x = 100 - x )( 4x = 100 )( x = 25 )So, for x <=25, the upper bound is 3x, and for x >25, the upper bound is 100 - x.Therefore, we can split the problem into two parts:1. x from 0 to 25: y ranges from ( lceil frac{x}{3} rceil ) to ( 3x )2. x from 26 to 75: y ranges from ( lceil frac{x}{3} rceil ) to ( 100 - x )Additionally, for each x, we need to ensure that ( lceil frac{x}{3} rceil leq min(3x, 100 - x) ). Since we already split at x=25, and for x <=25, 3x <=100 -x (since 3*25=75 and 100 -25=75). So, at x=25, both are equal.So, let's compute the number of y's for each x in 0 to 25 and 26 to75.First, for x from 0 to25:For each x, the number of y's is ( 3x - lceil frac{x}{3} rceil + 1 )Similarly, for x from26 to75:The number of y's is ( (100 - x) - lceil frac{x}{3} rceil + 1 )But let's compute these step by step.First, for x from 0 to25:Let me note that for x=0:- Lower y: ( lceil 0/3 rceil = 0 )- Upper y: 0- So, number of y's: 0 -0 +1=1For x=1:- Lower y: ( lceil 1/3 rceil =1 )- Upper y: 3*1=3- Number of y's: 3 -1 +1=3x=2:- Lower y: ( lceil 2/3 rceil =1 )- Upper y:6- Number of y's:6 -1 +1=6x=3:- Lower y:1- Upper y:9- Number of y's:9 -1 +1=9Wait, but wait, for x=3, ( frac{3}{3}=1 ), so lower y=1, upper y=9, so 9-1+1=9.Similarly, x=4:- Lower y: ( lceil 4/3 rceil =2 )- Upper y:12- Number of y's:12 -2 +1=11Wait, but hold on, for x=4, 3x=12, but 100 -x=96, which is larger, so upper y is 12.Wait, but in our initial split, for x<=25, upper y is 3x, which is correct.Wait, but 3x for x=4 is 12, which is less than 100 -4=96, so correct.So, proceeding:x=5:- Lower y: ( lceil 5/3 rceil =2 )- Upper y:15- Number of y's:15 -2 +1=14x=6:- Lower y:2- Upper y:18- Number of y's:18 -2 +1=17x=7:- Lower y:3- Upper y:21- Number of y's:21 -3 +1=19Wait, 21-3=18, +1=19.x=8:- Lower y:3 (since 8/3≈2.666, ceil is 3)- Upper y:24- Number of y's:24 -3 +1=22x=9:- Lower y:3- Upper y:27- Number of y's:27 -3 +1=25x=10:- Lower y:4 (10/3≈3.333, ceil is4)- Upper y:30- Number of y's:30 -4 +1=27x=11:- Lower y:4- Upper y:33- Number of y's:33 -4 +1=30x=12:- Lower y:4- Upper y:36- Number of y's:36 -4 +1=33x=13:- Lower y:5 (13/3≈4.333, ceil is5)- Upper y:39- Number of y's:39 -5 +1=35x=14:- Lower y:5- Upper y:42- Number of y's:42 -5 +1=38x=15:- Lower y:5- Upper y:45- Number of y's:45 -5 +1=41x=16:- Lower y:6 (16/3≈5.333, ceil is6)- Upper y:48- Number of y's:48 -6 +1=43x=17:- Lower y:6- Upper y:51- Number of y's:51 -6 +1=46x=18:- Lower y:6- Upper y:54- Number of y's:54 -6 +1=49x=19:- Lower y:7 (19/3≈6.333, ceil is7)- Upper y:57- Number of y's:57 -7 +1=51x=20:- Lower y:7- Upper y:60- Number of y's:60 -7 +1=54x=21:- Lower y:7- Upper y:63- Number of y's:63 -7 +1=57x=22:- Lower y:8 (22/3≈7.333, ceil is8)- Upper y:66- Number of y's:66 -8 +1=59x=23:- Lower y:8- Upper y:69- Number of y's:69 -8 +1=62x=24:- Lower y:8- Upper y:72- Number of y's:72 -8 +1=65x=25:- Lower y:9 (25/3≈8.333, ceil is9)- Upper y:75- Number of y's:75 -9 +1=67So, for x from 0 to25, the number of y's for each x is:x=0:1x=1:3x=2:6x=3:9x=4:11x=5:14x=6:17x=7:19x=8:22x=9:25x=10:27x=11:30x=12:33x=13:35x=14:38x=15:41x=16:43x=17:46x=18:49x=19:51x=20:54x=21:57x=22:59x=23:62x=24:65x=25:67Now, let's sum these up. Hmm, that's a lot, but let me try to compute the total.Let me list them:1, 3, 6, 9, 11, 14, 17, 19, 22, 25, 27, 30, 33, 35, 38, 41, 43, 46, 49, 51, 54, 57, 59, 62, 65, 67.Let me add them step by step:Start with 1.1 + 3 = 44 + 6 = 1010 + 9 = 1919 +11=3030 +14=4444 +17=6161 +19=8080 +22=102102 +25=127127 +27=154154 +30=184184 +33=217217 +35=252252 +38=290290 +41=331331 +43=374374 +46=420420 +49=469469 +51=520520 +54=574574 +57=631631 +59=690690 +62=752752 +65=817817 +67=884So, total number of points for x from0 to25 is 884.Now, moving on to x from26 to75.For x from26 to75, the upper y is ( 100 - x ), and the lower y is ( lceil frac{x}{3} rceil ).So, for each x, the number of y's is ( (100 - x) - lceil frac{x}{3} rceil + 1 ).Let me compute this for each x from26 to75.But that's 50 values, which is a lot. Maybe I can find a pattern or break it into chunks.Alternatively, notice that for x from26 to75, the lower bound is ( lceil x/3 rceil ), which increases roughly every 3 x's.Let me see:For x=26:- Lower y: ( lceil 26/3 rceil =9 ) (since 26/3≈8.666)- Upper y:100 -26=74- Number of y's:74 -9 +1=66x=27:- Lower y:9- Upper y:73- Number of y's:73 -9 +1=65x=28:- Lower y:10 (28/3≈9.333)- Upper y:72- Number of y's:72 -10 +1=63x=29:- Lower y:10- Upper y:71- Number of y's:71 -10 +1=62x=30:- Lower y:10- Upper y:70- Number of y's:70 -10 +1=61x=31:- Lower y:11 (31/3≈10.333)- Upper y:69- Number of y's:69 -11 +1=59x=32:- Lower y:11- Upper y:68- Number of y's:68 -11 +1=58x=33:- Lower y:11- Upper y:67- Number of y's:67 -11 +1=57x=34:- Lower y:12 (34/3≈11.333)- Upper y:66- Number of y's:66 -12 +1=55x=35:- Lower y:12- Upper y:65- Number of y's:65 -12 +1=54x=36:- Lower y:12- Upper y:64- Number of y's:64 -12 +1=53x=37:- Lower y:13 (37/3≈12.333)- Upper y:63- Number of y's:63 -13 +1=51x=38:- Lower y:13- Upper y:62- Number of y's:62 -13 +1=50x=39:- Lower y:13- Upper y:61- Number of y's:61 -13 +1=49x=40:- Lower y:14 (40/3≈13.333)- Upper y:60- Number of y's:60 -14 +1=47x=41:- Lower y:14- Upper y:59- Number of y's:59 -14 +1=46x=42:- Lower y:14- Upper y:58- Number of y's:58 -14 +1=45x=43:- Lower y:15 (43/3≈14.333)- Upper y:57- Number of y's:57 -15 +1=43x=44:- Lower y:15- Upper y:56- Number of y's:56 -15 +1=42x=45:- Lower y:15- Upper y:55- Number of y's:55 -15 +1=41x=46:- Lower y:16 (46/3≈15.333)- Upper y:54- Number of y's:54 -16 +1=39x=47:- Lower y:16- Upper y:53- Number of y's:53 -16 +1=38x=48:- Lower y:16- Upper y:52- Number of y's:52 -16 +1=37x=49:- Lower y:17 (49/3≈16.333)- Upper y:51- Number of y's:51 -17 +1=35x=50:- Lower y:17- Upper y:50- Number of y's:50 -17 +1=34x=51:- Lower y:17- Upper y:49- Number of y's:49 -17 +1=33x=52:- Lower y:18 (52/3≈17.333)- Upper y:48- Number of y's:48 -18 +1=31x=53:- Lower y:18- Upper y:47- Number of y's:47 -18 +1=30x=54:- Lower y:18- Upper y:46- Number of y's:46 -18 +1=29x=55:- Lower y:19 (55/3≈18.333)- Upper y:45- Number of y's:45 -19 +1=27x=56:- Lower y:19- Upper y:44- Number of y's:44 -19 +1=26x=57:- Lower y:19- Upper y:43- Number of y's:43 -19 +1=25x=58:- Lower y:20 (58/3≈19.333)- Upper y:42- Number of y's:42 -20 +1=23x=59:- Lower y:20- Upper y:41- Number of y's:41 -20 +1=22x=60:- Lower y:20- Upper y:40- Number of y's:40 -20 +1=21x=61:- Lower y:21 (61/3≈20.333)- Upper y:39- Number of y's:39 -21 +1=19x=62:- Lower y:21- Upper y:38- Number of y's:38 -21 +1=18x=63:- Lower y:21- Upper y:37- Number of y's:37 -21 +1=17x=64:- Lower y:22 (64/3≈21.333)- Upper y:36- Number of y's:36 -22 +1=15x=65:- Lower y:22- Upper y:35- Number of y's:35 -22 +1=14x=66:- Lower y:22- Upper y:34- Number of y's:34 -22 +1=13x=67:- Lower y:23 (67/3≈22.333)- Upper y:33- Number of y's:33 -23 +1=11x=68:- Lower y:23- Upper y:32- Number of y's:32 -23 +1=10x=69:- Lower y:23- Upper y:31- Number of y's:31 -23 +1=9x=70:- Lower y:24 (70/3≈23.333)- Upper y:30- Number of y's:30 -24 +1=7x=71:- Lower y:24- Upper y:29- Number of y's:29 -24 +1=6x=72:- Lower y:24- Upper y:28- Number of y's:28 -24 +1=5x=73:- Lower y:25 (73/3≈24.333)- Upper y:27- Number of y's:27 -25 +1=3x=74:- Lower y:25- Upper y:26- Number of y's:26 -25 +1=2x=75:- Lower y:25- Upper y:25 (100 -75=25)- Number of y's:25 -25 +1=1So, for x from26 to75, the number of y's for each x is:66,65,63,62,61,59,58,57,55,54,53,51,50,49,47,46,45,43,42,41,39,38,37,35,34,33,31,30,29,27,26,25,23,22,21,19,18,17,15,14,13,11,10,9,7,6,5,3,2,1Now, let's sum these up. Again, that's a lot, but let's try.List of numbers:66,65,63,62,61,59,58,57,55,54,53,51,50,49,47,46,45,43,42,41,39,38,37,35,34,33,31,30,29,27,26,25,23,22,21,19,18,17,15,14,13,11,10,9,7,6,5,3,2,1Let me add them step by step:Start with 66.66 +65=131131 +63=194194 +62=256256 +61=317317 +59=376376 +58=434434 +57=491491 +55=546546 +54=600600 +53=653653 +51=704704 +50=754754 +49=803803 +47=850850 +46=896896 +45=941941 +43=984984 +42=10261026 +41=10671067 +39=11061106 +38=11441144 +37=11811181 +35=12161216 +34=12501250 +33=12831283 +31=13141314 +30=13441344 +29=13731373 +27=14001400 +26=14261426 +25=14511451 +23=14741474 +22=14961496 +21=15171517 +19=15361536 +18=15541554 +17=15711571 +15=15861586 +14=16001600 +13=16131613 +11=16241624 +10=16341634 +9=16431643 +7=16501650 +6=16561656 +5=16611661 +3=16641664 +2=16661666 +1=1667So, the total number of points for x from26 to75 is1667.Therefore, the total number of integer points is the sum of the two parts:884 (from x=0 to25) +1667 (from x=26 to75)= 884 +1667=2551.Which matches the result from Pick's theorem.So, that's reassuring. So, the total number of integer points is2551.Therefore, the answer is boxed{2551}.**Final Answer**boxed{2551}