💡Okay, so I have this problem about an ellipse and a circle. Let me try to understand what it's asking. The ellipse is given by the equation x²/a² + y²/b² = 1, where a is greater than b, which is greater than 0. The left and right vertices of the ellipse are A₁ and A₂ respectively. Then, there's a circle with diameter A₁A₂, and this circle is tangent to the line bx - ay + 3ab = 0. I need to find the eccentricity of the ellipse C.First, let me recall some properties of ellipses. The standard form is x²/a² + y²/b² = 1, where a is the semi-major axis and b is the semi-minor axis. The vertices are at (±a, 0), so A₁ is (-a, 0) and A₂ is (a, 0). The distance between these two points is 2a, so the diameter of the circle is 2a, meaning the radius is a. The center of the circle would be the midpoint of A₁A₂, which is the origin (0, 0). So, the equation of the circle is x² + y² = a².Now, the circle is tangent to the line bx - ay + 3ab = 0. For a line to be tangent to a circle, the distance from the center of the circle to the line must be equal to the radius of the circle. The formula for the distance from a point (x₀, y₀) to the line Ax + By + C = 0 is |Ax₀ + By₀ + C| / sqrt(A² + B²). In this case, the center of the circle is (0, 0), so plugging into the distance formula, we get |b*0 - a*0 + 3ab| / sqrt(b² + (-a)²) = |3ab| / sqrt(a² + b²).Since the circle is tangent to the line, this distance must be equal to the radius of the circle, which is a. So, we have |3ab| / sqrt(a² + b²) = a. Because a and b are positive, we can drop the absolute value: 3ab / sqrt(a² + b²) = a.Let me write that equation down:3ab / sqrt(a² + b²) = aI can simplify this equation. First, multiply both sides by sqrt(a² + b²):3ab = a * sqrt(a² + b²)Then, divide both sides by a (since a ≠ 0):3b = sqrt(a² + b²)Now, square both sides to eliminate the square root:(3b)² = (sqrt(a² + b²))²9b² = a² + b²Subtract b² from both sides:8b² = a²So, a² = 8b². Therefore, a = sqrt(8)b = 2*sqrt(2)b.Now, I need to find the eccentricity of the ellipse. The formula for eccentricity e is e = c/a, where c is the distance from the center to a focus. For an ellipse, c² = a² - b².We already have a² = 8b², so c² = 8b² - b² = 7b². Therefore, c = sqrt(7b²) = b*sqrt(7).Now, the eccentricity e is c/a = (b*sqrt(7)) / (2*sqrt(2)b) = sqrt(7)/(2*sqrt(2)). Let me rationalize the denominator:sqrt(7)/(2*sqrt(2)) = (sqrt(7)*sqrt(2))/(2*2) = sqrt(14)/4.So, the eccentricity is sqrt(14)/4.Looking back at the options, option C is sqrt(14)/4, so that should be the correct answer.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Identified A₁ and A₂ as (-a, 0) and (a, 0). Correct.2. Circle with diameter A₁A₂ has center at (0,0) and radius a. Equation x² + y² = a². Correct.3. Distance from center (0,0) to the line bx - ay + 3ab = 0 is |3ab| / sqrt(b² + a²). Correct.4. Set this equal to radius a: 3ab / sqrt(a² + b²) = a. Correct.5. Simplified to 3b = sqrt(a² + b²). Squared both sides: 9b² = a² + b². So, 8b² = a². Correct.6. Eccentricity e = c/a, where c = sqrt(a² - b²) = sqrt(8b² - b²) = sqrt(7b²) = b*sqrt(7). So, e = (b*sqrt(7))/(2*sqrt(2)b) = sqrt(7)/(2*sqrt(2)) = sqrt(14)/4. Correct.Everything seems to check out. So, I'm confident that the answer is C: sqrt(14)/4.**Final Answer**boxed{C}

💡First, I'll assign variables to represent the work rates of A, B, C, and D. Let A, B, C, and D represent their respective work rates in units of work per hour.From the given information:1. A can complete the work in 5 hours, so A = 1/5 (work/hour).2. A and C together can complete the work in 2.5 hours, so A + C = 1/2.5 = 2/5 (work/hour).3. B and C together can complete the work in 4 hours, so B + C = 1/4 (work/hour).4. A and D together can complete the work in 3.6 hours, so A + D = 1/3.6 ≈ 0.2778 (work/hour).5. B, C, and D together can complete the work in 2 hours, so B + C + D = 1/2 = 0.5 (work/hour).Using equation 1, I know A = 1/5.Substituting A into equation 2:1/5 + C = 2/5C = 2/5 - 1/5C = 1/5 (work/hour).Now, using equation 3 to find B:B + 1/5 = 1/4B = 1/4 - 1/5B = 5/20 - 4/20B = 1/20 (work/hour).Therefore, B can complete the work alone in 20 hours.

💡First, I need to determine the number of ways to choose each position on the volleyball team. There are 10 team members, and each position must be filled by a different player.For the setter position, there are 10 possible choices since all members can play any position.After selecting the setter, there are 9 remaining members to choose from for the libero position.With the libero selected, 8 members are left for the outside hitter position.Next, 7 members remain for the middle blocker position.Finally, 6 members are left for the opposite hitter position.To find the total number of different lineups, I multiply the number of choices for each position together: 10 × 9 × 8 × 7 × 6.Calculating this gives 30,240 possible lineups.

💡To determine the number of 5-card hands that contain only queens and jacks from a standard 52-card deck, follow these steps:1. **Identify the relevant cards:** - There are 4 queens in the deck. - There are 4 jacks in the deck. - Total relevant cards = 4 queens + 4 jacks = 8 cards.2. **Determine the number of ways to choose 5 cards from these 8:** - This is a combination problem, where the order of selection does not matter. - The formula for combinations is: [ binom{n}{k} = frac{n!}{k!(n-k)!} ] where ( n ) is the total number of items, and ( k ) is the number of items to choose.3. **Apply the combination formula:** - Here, ( n = 8 ) and ( k = 5 ). [ binom{8}{5} = frac{8!}{5!(8-5)!} = frac{8!}{5! cdot 3!} ]4. **Calculate the factorials:** - ( 8! = 40320 ) - ( 5! = 120 ) - ( 3! = 6 )5. **Substitute the factorial values into the formula:** [ binom{8}{5} = frac{40320}{120 times 6} = frac{40320}{720} = 56 ]**Final Answer:** The number of 5-card hands that contain only queens and jacks is (boxed{56}).

💡Alright, so I have this problem here: Let ( m ) and ( n ) be positive integers with ( m > 2 ). I need to prove something about ( (2^m - 1) times (2^n + 1) ). Hmm, the problem statement just says "Prove: ( (2^m - 1) times (2^n + 1) )." That feels a bit incomplete. Maybe there's something missing? Like, what exactly am I supposed to prove about this product? Is it supposed to be a certain type of number, or maybe it's supposed to be divisible by something?Wait, looking back at the original problem, it just says "Prove: ( (2^m - 1) times (2^n + 1) )." That doesn't specify what to prove. Maybe it's a typo or something got cut off. But okay, maybe I can assume what needs to be proven. Often, with expressions like ( 2^m - 1 ) and ( 2^n + 1 ), they relate to Mersenne numbers or other number-theoretic concepts. Maybe I need to show that this product is composite or has certain properties.Alternatively, perhaps the problem is to show that ( 2^m - 1 ) divides ( 2^n + 1 ) under certain conditions, or something similar. Let me think. If ( m > 2 ), then ( 2^m - 1 ) is definitely greater than 3. For example, if ( m = 3 ), ( 2^3 - 1 = 7 ). So, ( 7 times (2^n + 1) ) would be the product. Maybe the problem is to show that this product is composite? But that seems too straightforward because both ( 2^m - 1 ) and ( 2^n + 1 ) are integers greater than 1, so their product is composite.Wait, but if ( m = 1 ), ( 2^1 - 1 = 1 ), but ( m > 2 ), so ( m ) is at least 3. So, ( 2^m - 1 ) is at least 7, and ( 2^n + 1 ) is at least 3 when ( n = 1 ). So, their product is at least 21, which is composite. So, maybe the problem is to show that ( (2^m - 1) times (2^n + 1) ) is composite? But that seems too simple. Maybe it's more about the factors or properties of this product.Alternatively, perhaps the problem is to show that ( 2^m - 1 ) divides ( 2^n + 1 ) under certain conditions. For example, if ( n ) is a multiple of ( m ), or something like that. Let me explore that.Suppose ( n = km ) for some integer ( k ). Then, ( 2^n + 1 = 2^{km} + 1 ). I know that ( 2^{km} - 1 ) is divisible by ( 2^m - 1 ) because ( 2^{km} - 1 = (2^m)^k - 1 ), which factors as ( (2^m - 1)(2^{m(k-1)} + 2^{m(k-2)} + dots + 2^m + 1) ). But here we have ( 2^{km} + 1 ), which is different.Wait, ( 2^{km} + 1 ) can be factored if ( k ) is odd. For example, ( 2^{3m} + 1 = (2^m + 1)(2^{2m} - 2^m + 1) ). So, if ( k ) is odd, ( 2^{km} + 1 ) can be factored. But does ( 2^m - 1 ) divide ( 2^{km} + 1 )?Let me test with specific numbers. Let ( m = 3 ), so ( 2^3 - 1 = 7 ). Let ( n = 6 ), which is ( 2 times 3 ). Then, ( 2^6 + 1 = 64 + 1 = 65 ). Does 7 divide 65? 65 divided by 7 is about 9.28, so no. So, in this case, ( 2^m - 1 ) does not divide ( 2^n + 1 ).What if ( n = 9 ), which is ( 3 times 3 ). Then, ( 2^9 + 1 = 512 + 1 = 513 ). Does 7 divide 513? 513 divided by 7 is approximately 73.28, so no. Hmm, so maybe ( 2^m - 1 ) doesn't divide ( 2^n + 1 ) when ( n ) is a multiple of ( m ).Wait, maybe it's when ( n ) is of the form ( 2m ) or something else. Let me try ( n = 4 ) with ( m = 3 ). Then, ( 2^4 + 1 = 16 + 1 = 17 ). Does 7 divide 17? No. Hmm.Alternatively, maybe ( n ) is related to ( m ) in another way. Let me think about the properties of ( 2^m - 1 ) and ( 2^n + 1 ). ( 2^m - 1 ) is a Mersenne number, and ( 2^n + 1 ) is similar to a Fermat number, but Fermat numbers are ( 2^{2^k} + 1 ).Wait, perhaps if ( n ) is of the form ( 2^k ), then ( 2^n + 1 ) is a Fermat number, but I don't see the connection immediately.Alternatively, maybe the problem is to show that ( (2^m - 1) times (2^n + 1) ) is not prime, which is obvious because it's a product of two numbers greater than 1. But that seems too trivial.Wait, maybe the original problem was to prove that ( 2^m - 1 ) divides ( 2^n + 1 ) under certain conditions, and the product is part of that. But in the initial problem statement, it just says "Prove: ( (2^m - 1) times (2^n + 1) )." Maybe it's supposed to be an identity or something.Wait, perhaps the problem is to show that ( (2^m - 1) times (2^n + 1) ) equals something else, but it's not specified. Maybe it's supposed to be equal to ( 2^{m+n} - 2^n + 2^m - 1 ). Let me check:( (2^m - 1)(2^n + 1) = 2^m times 2^n + 2^m - 2^n - 1 = 2^{m+n} + 2^m - 2^n - 1 ). Yeah, that's correct. So, maybe the problem is to expand the product, but that seems too straightforward.Alternatively, maybe it's to show that ( (2^m - 1) times (2^n + 1) ) is divisible by something, like 3 or another number. Let me see.If ( m ) is odd, ( 2^m - 1 ) is divisible by 3 because ( 2 equiv -1 mod 3 ), so ( 2^m equiv (-1)^m mod 3 ). If ( m ) is odd, ( (-1)^m = -1 ), so ( 2^m - 1 equiv -1 - 1 = -2 equiv 1 mod 3 ). Wait, that doesn't make sense. Wait, ( 2^1 - 1 = 1 ), which is not divisible by 3. ( 2^3 - 1 = 7 ), which is not divisible by 3. ( 2^5 - 1 = 31 ), not divisible by 3. Hmm, so maybe not.If ( m ) is even, ( 2^m - 1 ) is divisible by 3 because ( 2^2 = 4 equiv 1 mod 3 ), so ( 2^{2k} equiv 1^k = 1 mod 3 ), so ( 2^{2k} - 1 equiv 0 mod 3 ). So, if ( m ) is even, ( 2^m - 1 ) is divisible by 3.Similarly, ( 2^n + 1 ). If ( n ) is odd, ( 2^n equiv -1 mod 3 ), so ( 2^n + 1 equiv -1 + 1 = 0 mod 3 ). So, if ( n ) is odd, ( 2^n + 1 ) is divisible by 3.So, if ( m ) is even and ( n ) is odd, then both ( 2^m - 1 ) and ( 2^n + 1 ) are divisible by 3, so their product is divisible by 9.But I don't know if that's what the problem is asking.Wait, maybe the problem is to show that ( (2^m - 1) times (2^n + 1) ) is composite, which is obvious because both factors are greater than 1. But maybe it's more about the factors or properties.Alternatively, perhaps the problem is to show that ( 2^m - 1 ) divides ( 2^n + 1 ) when ( n ) is congruent to something modulo ( m ). Let me think about that.Suppose ( n equiv 1 mod m ). Then, ( n = km + 1 ). So, ( 2^n + 1 = 2^{km + 1} + 1 = 2 times (2^{km}) + 1 ). Now, ( 2^{km} equiv 1 mod (2^m - 1) ) because ( 2^m equiv 1 mod (2^m - 1) ), so ( 2^{km} = (2^m)^k equiv 1^k = 1 mod (2^m - 1) ). Therefore, ( 2 times (2^{km}) + 1 equiv 2 times 1 + 1 = 3 mod (2^m - 1) ). So, ( 2^n + 1 equiv 3 mod (2^m - 1) ). So, unless ( 2^m - 1 ) divides 3, which only happens when ( 2^m - 1 = 3 ), i.e., ( m = 2 ), but ( m > 2 ), so ( 2^m - 1 ) doesn't divide 3. So, in this case, ( 2^m - 1 ) does not divide ( 2^n + 1 ).Wait, maybe if ( n equiv m - 1 mod 2m ). Let me try with ( m = 3 ). Then, ( 2^3 - 1 = 7 ). Let ( n = 5 ), which is ( 2 times 3 - 1 = 5 ). Then, ( 2^5 + 1 = 32 + 1 = 33 ). Does 7 divide 33? 33 divided by 7 is about 4.714, so no.Alternatively, maybe ( n equiv m + 1 mod 2m ). Let me try ( n = 4 ) with ( m = 3 ). ( 2^4 + 1 = 16 + 1 = 17 ). 17 divided by 7 is about 2.428, so no.Hmm, maybe I'm approaching this the wrong way. Let me think about the original problem again. It just says "Prove: ( (2^m - 1) times (2^n + 1) )." Maybe it's supposed to be an identity or to show that this product equals something else, but it's not specified. Alternatively, maybe it's to show that this product is a certain type of number, like a perfect number or something else.Wait, perfect numbers are related to Mersenne primes. A perfect number is of the form ( 2^{p-1}(2^p - 1) ) where ( 2^p - 1 ) is a Mersenne prime. But here we have ( (2^m - 1) times (2^n + 1) ), which doesn't directly fit that form.Alternatively, maybe the problem is to show that this product is not a perfect square or something like that. Let me check with specific numbers. Let ( m = 3 ), ( n = 1 ). Then, ( (2^3 - 1) times (2^1 + 1) = 7 times 3 = 21 ), which is not a perfect square. ( m = 3 ), ( n = 2 ): ( 7 times 5 = 35 ), not a square. ( m = 4 ), ( n = 1 ): ( 15 times 3 = 45 ), not a square. ( m = 4 ), ( n = 2 ): ( 15 times 5 = 75 ), not a square. So, it seems like the product is rarely a perfect square, but that might not be what the problem is asking.Wait, maybe the problem is to show that ( (2^m - 1) times (2^n + 1) ) is always even. Let's see: ( 2^m - 1 ) is odd because ( 2^m ) is even, so subtracting 1 makes it odd. ( 2^n + 1 ) is also odd because ( 2^n ) is even, so adding 1 makes it odd. The product of two odd numbers is odd. So, the product is odd. So, maybe the problem is to show that it's odd, but that's straightforward.Alternatively, maybe it's to show that it's divisible by 3 under certain conditions. Earlier, I saw that if ( m ) is even, ( 2^m - 1 ) is divisible by 3, and if ( n ) is odd, ( 2^n + 1 ) is divisible by 3. So, if both ( m ) is even and ( n ) is odd, the product is divisible by 9. But if ( m ) is odd and ( n ) is even, neither ( 2^m - 1 ) nor ( 2^n + 1 ) is divisible by 3, so the product isn't divisible by 3.Wait, maybe the problem is to show that ( (2^m - 1) times (2^n + 1) ) is divisible by 3 if and only if ( m ) is even or ( n ) is odd. But that might be more involved.Alternatively, maybe the problem is to show that ( (2^m - 1) times (2^n + 1) ) is composite, which is obvious because it's a product of two integers greater than 1. But that seems too simple.Wait, maybe the problem is to show that ( (2^m - 1) times (2^n + 1) ) is a multiple of ( 2^{m+n} - 2^n + 2^m - 1 ), but that's just the expansion of the product, which is trivial.Alternatively, maybe it's to show that ( (2^m - 1) times (2^n + 1) ) is a multiple of ( 2^{gcd(m,n)} - 1 ) or something like that. Let me think.The gcd of ( 2^m - 1 ) and ( 2^n - 1 ) is ( 2^{gcd(m,n)} - 1 ). But here we have ( 2^n + 1 ), which is different. So, maybe the gcd of ( 2^m - 1 ) and ( 2^n + 1 ) is something else.Wait, let me try specific numbers. Let ( m = 3 ), ( n = 1 ): ( gcd(7, 3) = 1 ). ( m = 3 ), ( n = 2 ): ( gcd(7, 5) = 1 ). ( m = 3 ), ( n = 3 ): ( gcd(7, 9) = 1 ). ( m = 4 ), ( n = 1 ): ( gcd(15, 3) = 3 ). ( m = 4 ), ( n = 2 ): ( gcd(15, 5) = 5 ). ( m = 4 ), ( n = 3 ): ( gcd(15, 9) = 3 ). So, it seems like the gcd can vary.Hmm, I'm not sure. Maybe the problem is to show that ( (2^m - 1) times (2^n + 1) ) is always greater than something, but that seems vague.Wait, going back to the original problem: "Let ( m ) and ( n ) be positive integers with ( m > 2 ). Prove: ( (2^{m}-1) times (2^{n}+1) )." That's all. So, maybe it's just to compute or simplify the product, but that's straightforward.Wait, perhaps the problem is to show that ( (2^m - 1) times (2^n + 1) ) is equal to ( 2^{m+n} - 2^n + 2^m - 1 ), which is just the expansion. But that seems too simple.Alternatively, maybe the problem is to show that ( (2^m - 1) times (2^n + 1) ) is a multiple of ( 2^k - 1 ) for some ( k ), but I don't see a direct way.Wait, maybe it's to show that ( (2^m - 1) times (2^n + 1) ) is a multiple of ( 2^{gcd(m,n)} - 1 ). Let me test with ( m = 4 ), ( n = 2 ). ( gcd(4,2) = 2 ), so ( 2^2 - 1 = 3 ). ( (2^4 - 1) times (2^2 + 1) = 15 times 5 = 75 ), which is divisible by 3. Yes, 75 is divisible by 3. Another example: ( m = 6 ), ( n = 4 ). ( gcd(6,4) = 2 ), ( 2^2 - 1 = 3 ). ( (2^6 - 1) times (2^4 + 1) = 63 times 17 = 1071 ). 1071 divided by 3 is 357, so yes, divisible by 3.Another example: ( m = 5 ), ( n = 3 ). ( gcd(5,3) = 1 ), ( 2^1 - 1 = 1 ). So, any number is divisible by 1, which is trivial. So, in this case, it's true.But wait, is this always the case? Let me see. Let ( d = gcd(m,n) ). Then, ( m = d times k ), ( n = d times l ), where ( gcd(k,l) = 1 ). Then, ( 2^m - 1 = 2^{d k} - 1 ), which is divisible by ( 2^d - 1 ). Similarly, ( 2^n + 1 = 2^{d l} + 1 ). Now, does ( 2^d - 1 ) divide ( 2^{d l} + 1 )?Let me see. ( 2^{d l} + 1 ). If ( l ) is even, say ( l = 2p ), then ( 2^{d l} + 1 = 2^{2 d p} + 1 = (2^{d p})^2 + 1 ), which doesn't factor over integers. If ( l ) is odd, then ( 2^{d l} + 1 = (2^d)^l + 1 ). Since ( l ) is odd, this can be factored as ( (2^d + 1)(2^{d(l-1)} - 2^{d(l-2)} + dots - 2^d + 1) ). So, ( 2^d + 1 ) is a factor.But ( 2^d - 1 ) and ( 2^d + 1 ) are coprime because their gcd is ( gcd(2^d - 1, 2^d + 1) = gcd(2^d - 1, 2) = 1 ), since ( 2^d - 1 ) is odd. So, ( 2^d - 1 ) doesn't divide ( 2^{d l} + 1 ) unless ( 2^d - 1 ) divides 1, which is only when ( d = 1 ).Wait, so in general, ( 2^d - 1 ) divides ( 2^m - 1 ) and ( 2^n + 1 ) only if ( d = 1 ). Otherwise, ( 2^d - 1 ) doesn't divide ( 2^n + 1 ). So, the product ( (2^m - 1) times (2^n + 1) ) is divisible by ( 2^d - 1 ) only if ( d = 1 ), which is trivial.So, maybe the problem is not about divisibility by ( 2^d - 1 ).Wait, going back, maybe the problem is to show that ( (2^m - 1) times (2^n + 1) ) is a multiple of ( 2^{gcd(m,n)} - 1 ) only when ( gcd(m,n) = 1 ), but that seems more complicated.Alternatively, maybe the problem is to show that ( (2^m - 1) times (2^n + 1) ) is a multiple of ( 2^m - 1 ) and ( 2^n + 1 ), which is obvious because it's their product.Wait, I'm getting stuck here. Maybe the problem is just to compute the product, which is straightforward, but that seems too simple. Alternatively, maybe it's to show that this product is a certain type of number, but without more context, it's hard to tell.Wait, maybe the problem is to show that ( (2^m - 1) times (2^n + 1) ) is a multiple of ( 2^k - 1 ) for some ( k ), but I don't see a direct way.Alternatively, maybe it's to show that ( (2^m - 1) times (2^n + 1) ) is a multiple of ( 2^{gcd(m,n)} - 1 ) only when ( gcd(m,n) = 1 ), but that's not necessarily true.Wait, let me try with ( m = 4 ), ( n = 2 ). ( gcd(4,2) = 2 ), ( 2^2 - 1 = 3 ). The product is ( 15 times 5 = 75 ), which is divisible by 3. So, even though ( gcd(m,n) = 2 ), the product is still divisible by ( 2^2 - 1 = 3 ). So, maybe the product is always divisible by ( 2^{gcd(m,n)} - 1 ).Wait, let's test another example. ( m = 6 ), ( n = 4 ). ( gcd(6,4) = 2 ), ( 2^2 - 1 = 3 ). The product is ( 63 times 17 = 1071 ), which is divisible by 3. Another example: ( m = 5 ), ( n = 3 ). ( gcd(5,3) = 1 ), ( 2^1 - 1 = 1 ). The product is ( 31 times 9 = 279 ), which is divisible by 1, which is trivial.Another example: ( m = 6 ), ( n = 3 ). ( gcd(6,3) = 3 ), ( 2^3 - 1 = 7 ). The product is ( 63 times 9 = 567 ). Is 567 divisible by 7? 567 divided by 7 is 81, so yes. So, in this case, it is divisible by ( 2^{gcd(m,n)} - 1 ).Wait, so maybe the product ( (2^m - 1) times (2^n + 1) ) is divisible by ( 2^{gcd(m,n)} - 1 ). Let me see if that's always true.Let ( d = gcd(m,n) ), so ( m = d times k ), ( n = d times l ), with ( gcd(k,l) = 1 ). Then, ( 2^m - 1 = 2^{d k} - 1 ), which is divisible by ( 2^d - 1 ). Similarly, ( 2^n + 1 = 2^{d l} + 1 ). Now, does ( 2^d - 1 ) divide ( 2^{d l} + 1 )?If ( l ) is even, say ( l = 2p ), then ( 2^{d l} + 1 = 2^{2 d p} + 1 = (2^{d p})^2 + 1 ), which doesn't factor over integers, and ( 2^d - 1 ) doesn't divide it because ( 2^{d p} equiv 1 mod (2^d - 1) ), so ( (2^{d p})^2 + 1 equiv 1 + 1 = 2 mod (2^d - 1) ). So, unless ( 2^d - 1 ) divides 2, which only happens when ( d = 1 ), it doesn't divide.If ( l ) is odd, then ( 2^{d l} + 1 = (2^d)^l + 1 ). Since ( l ) is odd, this can be factored as ( (2^d + 1)(2^{d(l-1)} - 2^{d(l-2)} + dots - 2^d + 1) ). Now, ( 2^d - 1 ) and ( 2^d + 1 ) are coprime because ( gcd(2^d - 1, 2^d + 1) = gcd(2^d - 1, 2) = 1 ). So, ( 2^d - 1 ) doesn't divide ( 2^{d l} + 1 ) unless ( 2^d - 1 ) divides 1, which is only when ( d = 1 ).Therefore, in general, ( 2^d - 1 ) divides ( 2^m - 1 ) but doesn't divide ( 2^n + 1 ) unless ( d = 1 ). So, the product ( (2^m - 1) times (2^n + 1) ) is divisible by ( 2^d - 1 ) only if ( d = 1 ), which is trivial.Wait, but in the examples I tried earlier, like ( m = 4 ), ( n = 2 ), ( d = 2 ), and the product was divisible by ( 2^2 - 1 = 3 ). So, maybe there's something more here.Wait, in that case, ( 2^m - 1 = 15 ), which is divisible by 3, and ( 2^n + 1 = 5 ), which is not divisible by 3. But the product is 75, which is divisible by 3 because one of the factors is divisible by 3. So, in general, if ( d ) divides ( m ), then ( 2^d - 1 ) divides ( 2^m - 1 ), so if ( d ) divides ( m ), then ( 2^d - 1 ) divides the product ( (2^m - 1) times (2^n + 1) ) because it divides ( 2^m - 1 ).Similarly, if ( d ) divides ( n ), then ( 2^d - 1 ) divides ( 2^n - 1 ), but we have ( 2^n + 1 ). So, unless ( d ) divides ( 2n ), which is different.Wait, maybe the product is divisible by ( 2^d - 1 ) where ( d ) is the gcd of ( m ) and ( 2n ). Let me see.Let ( d = gcd(m, 2n) ). Then, ( d ) divides ( m ) and ( 2n ). So, ( 2^d - 1 ) divides ( 2^m - 1 ) because ( d ) divides ( m ). Also, ( 2^d - 1 ) divides ( 2^{2n} - 1 ), which is ( (2^n - 1)(2^n + 1) ). So, ( 2^d - 1 ) divides ( (2^n - 1)(2^n + 1) ). But ( 2^d - 1 ) and ( 2^n + 1 ) are coprime if ( d ) is odd, because ( 2^d - 1 ) is odd and ( 2^n + 1 ) is even only if ( n = 0 ), which it's not. Wait, no, ( 2^n + 1 ) is odd because ( 2^n ) is even, so ( 2^n + 1 ) is odd. So, ( 2^d - 1 ) and ( 2^n + 1 ) are both odd, so they could share common factors.Wait, this is getting too convoluted. Maybe the problem is simply to show that ( (2^m - 1) times (2^n + 1) ) is composite, which is obvious because it's a product of two integers greater than 1. But that seems too simple, and the problem specifies ( m > 2 ), which might be a hint.Wait, maybe the problem is to show that ( (2^m - 1) times (2^n + 1) ) is a multiple of ( 2^{gcd(m,n)} - 1 ). Let me test with ( m = 4 ), ( n = 2 ). ( gcd(4,2) = 2 ), ( 2^2 - 1 = 3 ). The product is ( 15 times 5 = 75 ), which is divisible by 3. Another example: ( m = 6 ), ( n = 4 ). ( gcd(6,4) = 2 ), ( 2^2 - 1 = 3 ). The product is ( 63 times 17 = 1071 ), which is divisible by 3. Another example: ( m = 5 ), ( n = 3 ). ( gcd(5,3) = 1 ), ( 2^1 - 1 = 1 ). The product is ( 31 times 9 = 279 ), which is divisible by 1, which is trivial.Wait, but in the case where ( d = gcd(m,n) ), ( 2^d - 1 ) divides ( 2^m - 1 ), so it divides the product. So, the product is always divisible by ( 2^d - 1 ). So, maybe the problem is to show that ( (2^m - 1) times (2^n + 1) ) is divisible by ( 2^{gcd(m,n)} - 1 ).Yes, that seems plausible. So, to prove that ( (2^m - 1) times (2^n + 1) ) is divisible by ( 2^{gcd(m,n)} - 1 ).To do this, let ( d = gcd(m,n) ). Then, ( m = d times k ) and ( n = d times l ) where ( gcd(k,l) = 1 ). Now, ( 2^m - 1 = 2^{d k} - 1 ), which is divisible by ( 2^d - 1 ) because ( 2^{d k} - 1 = (2^d)^k - 1 ), and ( a^k - 1 ) is divisible by ( a - 1 ). So, ( 2^d - 1 ) divides ( 2^m - 1 ), hence divides the product ( (2^m - 1) times (2^n + 1) ).Therefore, ( (2^m - 1) times (2^n + 1) ) is divisible by ( 2^{gcd(m,n)} - 1 ).So, maybe that's what the problem is asking to prove. It wasn't explicitly stated, but based on the structure and examples, it seems likely.

💡First, I'll define the number of pencils Reeta has as ( R ).According to the problem, Anika has 4 more than twice the number of pencils as Reeta. This can be expressed as ( 2R + 4 ).Together, they have 64 pencils. So, the equation becomes:[R + (2R + 4) = 64]Combining like terms:[3R + 4 = 64]Subtracting 4 from both sides:[3R = 60]Dividing both sides by 3:[R = 20]Therefore, Reeta has 20 pencils.