💡First, I need to determine the quotient when the particular number is divided by 5. The problem states that this quotient is equal to 320 divided by 4 plus 220.I start by calculating 320 divided by 4, which equals 80. Then, I add 220 to this result to find the total quotient: 80 plus 220 equals 300.Now that I know the quotient is 300, I can find the particular number by multiplying the quotient by 5: 300 multiplied by 5 equals 1500.Finally, to find the value of this particular number divided by 3, I divide 1500 by 3, which equals 500.

💡Okay, so I have this problem where I need to mark three points on graph paper such that when I form a triangle with them, the sum of the two smaller medians equals the semiperimeter of the triangle. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, I need to recall what a median of a triangle is. A median is a line segment joining a vertex to the midpoint of the opposite side. So, in any triangle, there are three medians, one from each vertex. Now, the problem is talking about the sum of the two smaller medians. That means I need to figure out which medians are the smaller ones and then add them up.Next, the semiperimeter of a triangle is half of its perimeter. So, if the sides of the triangle are a, b, and c, then the semiperimeter (s) is (a + b + c)/2. The problem states that the sum of the two smaller medians should equal this semiperimeter.Alright, so I need to find a triangle where this condition holds. Maybe it's easier if I start by considering a specific type of triangle, like a right-angled triangle, because I know some properties about medians in right-angled triangles.In a right-angled triangle, the median to the hypotenuse is half the hypotenuse. That's a useful property. So, if I have a right-angled triangle with legs of lengths a and b, and hypotenuse c, then the median to the hypotenuse is c/2. That might help.Let me try to assign some coordinates to the triangle to make it easier. Let's place the right angle at the origin (0,0), one vertex at (a,0), and the other at (0,b). Then, the hypotenuse would be from (a,0) to (0,b). The midpoint of the hypotenuse would be at (a/2, b/2), so the median from the right angle (0,0) to this midpoint is the line segment from (0,0) to (a/2, b/2). The length of this median is sqrt((a/2)^2 + (b/2)^2) = (1/2)sqrt(a^2 + b^2), which is indeed half the hypotenuse, as expected.Now, I need to find the other two medians. Let's find the median from (a,0) to the midpoint of the opposite side, which is from (a,0) to the midpoint of (0,0) and (0,b). The midpoint of (0,0) and (0,b) is (0, b/2). So, the median from (a,0) to (0, b/2). The length of this median can be calculated using the distance formula: sqrt((a - 0)^2 + (0 - b/2)^2) = sqrt(a^2 + (b/2)^2).Similarly, the median from (0,b) to the midpoint of (a,0) and (0,0) is sqrt((0 - a/2)^2 + (b - 0)^2) = sqrt((a/2)^2 + b^2).So, now I have expressions for all three medians:1. Median from (0,0): (1/2)sqrt(a^2 + b^2)2. Median from (a,0): sqrt(a^2 + (b/2)^2)3. Median from (0,b): sqrt((a/2)^2 + b^2)Now, I need to determine which of these are the two smaller medians. Let's compare them.First, let's compare the median from (a,0) and the median from (0,b). Since the expressions are symmetric in a and b, if a ≠ b, one will be longer than the other. Let's assume, without loss of generality, that a > b. Then, the median from (a,0) would be sqrt(a^2 + (b/2)^2) and the median from (0,b) would be sqrt((a/2)^2 + b^2). Since a > b, (a/2)^2 < (b/2)^2, so sqrt((a/2)^2 + b^2) < sqrt(a^2 + (b/2)^2). Therefore, the median from (0,b) is smaller than the median from (a,0).So, the two smaller medians are the median from (0,0) and the median from (0,b). Therefore, their sum should equal the semiperimeter.Let me write down the expressions:Sum of two smaller medians = (1/2)sqrt(a^2 + b^2) + sqrt((a/2)^2 + b^2)Semiperimeter = (a + b + sqrt(a^2 + b^2))/2So, setting them equal:(1/2)sqrt(a^2 + b^2) + sqrt((a/2)^2 + b^2) = (a + b + sqrt(a^2 + b^2))/2Let me simplify this equation.First, multiply both sides by 2 to eliminate the denominators:sqrt(a^2 + b^2) + 2*sqrt((a/2)^2 + b^2) = a + b + sqrt(a^2 + b^2)Subtract sqrt(a^2 + b^2) from both sides:2*sqrt((a/2)^2 + b^2) = a + bDivide both sides by 2:sqrt((a/2)^2 + b^2) = (a + b)/2Now, square both sides:((a/2)^2 + b^2) = ((a + b)/2)^2Expand both sides:(a^2)/4 + b^2 = (a^2 + 2ab + b^2)/4Multiply both sides by 4 to eliminate denominators:a^2 + 4b^2 = a^2 + 2ab + b^2Subtract a^2 + b^2 from both sides:3b^2 = 2abDivide both sides by b (assuming b ≠ 0):3b = 2aSo, 3b = 2a => a = (3/2)bSo, the ratio of a to b is 3:2.Therefore, if I choose a = 3k and b = 2k for some positive real number k, the condition will be satisfied.Let me choose k = 2 to make the coordinates integers. Then, a = 6 and b = 4.So, the coordinates of the triangle would be (0,0), (6,0), and (0,4).Let me verify this.First, calculate the medians.Median from (0,0): (1/2)sqrt(6^2 + 4^2) = (1/2)sqrt(36 + 16) = (1/2)sqrt(52) = sqrt(13)Median from (6,0): sqrt(6^2 + (4/2)^2) = sqrt(36 + 4) = sqrt(40) = 2*sqrt(10)Median from (0,4): sqrt((6/2)^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5So, the medians are sqrt(13) ≈ 3.605, 5, and 2*sqrt(10) ≈ 6.324.Wait, hold on, I thought the two smaller medians were sqrt(13) and 5, but 5 is larger than sqrt(13). Wait, no, sqrt(13) is approximately 3.605, which is smaller than 5. So, the two smaller medians are sqrt(13) and 5.Wait, but 5 is actually the median from (0,4), which is sqrt((6/2)^2 + 4^2) = 5.Wait, but according to my earlier assumption, if a = 6 and b = 4, then the median from (0,b) is 5, and the median from (0,0) is sqrt(13). So, the two smaller medians are sqrt(13) and 5.Wait, but 5 is larger than sqrt(13). So, actually, the two smaller medians are sqrt(13) and 2*sqrt(10). Wait, no, 2*sqrt(10) is approximately 6.324, which is larger than 5. So, the two smaller medians are sqrt(13) and 5.Wait, but 5 is larger than sqrt(13). So, actually, the two smaller medians are sqrt(13) and 5, but 5 is larger than sqrt(13). So, the two smaller medians are sqrt(13) and 5, but 5 is larger than sqrt(13). So, the two smaller medians are sqrt(13) and 5, but 5 is larger than sqrt(13). So, the two smaller medians are sqrt(13) and 5.Wait, I'm getting confused. Let me list the medians:1. From (0,0): sqrt(13) ≈ 3.6052. From (6,0): 2*sqrt(10) ≈ 6.3243. From (0,4): 5So, the two smaller medians are sqrt(13) and 5. Their sum is sqrt(13) + 5 ≈ 3.605 + 5 = 8.605.Now, the semiperimeter is (6 + 4 + sqrt(52))/2 = (10 + 2*sqrt(13))/2 = 5 + sqrt(13) ≈ 5 + 3.605 = 8.605.So, yes, the sum of the two smaller medians equals the semiperimeter.Therefore, the coordinates (0,0), (6,0), and (0,4) satisfy the condition.Wait, but I chose k = 2 to make the coordinates integers. What if I choose k = 1? Then, a = 3 and b = 2. Let me check.Median from (0,0): (1/2)sqrt(3^2 + 2^2) = (1/2)sqrt(13) ≈ 1.802Median from (3,0): sqrt(3^2 + (2/2)^2) = sqrt(9 + 1) = sqrt(10) ≈ 3.162Median from (0,2): sqrt((3/2)^2 + 2^2) = sqrt(2.25 + 4) = sqrt(6.25) = 2.5So, the medians are approximately 1.802, 3.162, and 2.5.The two smaller medians are 1.802 and 2.5. Their sum is approximately 4.302.The semiperimeter is (3 + 2 + sqrt(13))/2 ≈ (5 + 3.605)/2 ≈ 4.3025.So, that also works.Therefore, the ratio a:b is 3:2, and any triangle with sides in that ratio will satisfy the condition.So, to answer the original question, I can choose any three points on graph paper such that the triangle formed has sides in the ratio 3:2:sqrt(13). For simplicity, choosing integer coordinates like (0,0), (6,0), and (0,4) works.Alternatively, I could also choose other points, like (0,0), (3,0), and (0,2), which are smaller integers.So, in conclusion, marking the points (0,0), (6,0), and (0,4) on graph paper will form a triangle where the sum of the two smaller medians equals the semiperimeter.

💡Alright, so I have this problem: "How many odd three-digit integers have the property that their digits, read left to right, are in strictly decreasing order?" Hmm, okay. Let me try to wrap my head around this.First, let's break down the problem. We need to find three-digit numbers, which are odd, and their digits are in strictly decreasing order from left to right. So, for example, a number like 963 is strictly decreasing because 9 > 6 > 3, and it's odd because the last digit is 3. Got it.Since the number is a three-digit integer, it has hundreds, tens, and units places. Let's denote the digits as A (hundreds place), B (tens place), and C (units place). So, the number is ABC, where A > B > C, and C is an odd digit.Now, the number has to be odd, which means the units digit, C, must be one of the odd digits: 1, 3, 5, 7, or 9. So, C can be 1, 3, 5, 7, or 9. That gives us five cases to consider, each based on the value of C.For each case, we need to determine how many valid pairs of digits A and B exist such that A > B > C. Since A and B must be digits greater than C, we can think of this as choosing two digits from the set of digits greater than C and arranging them in decreasing order.Let me formalize this a bit. If C is fixed, say C = k, then A and B must be chosen from the digits {k+1, k+2, ..., 9}. The number of ways to choose two distinct digits from this set is given by the combination formula C(n, 2) = n(n-1)/2, where n is the number of available digits.So, for each value of C, I can calculate the number of possible (A, B) pairs as follows:1. **Case 1: C = 1** - Available digits for A and B: {2, 3, 4, 5, 6, 7, 8, 9} (8 digits) - Number of ways: C(8, 2) = 282. **Case 2: C = 3** - Available digits for A and B: {4, 5, 6, 7, 8, 9} (6 digits) - Number of ways: C(6, 2) = 153. **Case 3: C = 5** - Available digits for A and B: {6, 7, 8, 9} (4 digits) - Number of ways: C(4, 2) = 64. **Case 4: C = 7** - Available digits for A and B: {8, 9} (2 digits) - Number of ways: C(2, 2) = 15. **Case 5: C = 9** - Available digits for A and B: {} (0 digits) - Number of ways: C(0, 2) = 0Now, adding up all these possibilities: 28 (for C=1) + 15 (for C=3) + 6 (for C=5) + 1 (for C=7) + 0 (for C=9) = 50.Wait, let me double-check that. So, for each case, I'm calculating the number of ways to choose two digits greater than C and then arranging them in decreasing order. Since combinations inherently account for the order (as we're choosing without regard to order and then arranging them in a specific way), this should be correct.Is there any chance I might have missed something? Let's see. For example, when C=1, we have digits from 2 to 9, which is 8 digits. Choosing 2 gives us 28, which seems right. Similarly, for C=3, digits from 4 to 9, which is 6 digits, giving 15. That adds up.Wait, another thought: since the digits must be strictly decreasing, once we choose two digits, there's only one way to arrange them so that A > B. So, combinations are the right approach here because order doesn't matter in combinations, and we're effectively counting the number of such pairs.Yes, that makes sense. So, I think my calculation is correct. Therefore, the total number of such odd three-digit integers is 50.**Final Answer**The total number of such odd three-digit integers is boxed{50}.

💡Okay, so I have this problem where sticks fall on the sand and form a random shape, and I need to find the probability that they do not intersect. This will determine if the upcoming year will be successful and prosperous. Hmm, interesting. I guess I need to figure out the probability that these sticks don't cross each other when they land.First, I need to understand the setup. Are these sticks of equal length? The problem doesn't specify, but maybe I can assume they are for simplicity. Let's say there are three sticks, each of the same length, forming a polygon when they fall. Maybe a triangle or a quadrilateral? Wait, three sticks would form a triangle, but the mention of "rods" and "falling on the sand" makes me think of something like a polygon with four sides, maybe a quadrilateral. Hmm, not sure. Let me think.If it's a quadrilateral, then four sticks would form a four-sided figure. But the problem mentions "sticks" in general, so maybe it's about three sticks forming a triangle. But triangles can't intersect themselves, right? Because a triangle is a simple polygon. So maybe it's about four sticks forming a quadrilateral, which can potentially intersect itself if the sticks cross each other.Wait, the problem says "the sticks, which fall on the sand and form a random shape." So maybe it's not necessarily a polygon but just a random arrangement of sticks. So, if multiple sticks fall, they can form various shapes, and we need the probability that none of them intersect.But the problem doesn't specify how many sticks there are. Maybe it's a standard problem, like the probability that three sticks form a triangle without intersecting? Or maybe it's about four sticks forming a quadrilateral without intersecting.Wait, in the initial problem, the user wrote "the sticks, which fall on the sand and form a random shape," and then the assistant's response started talking about a polygon ABCD, which is a quadrilateral. So maybe the problem is about four sticks forming a quadrilateral, and we need the probability that they do not intersect.So, assuming that, let me try to understand the assistant's solution.The assistant named the polygon as ABCD. So, four points connected by four sticks. Then, they defined angles α and β. α is the angle at point B, between the first and second rod, and β is the angle at point C, between the second and third rod. They considered α between 0 and π, and β between 0 and 2π. So, the elementary outcomes are pairs (α, β), which fill a rectangle G on the coordinate plane.Then, the probability is proportional to the area within this rectangle. The assistant then drew a diagram, which I can't see, but they mentioned that the first and third segments AB and CD intersect at some point K if certain conditions are met. Specifically, the ray CD lies within the angle BCA, and the ray BA lies within the angle CBD. This translates to inequalities on α and β.Given that triangles ABC and CBD are isosceles, they derived the conditions:0 ≤ β < π/2 - α/2 and 0 ≤ α < π/2 - β/2.These inequalities define a quadrilateral F within the rectangle G. The area of F is 1/12 of the area of G, so the probability of self-intersection is 1/12. Therefore, the probability that the rods do not intersect is 1 - 1/12 = 11/12.Wait, but I need to verify this. Let me go through the steps again.First, the polygon is ABCD, a quadrilateral. The angles α and β are defined at points B and C, respectively. The assistant is considering the conditions under which the sticks AB and CD intersect. For them to intersect, the rays CD and BA must lie within certain angles relative to the other sticks.They mention that triangles ABC and CBD are isosceles. Hmm, why is that? Maybe because the sticks are of equal length, so the triangles formed are isosceles. If AB, BC, CD, and DA are all equal, then triangles ABC and CBD would indeed be isosceles.So, given that, the conditions for intersection are derived. The inequalities are:0 ≤ β < π/2 - α/2 and 0 ≤ α < π/2 - β/2.These define a region F within the rectangle G of all possible (α, β) pairs. The area of F is 1/12 of the area of G, so the probability of intersection is 1/12. Therefore, the probability of no intersection is 11/12.But wait, is this correct? Let me think about the area calculation.The rectangle G has sides of length π and 2π, so its area is 2π². The region F is defined by the inequalities above. Let me try to sketch this mentally.For 0 ≤ α ≤ π and 0 ≤ β < 2π, but the inequalities further restrict β and α. Specifically, β must be less than π/2 - α/2, and α must be less than π/2 - β/2.This seems like a region bounded by two lines in the α-β plane. Let me set up the inequalities:1. β < π/2 - α/22. α < π/2 - β/2These are two linear inequalities. Let's solve for the intersection points.Set β = π/2 - α/2 and α = π/2 - β/2.Substitute β from the first equation into the second:α = π/2 - (π/2 - α/2)/2= π/2 - π/4 + α/4= π/4 + α/4Multiply both sides by 4:4α = π + α3α = πα = π/3Then, β = π/2 - (π/3)/2 = π/2 - π/6 = π/3.So, the two lines intersect at (π/3, π/3). Therefore, the region F is a polygon bounded by α=0, β=0, β=π/2 - α/2, and α=π/2 - β/2.This is a quadrilateral with vertices at (0,0), (0, π/2), (π/3, π/3), and (π/2, 0). Wait, is that correct?Wait, when α=0, from the first inequality, β < π/2. So, the upper bound for β when α=0 is π/2. Similarly, when β=0, from the second inequality, α < π/2.But the intersection point is at (π/3, π/3). So, the region F is a quadrilateral with vertices at (0,0), (0, π/2), (π/3, π/3), and (π/2, 0). Wait, but when α=π/2, from the first inequality, β < π/2 - (π/2)/2 = π/2 - π/4 = π/4. So, actually, the point when α=π/2 is (π/2, π/4). Hmm, maybe I was wrong earlier.Wait, let's clarify. The region F is defined by both inequalities:β < π/2 - α/2 and α < π/2 - β/2.So, the boundaries are:1. β = π/2 - α/22. α = π/2 - β/2These two lines intersect at (π/3, π/3), as we found earlier.So, the region F is bounded by:- α from 0 to π/3- β from 0 to π/2 - α/2And- β from 0 to π/3- α from 0 to π/2 - β/2But actually, it's a single region bounded by both inequalities. So, it's a polygon with vertices at (0,0), (0, π/2), (π/3, π/3), and (π/2, 0). Wait, but when α=π/2, β must be less than π/2 - π/4 = π/4, so the point is (π/2, π/4). Hmm, conflicting.Wait, maybe I need to plot these lines.Line 1: β = π/2 - α/2. When α=0, β=π/2. When α=π, β=π/2 - π/2=0.Line 2: α = π/2 - β/2. When β=0, α=π/2. When β=π, α=π/2 - π/2=0.So, these two lines intersect at (π/3, π/3), as before.So, the region F is the area below both lines. So, it's a quadrilateral with vertices at (0,0), (0, π/2), (π/3, π/3), and (π/2, 0). Wait, but when α=π/2, β=0, but according to the first inequality, β < π/2 - π/4 = π/4. So, actually, the point is (π/2, 0). Hmm, but β can't be negative, so the region is bounded by β=0.Wait, maybe it's a triangle? Because the lines intersect at (π/3, π/3), and the region is bounded by β=0, α=0, and the two lines.Wait, no, because when α increases beyond π/3, the upper bound for β decreases, and when β increases beyond π/3, the upper bound for α decreases.So, actually, the region F is a polygon with vertices at (0,0), (0, π/2), (π/3, π/3), and (π/2, 0). So, it's a quadrilateral.To find the area of F, we can integrate over the region or use geometry.Let me try to calculate the area.First, the area under line 1: β = π/2 - α/2 from α=0 to α=π/3.At α=0, β=π/2; at α=π/3, β=π/2 - π/6=π/3.So, this is a straight line from (0, π/2) to (π/3, π/3).Similarly, the area under line 2: α = π/2 - β/2 from β=0 to β=π/3.At β=0, α=π/2; at β=π/3, α=π/2 - π/6=π/3.So, this is a straight line from (π/2, 0) to (π/3, π/3).So, the region F is bounded by these two lines and the axes.Therefore, F is a quadrilateral with vertices at (0,0), (0, π/2), (π/3, π/3), and (π/2, 0).To find the area of F, we can divide it into two parts: the triangle from (0,0) to (0, π/2) to (π/3, π/3), and the triangle from (0,0) to (π/3, π/3) to (π/2, 0). Wait, but actually, it's a single quadrilateral, so maybe we can use the shoelace formula.Using the shoelace formula for the vertices (0,0), (0, π/2), (π/3, π/3), (π/2, 0):Area = 1/2 | (0*π/2 + 0*π/3 + π/3*0 + π/2*0) - (0*0 + π/2*π/3 + π/3*π/2 + 0*0) |Wait, that seems complicated. Let me list the coordinates:1. (0,0)2. (0, π/2)3. (π/3, π/3)4. (π/2, 0)Using shoelace:Sum1 = (0*π/2) + (0*π/3) + (π/3*0) + (π/2*0) = 0 + 0 + 0 + 0 = 0Sum2 = (0*0) + (π/2*π/3) + (π/3*π/2) + (0*0) = 0 + π²/6 + π²/6 + 0 = π²/3Area = 1/2 |Sum1 - Sum2| = 1/2 |0 - π²/3| = π²/6Wait, but the area of the rectangle G is π * 2π = 2π². So, the area of F is π²/6, which is (π²/6)/(2π²) = 1/12 of the area of G. So, the probability of intersection is 1/12, hence the probability of no intersection is 11/12.Okay, that makes sense now. The area of the region where the sticks intersect is 1/12 of the total possible outcomes, so the probability of no intersection is 11/12.But wait, let me think again. The rectangle G has sides of length π and 2π, so area 2π². The region F has area π²/6, so the ratio is (π²/6)/(2π²) = 1/12. Yes, that's correct.Therefore, the probability that the sticks do not intersect is 1 - 1/12 = 11/12.I think that's the answer. But let me just make sure I didn't miss anything.Wait, the problem didn't specify the number of sticks. I assumed four sticks forming a quadrilateral. If it's three sticks forming a triangle, then since triangles can't intersect themselves, the probability would be 1. But the assistant's solution was about a quadrilateral, so maybe the problem is about four sticks.Alternatively, if it's about three sticks forming a triangle, but considering them as three separate sticks that might intersect when falling randomly, then the probability would be different.Wait, the problem says "sticks, which fall on the sand and form a random shape." So, it could be any number of sticks, but the assistant's solution was about four sticks. Maybe the problem is about four sticks.Alternatively, maybe it's about three sticks forming a triangle, but the assistant considered four sticks. Hmm.Wait, in the assistant's solution, they mentioned angles at B and C, implying a quadrilateral. So, I think the problem is about four sticks forming a quadrilateral, and the probability that they do not intersect is 11/12.Therefore, the final answer is 11/12.

💡First, recognize that Jace earned 1000.He gave away 358 cents, which is equivalent to 3.58.After giving away 3.58, Jace has 642 left.To find the total amount he had before giving away the money, add the amount he has left to the amount he gave away:642 + 3.58 = 645.58Finally, subtract this total from the amount he earned to determine his debt:1000 - 645.58 = 354.42

💡First, I need to determine the area of a square field with a side length of 16 meters.I recall that the area of a square is calculated by squaring the length of one of its sides.So, I will square the side length:16 meters multiplied by 16 meters equals 256 square meters.Therefore, the area of the square field is 256 square meters.