💡Okay, so I need to figure out the maximum value of ( k ) such that when I divide the product of all numbers from 1001 to 2006 by ( 11^k ), the result is still an integer. Hmm, that sounds like I need to find how many times the prime number 11 divides into that big product. First, I remember that to find the exponent of a prime ( p ) in the factorial of a number ( n! ), there's a formula. It's something like adding up the integer divisions of ( n ) by ( p ), ( p^2 ), ( p^3 ), and so on until the division result is zero. So, for example, the exponent of 11 in ( 2006! ) would be calculated by adding ( lfloor frac{2006}{11} rfloor + lfloor frac{2006}{11^2} rfloor + lfloor frac{2006}{11^3} rfloor + dots ).But wait, the product from 1001 to 2006 isn't exactly ( 2006! ). It's actually ( frac{2006!}{1000!} ). So, to find the exponent of 11 in this product, I need to subtract the exponent of 11 in ( 1000! ) from the exponent of 11 in ( 2006! ). That makes sense because when you divide factorials, the exponents of the primes subtract.Alright, let me start by calculating the exponent of 11 in ( 2006! ). First, ( lfloor frac{2006}{11} rfloor ). Let me compute that. 2006 divided by 11 is approximately 182.36. So, taking the floor of that gives me 182. Next, ( lfloor frac{2006}{11^2} rfloor ). 11 squared is 121. So, 2006 divided by 121 is approximately 16.578. Taking the floor gives me 16.Then, ( lfloor frac{2006}{11^3} rfloor ). 11 cubed is 1331. Dividing 2006 by 1331 gives approximately 1.507. Taking the floor, that's 1.What about ( 11^4 )? That's 14641, which is way larger than 2006, so ( lfloor frac{2006}{14641} rfloor ) is 0. So, I can stop here.Adding them up: 182 + 16 + 1 = 199. So, the exponent of 11 in ( 2006! ) is 199.Now, I need to find the exponent of 11 in ( 1000! ). Let's do the same process.First, ( lfloor frac{1000}{11} rfloor ). 1000 divided by 11 is approximately 90.909. Taking the floor gives me 90.Next, ( lfloor frac{1000}{11^2} rfloor ). That's 1000 divided by 121, which is approximately 8.264. Taking the floor gives me 8.What about ( lfloor frac{1000}{11^3} rfloor )? 1000 divided by 1331 is less than 1, so that's 0. So, I can stop here.Adding them up: 90 + 8 = 98. So, the exponent of 11 in ( 1000! ) is 98.Therefore, the exponent of 11 in the product from 1001 to 2006 is ( 199 - 98 = 101 ). So, the maximum value of ( k ) is 101.Wait, let me double-check my calculations to make sure I didn't make a mistake. For ( 2006! ):- ( 2006 / 11 = 182.36 ) → 182- ( 2006 / 121 = 16.578 ) → 16- ( 2006 / 1331 = 1.507 ) → 1Total: 182 + 16 + 1 = 199. That seems correct.For ( 1000! ):- ( 1000 / 11 = 90.909 ) → 90- ( 1000 / 121 = 8.264 ) → 8Total: 90 + 8 = 98. That also seems correct.Subtracting them: 199 - 98 = 101. Yeah, that makes sense. I think I got it right. So, the maximum ( k ) is 101.

💡Okay, so I have this problem about an even function, and I need to figure out which of the statements A, B, C, or D must be true. Let me start by recalling what an even function is. An even function satisfies the condition f(-x) = f(x) for all real numbers x. That means the graph of the function is symmetric with respect to the y-axis. Alright, let me go through each option one by one.Option A says: For all x in R, f(x) > f(-x). Hmm, since f is even, f(-x) is equal to f(x). So, substituting that in, this statement becomes f(x) > f(x) for all x. But wait, that's not possible because a number can't be greater than itself. So, this statement is always false. Therefore, option A is incorrect.Moving on to option B: There exists an x in R such that f(x) > f(-x). Again, since f is even, f(-x) = f(x), so this would mean there exists an x where f(x) > f(x). But that's like saying there's an x where a number is greater than itself, which is impossible. So, option B is also incorrect.Now, option C: For all x in R, f(x)f(-x) ≥ 0. Let's think about this. Since f is even, f(-x) = f(x). So, f(x)f(-x) becomes f(x)*f(x), which is [f(x)]². Now, the square of any real number is always non-negative, right? So, [f(x)]² is always greater than or equal to zero. Therefore, this statement is always true for any even function. So, option C must be true.Lastly, option D: There exists an x in R such that f(x)f(-x) < 0. Again, substituting f(-x) with f(x), this becomes f(x)*f(x) < 0, which is [f(x)]² < 0. But as I just thought earlier, the square of a real number can't be negative. So, this statement is always false. Therefore, option D is incorrect.Wait, let me double-check if there's any case where option C might not hold. Suppose f(x) is zero for some x, then [f(x)]² would be zero, which is still non-negative. If f(x) is positive, then [f(x)]² is positive. If f(x) is negative, since it's squared, it becomes positive. So, in all cases, [f(x)]² is non-negative. Therefore, option C is indeed always true.I think that's solid. So, the only statement that must be true is option C.**Final Answer**boxed{C}

💡Alright, so I have this problem here: I need to find two positive 4-digit integers, k and ℓ, such that their greatest common divisor (gcd) is 5. Then, I have to find the smallest possible value for their least common multiple (lcm). Hmm, okay, let's break this down step by step.First, I remember that there's a relationship between gcd and lcm of two numbers. Specifically, the product of the gcd and lcm of two numbers equals the product of the numbers themselves. So, mathematically, that's:[gcd(k, ell) times mathop{text{lcm}}[k, ell] = k times ell]Given that the gcd is 5, this formula becomes:[5 times mathop{text{lcm}}[k, ell] = k times ell]Which means:[mathop{text{lcm}}[k, ell] = frac{k times ell}{5}]So, to minimize the lcm, I need to minimize the product (k times ell). Since both k and ℓ are 4-digit numbers, the smallest 4-digit number is 1000. But wait, they also have to be multiples of 5 because their gcd is 5. So, the smallest 4-digit multiples of 5 are 1000 and 1005.Let me check if 1000 and 1005 have a gcd of 5. To do that, I can factorize both numbers:- 1000 can be factored into (2^3 times 5^3).- 1005 can be factored into (3 times 5 times 67).Looking at these factorizations, the only common prime factor is 5, and the lowest power of 5 in both is (5^1). So, yes, the gcd is indeed 5.Now, let's compute the lcm using the formula:[mathop{text{lcm}}[1000, 1005] = frac{1000 times 1005}{5} = frac{1000 times 1005}{5}]Simplifying this, 1000 divided by 5 is 200, so:[200 times 1005 = 201,000]So, the lcm is 201,000. Is this the smallest possible? Well, since 1000 and 1005 are the smallest 4-digit multiples of 5, their product should give the smallest possible lcm when divided by 5. I don't think there are smaller 4-digit numbers that are multiples of 5 with a gcd of 5, so this should be the minimum.Wait, just to make sure, are there any other pairs of 4-digit numbers that are multiples of 5, have a gcd of 5, and result in a smaller lcm? Let's think about it.If I take the next pair, say 1005 and 1010. Their gcd would still be 5 because 1005 is (3 times 5 times 67) and 1010 is (2 times 5 times 101). So, gcd is 5. Then, their lcm would be:[frac{1005 times 1010}{5} = frac{1005 times 1010}{5} = 201 times 1010 = 203,010]Which is larger than 201,000. So, 201,000 is indeed smaller.What if I take 1000 and 1010? Let's check their gcd. 1000 is (2^3 times 5^3) and 1010 is (2 times 5 times 101). So, the gcd would be (2 times 5 = 10), which is higher than 5. So, that doesn't satisfy the condition of gcd being exactly 5.Similarly, if I take 1000 and 1015, let's see. 1015 is (5 times 7 times 29). So, gcd(1000, 1015) is 5, since 1000 has (5^3) and 1015 has (5^1). So, their lcm would be:[frac{1000 times 1015}{5} = 200 times 1015 = 203,000]Which is again larger than 201,000.So, it seems like 1000 and 1005 give the smallest possible lcm of 201,000. Let me just confirm if there are any other pairs closer to 1000 that might give a smaller product.Wait, 1000 and 1005 are consecutive multiples of 5, so their product is as small as possible. The next multiple after 1005 is 1010, which we already saw gives a larger lcm. So, I think 201,000 is indeed the smallest possible lcm.Just to recap:1. The problem requires two 4-digit numbers with gcd 5.2. The smallest 4-digit multiples of 5 are 1000 and 1005.3. Their gcd is 5, as their only common factor is 5.4. Calculating their lcm gives 201,000.5. Checking other nearby multiples of 5 results in larger lcms.Therefore, the smallest possible lcm is 201,000.**Final Answer**The smallest possible value for the least common multiple is boxed{201000}.

💡First, I need to identify the unique digits in the number 3033. The digits are 3, 0, 3, and 3. Here, the digit 3 appears three times, and the digit 0 appears once.To form a valid four-digit number, the first digit (thousands place) must be 3 because 0 cannot be the leading digit. This leaves me with the remaining digits: 0, 3, and 3.Next, I need to determine the number of unique arrangements for the last three digits. Since there are two identical 3s, the number of unique permutations is calculated by dividing the total permutations by the factorial of the number of identical digits. This gives me 3! / 2! = 3 unique arrangements.Therefore, there are 3 different four-digit numbers that can be formed using the digits in 3033.

💡Okay, so I have this geometry problem here, and I need to prove that point N bisects the segment AH in an acute triangle ABC with orthocenter H. Let me try to visualize and break down the problem step by step.First, let me recall some basic definitions:- **Orthocenter (H):** The point where the three altitudes of a triangle intersect.- **Feet of the altitudes (D, E, F):** The points where the altitudes from A, B, and C meet the opposite sides.Given that ABC is an acute triangle, all the feet of the altitudes lie on the sides, and the orthocenter H lies inside the triangle.Now, let me parse the problem:1. **DF intersects the altitude from B at P.** - So, DF is a line connecting the feet of the altitudes from A and C. - The altitude from B is the line BH, which goes from B to E (the foot on AC). - So, P is the intersection point of DF and BH.2. **The normal to BC through P intersects AB at Q.** - A normal to BC through P is a line perpendicular to BC passing through P. - This normal intersects side AB at point Q.3. **EQ intersects the altitude from A at N.** - EQ is the line connecting E (foot from B) and Q. - The altitude from A is the line AH, which goes from A to D (the foot on BC). - So, N is the intersection point of EQ and AH.The goal is to prove that N bisects AH, meaning that AN = NH.Hmm, okay. Let me try to draw this triangle and mark all these points to get a better idea.[Imagining the triangle ABC with all the points D, E, F, H, P, Q, N marked.]Now, to prove that N bisects AH, I need to show that N is the midpoint of AH. That is, AN = NH.I wonder if there are any properties or theorems related to orthocenters, altitudes, and midpoints that I can use here. Maybe something related to similar triangles, harmonic divisions, or projective geometry?Let me think about the properties of the orthocenter and the feet of the altitudes. Since D, E, F are feet of the altitudes, they form the orthic triangle of ABC. The orthic triangle has some interesting properties, especially in acute triangles.Also, since P is the intersection of DF and BH, and Q is the intersection of the normal from P to BC with AB, maybe there's a way to relate these points through some projective transformations or harmonic conjugates.Wait, harmonic conjugates might be useful here. If I can show that N is the harmonic conjugate of H with respect to A and some other point, that might help me establish that N is the midpoint.Alternatively, maybe using coordinate geometry could be a straightforward approach. Assign coordinates to the triangle and compute the coordinates of N to check if it's the midpoint.Let me try that. Let's assign coordinates to triangle ABC.Let me place point A at (0, 0), point B at (2b, 0), and point C at (2c, 2d), ensuring that the triangle is acute. The coordinates are chosen to be even numbers to make calculations easier, but I can adjust later if needed.Now, let's find the coordinates of the orthocenter H.First, I need the equations of the altitudes.1. **Altitude from A to BC:** - The side BC goes from (2b, 0) to (2c, 2d). - The slope of BC is (2d - 0)/(2c - 2b) = d/(c - b). - Therefore, the slope of the altitude from A is the negative reciprocal: -(c - b)/d. - Since it passes through A(0, 0), its equation is y = [-(c - b)/d]x.2. **Altitude from B to AC:** - The side AC goes from (0, 0) to (2c, 2d). - The slope of AC is (2d - 0)/(2c - 0) = d/c. - Therefore, the slope of the altitude from B is the negative reciprocal: -c/d. - Since it passes through B(2b, 0), its equation is y = [-c/d](x - 2b).3. **Altitude from C to AB:** - The side AB is horizontal, from (0, 0) to (2b, 0). - The slope of AB is 0, so the altitude from C is vertical. - Therefore, the equation is x = 2c.Wait, but if AB is horizontal, the altitude from C should be vertical, yes. So, the altitude from C is x = 2c.But wait, in an acute triangle, the orthocenter lies inside the triangle. Let me check if with these coordinates, H is inside.Let me compute H by solving the intersection of two altitudes, say from A and from B.From A: y = [-(c - b)/d]x.From B: y = [-c/d](x - 2b).Set them equal:[-(c - b)/d]x = [-c/d](x - 2b)Multiply both sides by d:-(c - b)x = -c(x - 2b)Simplify:-(c - b)x = -cx + 2bcMultiply both sides by -1:(c - b)x = cx - 2bcExpand left side:cx - bx = cx - 2bcSubtract cx from both sides:-bx = -2bcDivide both sides by -b (assuming b ≠ 0):x = 2cWait, that's interesting. So x = 2c, which is the same as the altitude from C. So, the orthocenter H is at (2c, y). Let me plug x = 2c into the equation from A:y = [-(c - b)/d](2c) = [-2c(c - b)]/dSo, H is at (2c, [-2c(c - b)]/d)Wait, but in my coordinate system, point C is at (2c, 2d). So, H is at (2c, [-2c(c - b)]/d). Hmm, that seems a bit odd because the y-coordinate is negative if c > b, but since the triangle is acute, H should be inside the triangle, so y should be positive.Wait, maybe I made a mistake in the slope of the altitude from A.Let me double-check.The slope of BC is (2d - 0)/(2c - 2b) = d/(c - b). So, the slope of the altitude from A should be the negative reciprocal, which is -(c - b)/d. That seems correct.Then, plugging into the equation, y = [-(c - b)/d]x.From B, the altitude has slope -c/d, passing through (2b, 0), so y = [-c/d](x - 2b).Setting them equal:[-(c - b)/d]x = [-c/d](x - 2b)Multiply both sides by d:-(c - b)x = -c(x - 2b)Simplify:-(c - b)x = -cx + 2bcMultiply both sides by -1:(c - b)x = cx - 2bcExpand:cx - bx = cx - 2bcSubtract cx:-bx = -2bcDivide by -b:x = 2cSo, x = 2c, which is the same as the altitude from C, which is vertical at x = 2c.So, H is at (2c, y). Plugging back into the equation from A:y = [-(c - b)/d](2c) = [-2c(c - b)]/dHmm, so if c > b, then y is negative, which would place H below the x-axis, but in an acute triangle, H should be inside the triangle, so y should be positive. Therefore, perhaps my coordinate system is not suitable.Maybe I should choose coordinates such that H is inside the triangle. Let me instead place the triangle in a way that all coordinates are positive and H is inside.Let me try a different coordinate system. Let me set A at (0, 0), B at (2, 0), and C at (0, 2). So, ABC is a right-angled triangle at A, but wait, it needs to be acute. So, maybe C at (1, 2). Let me check.Wait, if I set A at (0, 0), B at (2, 0), and C at (1, 2), then ABC is acute.Let me compute the orthocenter H.First, find the equations of the altitudes.1. **Altitude from A to BC:** - BC goes from (2, 0) to (1, 2). - Slope of BC: (2 - 0)/(1 - 2) = 2/(-1) = -2. - Therefore, slope of altitude from A is 1/2. - Equation: y = (1/2)x.2. **Altitude from B to AC:** - AC goes from (0, 0) to (1, 2). - Slope of AC: (2 - 0)/(1 - 0) = 2. - Therefore, slope of altitude from B is -1/2. - Equation: passes through B(2, 0): y - 0 = (-1/2)(x - 2) => y = (-1/2)x + 1.3. **Altitude from C to AB:** - AB is horizontal from (0, 0) to (2, 0). - So, altitude from C is vertical, x = 1.Now, find H by solving the intersection of two altitudes.First, solve altitude from A and altitude from B:y = (1/2)x and y = (-1/2)x + 1.Set equal:(1/2)x = (-1/2)x + 1Add (1/2)x to both sides:x = 1Then y = (1/2)(1) = 1/2.So, H is at (1, 1/2).Good, that's inside the triangle.Now, let's find D, E, F.- D is the foot from A to BC.- E is the foot from B to AC.- F is the foot from C to AB.Compute D:Equation of BC: from (2, 0) to (1, 2). Slope is -2, as before. Equation: y - 0 = -2(x - 2) => y = -2x + 4.Altitude from A is y = (1/2)x.Intersection D is where y = (1/2)x and y = -2x + 4.Set equal:(1/2)x = -2x + 4Multiply both sides by 2:x = -4x + 8Add 4x:5x = 8 => x = 8/5 = 1.6Then y = (1/2)(8/5) = 4/5 = 0.8So, D is at (8/5, 4/5).Compute E:Foot from B to AC.Equation of AC: from (0, 0) to (1, 2). Slope is 2, equation: y = 2x.Altitude from B has slope -1/2, equation: y = (-1/2)x + 1.Intersection E is where y = 2x and y = (-1/2)x + 1.Set equal:2x = (-1/2)x + 1Multiply both sides by 2:4x = -x + 2Add x:5x = 2 => x = 2/5 = 0.4Then y = 2*(2/5) = 4/5 = 0.8So, E is at (2/5, 4/5).Compute F:Foot from C to AB.AB is horizontal, y = 0. So, the foot from C(1, 2) is (1, 0).So, F is at (1, 0).Now, let's find P: intersection of DF and BH.First, find equation of DF.Points D(8/5, 4/5) and F(1, 0).Slope of DF: (0 - 4/5)/(1 - 8/5) = (-4/5)/(-3/5) = (4/5)/(3/5) = 4/3.Equation of DF: Using point F(1, 0):y - 0 = (4/3)(x - 1) => y = (4/3)x - 4/3.Now, find equation of BH.Point B is at (2, 0), and H is at (1, 1/2).Slope of BH: (1/2 - 0)/(1 - 2) = (1/2)/(-1) = -1/2.Equation of BH: y - 0 = (-1/2)(x - 2) => y = (-1/2)x + 1.Intersection P is where y = (4/3)x - 4/3 and y = (-1/2)x + 1.Set equal:(4/3)x - 4/3 = (-1/2)x + 1Multiply both sides by 6 to eliminate denominators:8x - 8 = -3x + 6Add 3x:11x - 8 = 6Add 8:11x = 14 => x = 14/11 ≈ 1.2727Then y = (-1/2)(14/11) + 1 = (-7/11) + 1 = 4/11 ≈ 0.3636So, P is at (14/11, 4/11).Next, find Q: the normal to BC through P intersects AB at Q.First, find the normal to BC through P.Slope of BC is -2, so the normal has slope 1/2.Equation of normal: y - 4/11 = (1/2)(x - 14/11)Simplify:y = (1/2)x - (14/11)(1/2) + 4/11= (1/2)x - 7/11 + 4/11= (1/2)x - 3/11This normal intersects AB at Q. AB is y = 0.Set y = 0:0 = (1/2)x - 3/11=> (1/2)x = 3/11=> x = 6/11 ≈ 0.5455So, Q is at (6/11, 0).Now, find EQ: line from E(2/5, 4/5) to Q(6/11, 0).Compute the slope:(0 - 4/5)/(6/11 - 2/5) = (-4/5)/(30/55 - 22/55) = (-4/5)/(8/55) = (-4/5)*(55/8) = (-4*11)/8 = (-44)/8 = -11/2.Equation of EQ: Using point Q(6/11, 0):y - 0 = (-11/2)(x - 6/11)Simplify:y = (-11/2)x + (11/2)*(6/11) = (-11/2)x + 3Now, find N: intersection of EQ and AH.AH is the altitude from A to BC, which we already have as y = (1/2)x.Set y = (1/2)x equal to y = (-11/2)x + 3.(1/2)x = (-11/2)x + 3Add (11/2)x to both sides:(12/2)x = 36x = 3 => x = 1/2Then y = (1/2)(1/2) = 1/4So, N is at (1/2, 1/4).Now, let's check if N bisects AH.Point A is at (0, 0), H is at (1, 1/2).Midpoint of AH should be at ((0 + 1)/2, (0 + 1/2)/2) = (0.5, 0.25), which is exactly N(1/2, 1/4).Therefore, N is indeed the midpoint of AH, so N bisects AH.Wow, that worked out! By assigning coordinates and computing each point step by step, I was able to verify that N is the midpoint of AH. This approach, while computational, confirms the result. I think this is a solid way to prove the statement, especially since it's an acute triangle and all the points lie where they should.I wonder if there's a more elegant synthetic proof without coordinates, maybe using properties of harmonic division or projective geometry as I initially thought. But for now, the coordinate method suffices to show that N bisects AH.

💡Let's denote the two positive numbers as ( x ) and ( y ), where ( x > y ).We are given that the sum of the two numbers is 6 times their difference. This can be written as:[x + y = 6(x - y)]Expanding and simplifying the equation:[x + y = 6x - 6y][x + y - 6x + 6y = 0][-5x + 7y = 0]Rearranging the equation to find the ratio of ( x ) to ( y ):[7y = 5x][frac{x}{y} = frac{7}{5}]Thus, the ratio of the larger number to the smaller number is ( frac{7}{5} ).