💡Okay, so I have this problem: What is the remainder when 7 to the power of 51 is divided by 6? Hmm, that sounds a bit tricky at first, but maybe I can figure it out step by step.First, I remember that when dealing with remainders and exponents, modular arithmetic can be really helpful. I think modular arithmetic is like a system of arithmetic for integers where numbers "wrap around" upon reaching a certain value—the modulus. In this case, the modulus is 6 because we're dividing by 6.So, let me recall that 7 divided by 6 is 1 with a remainder of 1. That means 7 is congruent to 1 modulo 6. In mathematical terms, that's written as 7 ≡ 1 (mod 6). Okay, so 7 is just 1 more than a multiple of 6. That might be useful.Now, if 7 is congruent to 1 modulo 6, then 7 raised to any power should also be congruent to 1 raised to that power modulo 6. Let me test this with smaller exponents to see if it holds.For example, 7^1 is 7, which we already know is 1 modulo 6. 7^2 is 49. If I divide 49 by 6, 6 times 8 is 48, so the remainder is 1. So, 49 ≡ 1 (mod 6). Interesting, same remainder.How about 7^3? That's 343. Dividing 343 by 6: 6 times 57 is 342, so the remainder is 1 again. So, 343 ≡ 1 (mod 6). Hmm, it seems like every time I raise 7 to a power, the remainder when divided by 6 is always 1.Wait, is that always the case? Let me think. Since 7 ≡ 1 (mod 6), then 7^n ≡ 1^n (mod 6). And 1 raised to any power is still 1. So, 7^n ≡ 1 (mod 6) for any positive integer n.Therefore, applying this to 7^51, it should also be congruent to 1 modulo 6. So, the remainder when 7^51 is divided by 6 is 1.But just to make sure I'm not missing something, let me try another approach. Maybe using the concept of cyclicity in remainders. Sometimes, powers of numbers modulo a certain number repeat after a certain period.Let's see the pattern of 7^n modulo 6:- 7^1 ≡ 1 (mod 6)- 7^2 ≡ 1 (mod 6)- 7^3 ≡ 1 (mod 6)- ...It seems like the remainder is always 1, so there's no cycle here—it's just consistently 1. That reinforces my earlier conclusion.Another way to think about it is to express 7 as 6 + 1. So, 7 = 6 + 1. Then, 7^51 = (6 + 1)^51. If I expand this using the binomial theorem, every term except the last one will have a factor of 6, which means they will be divisible by 6. The last term is 1^51, which is 1. Therefore, when you divide (6 + 1)^51 by 6, all the terms with 6 will leave no remainder, and only the 1 will remain. So, the remainder is 1.This seems consistent with all the methods I've tried. I don't see any flaws in this reasoning. It looks like no matter how I approach it, the remainder is 1.Just to be absolutely thorough, let me compute 7^51 modulo 6 using another perspective. Since 7 ≡ 1 (mod 6), multiplying 1 by itself any number of times will still give 1. So, 7^51 is just 1 multiplied by itself 51 times, which is still 1. Therefore, modulo 6, it's 1.I think I've covered all the angles here. Every method points to the same conclusion: the remainder is 1. I don't think I've missed anything, and all the different approaches confirm each other. So, I'm pretty confident that the remainder when 7^51 is divided by 6 is indeed 1.

💡Alright, so I have this problem with two parts. Let me try to tackle them one by one.**Problem (1):** When ( b = 0 ), we need to find the range of ( k ) such that for any ( x in (0, +infty) ), the inequality ( f(x) geq h(x) geq g(x) ) holds. Here, ( f(x) = e^x ), ( g(x) = ln x ), and ( h(x) = kx ).Okay, so ( h(x) = kx ) is a straight line through the origin. The inequalities mean that this line must lie between ( e^x ) and ( ln x ) for all positive ( x ). That is, ( e^x geq kx geq ln x ) for all ( x > 0 ).To find the range of ( k ), I think I need to find the minimum value of ( frac{e^x}{x} ) and the maximum value of ( frac{ln x}{x} ) over ( x > 0 ). Because if ( k ) is between these two values, the inequalities will hold.Let me define two functions:1. ( m(x) = frac{e^x}{x} )2. ( n(x) = frac{ln x}{x} )I need to find the minimum of ( m(x) ) and the maximum of ( n(x) ).Starting with ( m(x) = frac{e^x}{x} ). To find its minimum, I can take the derivative.( m'(x) = frac{d}{dx} left( frac{e^x}{x} right) = frac{e^x cdot x - e^x cdot 1}{x^2} = frac{e^x (x - 1)}{x^2} )Setting ( m'(x) = 0 ):( frac{e^x (x - 1)}{x^2} = 0 )Since ( e^x ) is never zero, the critical point is at ( x = 1 ).Now, let's check the behavior around ( x = 1 ):- For ( x < 1 ), say ( x = 0.5 ), ( m'(0.5) = frac{e^{0.5}(0.5 - 1)}{(0.5)^2} = frac{e^{0.5}(-0.5)}{0.25} = -2 e^{0.5} < 0 )- For ( x > 1 ), say ( x = 2 ), ( m'(2) = frac{e^2(2 - 1)}{4} = frac{e^2}{4} > 0 )So, ( m(x) ) has a minimum at ( x = 1 ). The minimum value is ( m(1) = frac{e^1}{1} = e ).Now, moving on to ( n(x) = frac{ln x}{x} ). Let's find its maximum.Compute the derivative:( n'(x) = frac{d}{dx} left( frac{ln x}{x} right) = frac{1 cdot x - ln x cdot 1}{x^2} = frac{1 - ln x}{x^2} )Set ( n'(x) = 0 ):( frac{1 - ln x}{x^2} = 0 )The numerator must be zero: ( 1 - ln x = 0 ) => ( ln x = 1 ) => ( x = e )Check the behavior around ( x = e ):- For ( x < e ), say ( x = 1 ), ( n'(1) = frac{1 - 0}{1} = 1 > 0 )- For ( x > e ), say ( x = e^2 ), ( n'(e^2) = frac{1 - 2}{(e^2)^2} = frac{-1}{e^4} < 0 )So, ( n(x) ) has a maximum at ( x = e ). The maximum value is ( n(e) = frac{ln e}{e} = frac{1}{e} ).Therefore, for ( e^x geq kx geq ln x ) to hold for all ( x > 0 ), ( k ) must be at least the minimum of ( m(x) ) and at most the maximum of ( n(x) ). Wait, no, actually, since ( e^x geq kx ) requires ( k leq frac{e^x}{x} ) for all ( x ), so ( k ) must be less than or equal to the minimum of ( frac{e^x}{x} ), which is ( e ). Similarly, ( kx geq ln x ) requires ( k geq frac{ln x}{x} ) for all ( x ), so ( k ) must be greater than or equal to the maximum of ( frac{ln x}{x} ), which is ( frac{1}{e} ).Hence, the range of ( k ) is ( frac{1}{e} leq k leq e ).**Problem (2):** The line ( h(x) ) is tangent to both ( f(x) ) and ( g(x) ) at points ( A(x_1, f(x_1)) ) and ( B(x_2, g(x_2)) ), respectively, where ( x_1 < 0 ).**(i)** Prove that ( x_2 > e ).Okay, so ( h(x) ) is tangent to ( f(x) ) at ( x = x_1 ) and to ( g(x) ) at ( x = x_2 ). Since it's tangent, the slope of ( h(x) ) must be equal to the derivative of ( f(x) ) at ( x_1 ) and also equal to the derivative of ( g(x) ) at ( x_2 ).Given ( f(x) = e^x ), so ( f'(x) = e^x ). Therefore, the slope ( k = f'(x_1) = e^{x_1} ).Similarly, ( g(x) = ln x ), so ( g'(x) = frac{1}{x} ). Therefore, the slope ( k = g'(x_2) = frac{1}{x_2} ).So, ( e^{x_1} = frac{1}{x_2} ) => ( x_2 = e^{-x_1} ).Also, since ( h(x) ) is tangent to ( f(x) ) at ( x_1 ), the equation of the tangent line is:( h(x) = f(x_1) + f'(x_1)(x - x_1) )= ( e^{x_1} + e^{x_1}(x - x_1) )= ( e^{x_1} + e^{x_1}x - e^{x_1}x_1 )= ( e^{x_1}(1 - x_1) + e^{x_1}x )= ( e^{x_1}x + e^{x_1}(1 - x_1) )Similarly, the tangent line at ( g(x) ) at ( x_2 ) is:( h(x) = g(x_2) + g'(x_2)(x - x_2) )= ( ln x_2 + frac{1}{x_2}(x - x_2) )= ( ln x_2 + frac{x}{x_2} - 1 )Since both expressions represent ( h(x) ), their slopes and intercepts must be equal.We already have the slope equality: ( e^{x_1} = frac{1}{x_2} ).Now, equate the intercepts:( e^{x_1}(1 - x_1) = ln x_2 - 1 )But since ( x_2 = e^{-x_1} ), substitute:( e^{x_1}(1 - x_1) = ln(e^{-x_1}) - 1 )= ( -x_1 - 1 )So, we have:( e^{x_1}(1 - x_1) = -x_1 - 1 )Let me rearrange this:( e^{x_1}(1 - x_1) + x_1 + 1 = 0 )Hmm, this seems a bit complicated. Maybe I can express everything in terms of ( x_2 ) since ( x_2 = e^{-x_1} ), so ( x_1 = -ln x_2 ).Substituting ( x_1 = -ln x_2 ) into the equation:( e^{-ln x_2}(1 - (-ln x_2)) + (-ln x_2) + 1 = 0 )Simplify ( e^{-ln x_2} = frac{1}{x_2} ), so:( frac{1}{x_2}(1 + ln x_2) - ln x_2 + 1 = 0 )Multiply through by ( x_2 ):( (1 + ln x_2) - x_2 ln x_2 + x_2 = 0 )Simplify:( 1 + ln x_2 - x_2 ln x_2 + x_2 = 0 )Factor terms with ( ln x_2 ):( 1 + x_2 + ln x_2 (1 - x_2) = 0 )Hmm, this is a transcendental equation in ( x_2 ). Maybe I can analyze the function ( F(x) = 1 + x + ln x (1 - x) ) and see where it equals zero.But perhaps another approach is better. Let's recall that ( x_1 < 0 ), so ( x_1 = -ln x_2 < 0 ) implies ( ln x_2 > 0 ), so ( x_2 > 1 ).But we need to show ( x_2 > e ).Let me consider the equation ( e^{x_1}(1 - x_1) = -x_1 - 1 ). Let me denote ( x_1 = -t ) where ( t > 0 ) because ( x_1 < 0 ).So, substituting ( x_1 = -t ):( e^{-t}(1 + t) = t - 1 )Multiply both sides by ( e^{t} ):( (1 + t) = (t - 1)e^{t} )So, ( (1 + t) = (t - 1)e^{t} )Let me rearrange:( (t - 1)e^{t} - (t + 1) = 0 )Let me define ( G(t) = (t - 1)e^{t} - (t + 1) ). We need to find ( t > 0 ) such that ( G(t) = 0 ).Compute ( G(1) = (0)e^{1} - (2) = -2 < 0 )Compute ( G(2) = (1)e^{2} - 3 ≈ 7.389 - 3 = 4.389 > 0 )So, by Intermediate Value Theorem, there is a root between ( t = 1 ) and ( t = 2 ).Thus, ( t ) is between 1 and 2, so ( x_1 = -t ) is between -2 and -1.Therefore, ( x_2 = e^{-x_1} = e^{t} ). Since ( t > 1 ), ( x_2 = e^{t} > e^{1} = e ).Hence, ( x_2 > e ). So, part (i) is proved.**(ii)** When ( x geq x_2 ), the inequality ( a(x_1 - 1) + x ln x - x geq 0 ) always holds for ( x ). Find the range of the real number ( a ).Alright, so we need to find the range of ( a ) such that for all ( x geq x_2 ), ( a(x_1 - 1) + x ln x - x geq 0 ).Let me rearrange the inequality:( a(x_1 - 1) geq x - x ln x )Let me denote ( G(x) = x - x ln x ). So, the inequality becomes:( a(x_1 - 1) geq G(x) ) for all ( x geq x_2 ).We need to find the maximum of ( G(x) ) on ( [x_2, +infty) ) because ( a(x_1 - 1) ) must be greater than or equal to the maximum of ( G(x) ) in that interval.First, let's analyze ( G(x) = x - x ln x ).Compute its derivative:( G'(x) = 1 - (ln x + 1) = -ln x )So, ( G'(x) = -ln x ). Setting ( G'(x) = 0 ):( -ln x = 0 ) => ( x = 1 )But since ( x geq x_2 ) and ( x_2 > e > 1 ), the critical point at ( x = 1 ) is not in our interval. Therefore, on ( [x_2, +infty) ), ( G'(x) = -ln x < 0 ) because ( x > e ) implies ( ln x > 1 ), so ( G'(x) < 0 ).Thus, ( G(x) ) is decreasing on ( [x_2, +infty) ). Therefore, the maximum of ( G(x) ) on this interval is at ( x = x_2 ).So, ( G(x) leq G(x_2) ) for all ( x geq x_2 ).Thus, the inequality ( a(x_1 - 1) geq G(x) ) for all ( x geq x_2 ) reduces to:( a(x_1 - 1) geq G(x_2) )Compute ( G(x_2) = x_2 - x_2 ln x_2 )But from part (i), we have ( x_2 = e^{-x_1} ) and from the tangent condition, we had:( e^{x_1}(1 - x_1) = -x_1 - 1 )Let me recall that ( x_2 = e^{-x_1} ), so ( ln x_2 = -x_1 ). Therefore, ( G(x_2) = x_2 - x_2 (-x_1) = x_2 + x_1 x_2 )But ( x_2 = e^{-x_1} ), so ( G(x_2) = e^{-x_1} + x_1 e^{-x_1} = e^{-x_1}(1 + x_1) )From the equation in part (i):( e^{x_1}(1 - x_1) = -x_1 - 1 )Multiply both sides by ( e^{-x_1} ):( (1 - x_1) = (-x_1 - 1)e^{-x_1} )Thus, ( (-x_1 - 1)e^{-x_1} = 1 - x_1 )But ( G(x_2) = e^{-x_1}(1 + x_1) ). Let me see:( G(x_2) = e^{-x_1}(1 + x_1) = (1 + x_1)e^{-x_1} )From the equation above:( (-x_1 - 1)e^{-x_1} = 1 - x_1 )Multiply both sides by -1:( (x_1 + 1)e^{-x_1} = x_1 - 1 )So, ( G(x_2) = x_1 - 1 )Wait, that's interesting. So, ( G(x_2) = x_1 - 1 )Therefore, going back to our inequality:( a(x_1 - 1) geq G(x_2) = x_1 - 1 )So, ( a(x_1 - 1) geq x_1 - 1 )Now, since ( x_1 < 0 ), ( x_1 - 1 < 0 - 1 = -1 < 0 ). So, ( x_1 - 1 ) is negative.Therefore, when we divide both sides of the inequality ( a(x_1 - 1) geq x_1 - 1 ) by ( x_1 - 1 ), which is negative, the inequality sign flips.So, ( a leq 1 )Hence, the range of ( a ) is ( a leq 1 ).**Final Answer**(1) The range of ( k ) is (boxed{left[ dfrac{1}{e}, e right]}).(2)(i) Proved that ( x_2 > e ).(2)(ii) The range of ( a ) is (boxed{(-infty, 1]}).

💡Alright, so I'm trying to solve this problem about the sum of squares of consecutive integers and whether that sum can be a perfect square. The problem has two parts: first, to prove that for m = 3, 4, 5, 6, the sum of the squares of m consecutive integers can't be a perfect square, and second, to find an example where the sum of the squares of 11 consecutive integers is a perfect square.Starting with part (1), I need to think about how to approach proving that for these specific values of m, the sum can't be a perfect square. Maybe I can use modular arithmetic to show that the sum can't be congruent to a square modulo some number, which would imply it can't be a square.For m = 3, let's consider three consecutive integers: x-1, x, x+1. Their squares are (x-1)^2, x^2, and (x+1)^2. Adding these up:(x-1)^2 + x^2 + (x+1)^2 = (x^2 - 2x + 1) + x^2 + (x^2 + 2x + 1) = 3x^2 + 2.Now, I need to see if 3x^2 + 2 can be a perfect square. Maybe looking at this modulo 3 would help. Squares modulo 3 are either 0 or 1 because:- If x ≡ 0 mod 3, then x^2 ≡ 0 mod 3.- If x ≡ 1 mod 3, then x^2 ≡ 1 mod 3.- If x ≡ 2 mod 3, then x^2 ≡ 4 ≡ 1 mod 3.So, 3x^2 is always 0 mod 3, and adding 2 gives 2 mod 3. But squares can't be 2 mod 3, so 3x^2 + 2 can't be a perfect square. That takes care of m = 3.Moving on to m = 4. Let's take four consecutive integers: x-1, x, x+1, x+2. Their squares are (x-1)^2, x^2, (x+1)^2, (x+2)^2. Adding these up:(x-1)^2 + x^2 + (x+1)^2 + (x+2)^2 = (x^2 - 2x + 1) + x^2 + (x^2 + 2x + 1) + (x^2 + 4x + 4) = 4x^2 + 4x + 6.Simplify this: 4x^2 + 4x + 6. Let's see modulo 4. Squares modulo 4 are 0 or 1:- If x is even, x = 2k, then x^2 = 4k^2 ≡ 0 mod 4.- If x is odd, x = 2k+1, then x^2 = 4k^2 + 4k + 1 ≡ 1 mod 4.So, 4x^2 is always 0 mod 4. Then 4x is 0 mod 4 if x is even, and 4 mod 4 if x is odd. Adding 6:If x is even: 0 + 0 + 6 ≡ 6 mod 4 ≡ 2 mod 4.If x is odd: 0 + 4 + 6 ≡ 10 mod 4 ≡ 2 mod 4.But squares modulo 4 are only 0 or 1, so 2 mod 4 can't be a square. Therefore, the sum for m = 4 can't be a perfect square.Next, m = 5. Let's take five consecutive integers: x-2, x-1, x, x+1, x+2. Their squares are (x-2)^2, (x-1)^2, x^2, (x+1)^2, (x+2)^2. Adding these up:(x-2)^2 + (x-1)^2 + x^2 + (x+1)^2 + (x+2)^2 = (x^2 - 4x + 4) + (x^2 - 2x + 1) + x^2 + (x^2 + 2x + 1) + (x^2 + 4x + 4) = 5x^2 + 10.So, the sum is 5x^2 + 10. Let's check modulo 4 again. 5x^2 is equivalent to x^2 mod 4 because 5 ≡ 1 mod 4. So, x^2 is 0 or 1 mod 4, and 10 ≡ 2 mod 4. Therefore, the sum is either 0 + 2 ≡ 2 mod 4 or 1 + 2 ≡ 3 mod 4. Neither 2 nor 3 are squares modulo 4, so the sum can't be a perfect square.For m = 6, let's take six consecutive integers: x-2, x-1, x, x+1, x+2, x+3. Their squares are (x-2)^2, (x-1)^2, x^2, (x+1)^2, (x+2)^2, (x+3)^2. Adding these up:(x-2)^2 + (x-1)^2 + x^2 + (x+1)^2 + (x+2)^2 + (x+3)^2 = (x^2 - 4x + 4) + (x^2 - 2x + 1) + x^2 + (x^2 + 2x + 1) + (x^2 + 4x + 4) + (x^2 + 6x + 9) = 6x^2 + 12x + 19.Hmm, that seems a bit messy. Maybe there's a pattern or a better way to compute this. Alternatively, maybe I can use a different modulus or find another property that shows this sum can't be a square.Alternatively, maybe I can consider the sum of squares formula for consecutive integers. The sum of squares from x = a to x = b is given by:Sum = (b(b+1)(2b+1) - (a-1)a(2a-1))/6.But since we're dealing with consecutive integers, maybe it's better to express them as centered around some integer x. For m consecutive integers, the sum of squares can be expressed in terms of x and m.Wait, maybe I can generalize this. For m consecutive integers starting at x, the sum of squares is:Sum = x^2 + (x+1)^2 + (x+2)^2 + ... + (x+m-1)^2.This can be written as:Sum = m x^2 + 2x(0 + 1 + 2 + ... + (m-1)) + (0^2 + 1^2 + 2^2 + ... + (m-1)^2).Simplifying, we get:Sum = m x^2 + 2x * [m(m-1)/2] + [m(m-1)(2m-1)/6].So, Sum = m x^2 + m(m-1)x + m(m-1)(2m-1)/6.Now, for m = 3, 4, 5, 6, we can plug these values in and see if the sum can be a perfect square.For m = 3:Sum = 3x^2 + 3*2x + 3*2*5/6 = 3x^2 + 6x + 5.Wait, earlier I had 3x^2 + 2, but that was for centered around x. Maybe I need to adjust for starting at x.Wait, maybe I should consider the sum starting at x, so for m = 3, it's x^2 + (x+1)^2 + (x+2)^2.Calculating that:x^2 + (x+1)^2 + (x+2)^2 = x^2 + x^2 + 2x + 1 + x^2 + 4x + 4 = 3x^2 + 6x + 5.Hmm, earlier I had 3x^2 + 2 when centered around x, but starting at x, it's 3x^2 + 6x + 5. So, maybe I need to adjust my approach.But regardless, I can still use modular arithmetic. Let's take m = 3, sum = 3x^2 + 6x + 5.Let's see modulo 3:3x^2 ≡ 0 mod 3,6x ≡ 0 mod 3,5 ≡ 2 mod 3.So, sum ≡ 2 mod 3. Since squares mod 3 are 0 or 1, sum can't be a square.Similarly, for m = 4, sum = 4x^2 + 12x + 14.Modulo 4:4x^2 ≡ 0 mod 4,12x ≡ 0 mod 4,14 ≡ 2 mod 4.So, sum ≡ 2 mod 4. Squares mod 4 are 0 or 1, so sum can't be a square.For m = 5, sum = 5x^2 + 20x + 30.Modulo 4:5x^2 ≡ x^2 mod 4,20x ≡ 0 mod 4,30 ≡ 2 mod 4.So, sum ≡ x^2 + 2 mod 4.If x is even, x^2 ≡ 0, so sum ≡ 2 mod 4.If x is odd, x^2 ≡ 1, so sum ≡ 3 mod 4.Neither 2 nor 3 are squares mod 4, so sum can't be a square.For m = 6, sum = 6x^2 + 30x + 55.Modulo 4:6x^2 ≡ 2x^2 mod 4,30x ≡ 2x mod 4,55 ≡ 3 mod 4.So, sum ≡ 2x^2 + 2x + 3 mod 4.Let's consider possible values of x mod 4:If x ≡ 0 mod 4:sum ≡ 0 + 0 + 3 ≡ 3 mod 4.If x ≡ 1 mod 4:sum ≡ 2*1 + 2*1 + 3 ≡ 2 + 2 + 3 ≡ 7 ≡ 3 mod 4.If x ≡ 2 mod 4:sum ≡ 2*0 + 2*2 + 3 ≡ 0 + 4 + 3 ≡ 7 ≡ 3 mod 4.If x ≡ 3 mod 4:sum ≡ 2*1 + 2*3 + 3 ≡ 2 + 6 + 3 ≡ 11 ≡ 3 mod 4.So, sum ≡ 3 mod 4 in all cases. But squares mod 4 are 0 or 1, so sum can't be a square.Okay, so that covers m = 3, 4, 5, 6.Now, part (2): Find an example where the sum of the squares of 11 consecutive integers is equal to the square of another integer.Let me think about this. Maybe I can use a similar approach, but this time find specific integers where the sum is a perfect square.Let's consider 11 consecutive integers. Let's denote them as x-5, x-4, ..., x, ..., x+4, x+5. This way, they are symmetric around x.The sum of their squares is:(x-5)^2 + (x-4)^2 + ... + x^2 + ... + (x+4)^2 + (x+5)^2.Calculating this sum:Each pair (x-k)^2 + (x+k)^2 = 2x^2 + 2k^2.So, for k = 1 to 5, we have:2x^2 + 2(1)^2 + 2x^2 + 2(2)^2 + ... + 2x^2 + 2(5)^2 + x^2.Wait, actually, since there are 11 terms, the middle term is x^2, and the rest are symmetric.So, sum = x^2 + 2[(x-1)^2 + (x-2)^2 + (x-3)^2 + (x-4)^2 + (x-5)^2].But wait, that's not quite right. Let me count:From x-5 to x+5, inclusive, that's 11 terms. So, the sum is:Sum = (x-5)^2 + (x-4)^2 + (x-3)^2 + (x-2)^2 + (x-1)^2 + x^2 + (x+1)^2 + (x+2)^2 + (x+3)^2 + (x+4)^2 + (x+5)^2.Now, group the terms:Sum = [ (x-5)^2 + (x+5)^2 ] + [ (x-4)^2 + (x+4)^2 ] + [ (x-3)^2 + (x+3)^2 ] + [ (x-2)^2 + (x+2)^2 ] + [ (x-1)^2 + (x+1)^2 ] + x^2.Each pair (x-k)^2 + (x+k)^2 = 2x^2 + 2k^2.So, Sum = 5*(2x^2) + 2*(1^2 + 2^2 + 3^2 + 4^2 + 5^2) + x^2.Wait, no, that's not quite right. Let's see:Each pair contributes 2x^2 + 2k^2, and there are 5 pairs, plus the middle term x^2.So, Sum = 5*(2x^2) + 2*(1^2 + 2^2 + 3^2 + 4^2 + 5^2) + x^2.Simplify:Sum = 10x^2 + 2*(1 + 4 + 9 + 16 + 25) + x^2 = 11x^2 + 2*55 = 11x^2 + 110.So, Sum = 11(x^2 + 10).We need this to be a perfect square. Let's set 11(x^2 + 10) = y^2.So, y^2 = 11(x^2 + 10).We need to find integers x and y such that this equation holds.This is a Diophantine equation. Let's see if we can find integer solutions.Let me try small values of x:If x = 1:y^2 = 11(1 + 10) = 11*11 = 121 ⇒ y = 11.So, x = 1, y = 11.Therefore, the sum of squares of 11 consecutive integers centered at x = 1 is 121, which is 11^2.Let's verify:The integers are -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5.Wait, no, if x = 1, the integers are 1-5= -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6.Wait, that's 11 integers: -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6.Calculating their squares:(-4)^2 = 16,(-3)^2 = 9,(-2)^2 = 4,(-1)^2 = 1,0^2 = 0,1^2 = 1,2^2 = 4,3^2 = 9,4^2 = 16,5^2 = 25,6^2 = 36.Sum = 16 + 9 + 4 + 1 + 0 + 1 + 4 + 9 + 16 + 25 + 36.Calculating step by step:16 + 9 = 25,25 + 4 = 29,29 + 1 = 30,30 + 0 = 30,30 + 1 = 31,31 + 4 = 35,35 + 9 = 44,44 + 16 = 60,60 + 25 = 85,85 + 36 = 121.Yes, the sum is 121, which is 11^2.So, the example is the sum of squares of the 11 consecutive integers from -4 to 6, which equals 121, a perfect square.Alternatively, if x = -1, the integers would be -6 to 4, and the sum would still be 121.So, that's the example.

💡I need to solve the equation ((3b + 7)(b - 2) = 9b) to find the largest possible value of (b).First, I'll expand the left side of the equation:[(3b + 7)(b - 2) = 3b cdot b + 3b cdot (-2) + 7 cdot b + 7 cdot (-2) = 3b^2 - 6b + 7b - 14 = 3b^2 + b - 14]Now, the equation becomes:[3b^2 + b - 14 = 9b]I'll move all terms to one side to set the equation to zero:[3b^2 + b - 14 - 9b = 0 implies 3b^2 - 8b - 14 = 0]Next, I'll attempt to factor the quadratic equation (3b^2 - 8b - 14 = 0). However, factoring doesn't seem straightforward, so I'll use the quadratic formula:[b = frac{-B pm sqrt{B^2 - 4AC}}{2A}]where (A = 3), (B = -8), and (C = -14).Plugging in the values:[b = frac{-(-8) pm sqrt{(-8)^2 - 4 cdot 3 cdot (-14)}}{2 cdot 3} = frac{8 pm sqrt{64 + 168}}{6} = frac{8 pm sqrt{232}}{6}]Simplifying (sqrt{232}):[sqrt{232} = sqrt{4 cdot 58} = 2sqrt{58}]So, the solutions are:[b = frac{8 pm 2sqrt{58}}{6} = frac{4 pm sqrt{58}}{3}]The two possible values for (b) are (frac{4 + sqrt{58}}{3}) and (frac{4 - sqrt{58}}{3}). Since (sqrt{58}) is approximately 7.62, the largest value is:[frac{4 + 7.62}{3} approx frac{11.62}{3} approx 3.87]Therefore, the largest possible value of (b) is (frac{4 + sqrt{58}}{3}).

💡Alright, so I have this problem where I need to find the smallest positive integer ( k ) such that among any ( k ) positive integers, there's always a way to pick a positive even number of them whose sum is divisible by ( n ). Hmm, okay. Let me try to break this down.First, I should understand what the problem is asking. It's saying that no matter what ( k ) positive integers you give me, I can always find an even number of them (like 2, 4, 6, etc.) that add up to a multiple of ( n ). And I need to find the smallest such ( k ) depending on ( n ).I think this might relate to the Pigeonhole Principle or maybe some modular arithmetic. Let me recall: the Pigeonhole Principle says that if you have more pigeons than pigeonholes, at least two pigeons must share a hole. Maybe I can apply that here somehow.Let me consider some small cases to get a feel for the problem.**Case 1: ( n = 1 )**If ( n = 1 ), any sum is divisible by 1. So, the smallest ( k ) would be 2 because I need at least two numbers to form an even count. But wait, actually, even one number is trivially divisible by 1, but the problem specifies a positive even number of them. So, I need at least two numbers. So, ( k = 2 ) for ( n = 1 ).**Case 2: ( n = 2 )**For ( n = 2 ), I need to find the smallest ( k ) such that among any ( k ) numbers, there's an even number of them adding up to a multiple of 2. Since 2 is even, maybe the required ( k ) is smaller.Let me think: if I have two numbers, say 1 and 1. Their sum is 2, which is divisible by 2. If I have three numbers, say 1, 1, 1. Then, any two of them sum to 2, which is divisible by 2. Wait, so maybe ( k = 3 ) is sufficient for ( n = 2 )?But hold on, if I have two numbers, 1 and 3. Their sum is 4, which is divisible by 2. But if I have only two numbers, say 1 and 2. Their sum is 3, which is not divisible by 2. Wait, so in that case, I can't find an even number of them (which is just two) that sum to a multiple of 2. So, ( k = 2 ) is not enough for ( n = 2 ). Therefore, ( k = 3 ) is needed.Wait, but if ( k = 3 ), can I always find two numbers whose sum is divisible by 2? Let's see. If I have three numbers, there are three possible pairs. If any pair sums to an even number, which is divisible by 2, then we're done. If all pairs sum to odd numbers, that would mean all three numbers are odd. Because odd + odd = even, so if all pairs are odd, that's impossible. Wait, no. If all three numbers are odd, then any two would sum to even. So, actually, if you have three numbers, either at least two are even, or all are odd. If all are odd, then any two sum to even. So, yes, ( k = 3 ) works for ( n = 2 ).**Case 3: ( n = 3 )**Now, for ( n = 3 ), which is odd. Let me try to see what ( k ) would be. I think the answer might be ( 2n ), so 6 in this case. Let me test it.If I have six numbers, can I always find an even number of them that sum to a multiple of 3? Let's see. Maybe using the Pigeonhole Principle on partial sums.Suppose I have six numbers ( a_1, a_2, a_3, a_4, a_5, a_6 ). Let's look at the partial sums modulo 3:( s_1 = a_1 mod 3 )( s_2 = (a_1 + a_2) mod 3 )( s_3 = (a_1 + a_2 + a_3) mod 3 )...( s_6 = (a_1 + a_2 + a_3 + a_4 + a_5 + a_6) mod 3 )There are six partial sums and three possible residues modulo 3. By the Pigeonhole Principle, at least two of these partial sums must be congruent modulo 3. Let's say ( s_i equiv s_j mod 3 ) for some ( i < j ). Then, the sum from ( a_{i+1} ) to ( a_j ) is ( s_j - s_i equiv 0 mod 3 ). The number of terms in this sum is ( j - i ), which is even or odd?Wait, ( j - i ) could be either. Hmm, but we need an even number of terms. So, maybe this approach isn't directly giving me an even number. Maybe I need to adjust it.Alternatively, perhaps I should consider pairs of numbers. If I have six numbers, I can form three pairs. If any pair sums to a multiple of 3, we're done. If not, then each pair has a sum that is either 1 or 2 modulo 3. Maybe by some combinatorial argument, we can find four numbers whose total sum is 0 modulo 3.Wait, this is getting a bit tangled. Maybe I should look for a general approach rather than specific cases.I recall that for similar problems, like finding a subset with a sum divisible by ( n ), the answer is often ( n ) or ( 2n ). But here, it's an even number of elements, so maybe it's different.Let me think about the Erdős–Ginzburg–Ziv theorem. I think that theorem states that for any ( 2n - 1 ) integers, there exists a subset of ( n ) integers whose sum is divisible by ( n ). But in our case, we need an even number of integers, so maybe the required ( k ) is different.Wait, actually, the Erdős–Ginzburg–Ziv theorem is about any ( 2n - 1 ) integers, there exists a subset of size ( n ) with sum divisible by ( n ). So, for our problem, if ( n ) is even, maybe ( k = n + 1 ) suffices? Because if ( n ) is even, then ( n + 1 ) is odd, and perhaps we can apply some similar reasoning.But I need to be careful. Let me try to think step by step.Suppose ( n ) is odd. Let me try to see what ( k ) should be. If ( k = 2n ), then among any ( 2n ) integers, can I always find an even number of them whose sum is divisible by ( n )?Let me try to construct a scenario where this is true. Consider the partial sums as before, but this time, group them into pairs. So, for ( 2n ) numbers, I can form ( n ) pairs. If any pair sums to 0 modulo ( n ), we're done. If not, then each pair has a sum that is non-zero modulo ( n ). But since there are ( n ) pairs and ( n - 1 ) non-zero residues modulo ( n ), by the Pigeonhole Principle, at least two pairs must have the same sum modulo ( n ). Then, combining these two pairs would give a sum that is ( 2 times text{sum} mod n ). Hmm, but I need the sum to be 0 modulo ( n ), not necessarily twice something.Wait, maybe that's not the right approach. Let me think differently.Suppose I have ( 2n ) numbers. Let me consider all possible subsets of even size. There are ( 2^{2n - 1} ) such subsets. Each subset has a sum modulo ( n ). There are ( n ) possible residues. So, by the Pigeonhole Principle, at least ( 2^{2n - 1} / n ) subsets have the same residue. But this seems too vague.Alternatively, maybe I can use induction. Suppose for ( n ), the required ( k ) is ( 2n ). Let me see if that holds.Wait, I think I need to look for a more structured approach. Let me consider the residues of the numbers modulo ( n ). If any two numbers have the same residue, their sum is ( 2 times text{residue} mod n ). If that's 0, then we're done. If not, maybe we can find four numbers whose combined sum is 0 modulo ( n ).But this is getting complicated. Maybe I should look for a pattern or recall similar problems.Wait, I think the answer might be ( 2n ) when ( n ) is odd and ( n + 1 ) when ( n ) is even. Let me see if that makes sense.For ( n = 2 ), which is even, ( k = 3 ) works, as we saw earlier. For ( n = 3 ), which is odd, ( k = 6 ) might be necessary. Let me test that.If I have six numbers, can I always find an even number of them summing to a multiple of 3? Let me try to construct a set where this is not possible. Suppose all six numbers are 1 modulo 3. Then, any even number of them would sum to ( 2, 4, 6, 8, 10 ) modulo 3, which are ( 2, 1, 0, 2, 1 ) respectively. So, when taking four numbers, the sum is ( 4 mod 3 = 1 ), and six numbers sum to ( 6 mod 3 = 0 ). So, in this case, six numbers do sum to 0 modulo 3, but that's an even number (six). So, even if all numbers are 1 modulo 3, six of them would sum to 0 modulo 3.Wait, but what if I have five numbers? Let me see. If I have five numbers, all 1 modulo 3. Then, the possible even subset sums are:- 2 numbers: 2 mod 3- 4 numbers: 4 mod 3 = 1 mod 3Neither of which is 0 modulo 3. So, with five numbers, it's possible to have no even subset summing to 0 modulo 3. Therefore, ( k = 6 ) is necessary for ( n = 3 ).Similarly, for ( n = 4 ), which is even, let's see if ( k = 5 ) works. Suppose I have five numbers. If any two sum to 0 modulo 4, we're done. If not, then each pair has a sum of 1, 2, or 3 modulo 4. Since there are ( binom{5}{2} = 10 ) pairs and only 3 non-zero residues, by the Pigeonhole Principle, at least four pairs must have the same residue. Maybe from these, we can find four numbers whose total sum is 0 modulo 4.Wait, this is getting a bit too vague. Let me think of a specific example. Suppose all five numbers are 1 modulo 4. Then, any two sum to 2 modulo 4, any four sum to 4 modulo 4 = 0. So, in this case, four numbers sum to 0 modulo 4. So, even if all numbers are 1 modulo 4, five numbers ensure that four of them sum to 0 modulo 4.But what if the numbers have different residues? For example, suppose I have numbers with residues 1, 1, 1, 1, 2 modulo 4. Then, the pairs:- 1 + 1 = 2 mod 4- 1 + 2 = 3 mod 4So, pairs can sum to 2 or 3. But if I take four 1's, their sum is 4 mod 4 = 0. So, again, we can find an even number (four) whose sum is 0 modulo 4.Alternatively, if I have numbers with residues 1, 1, 1, 3, 3 modulo 4. Then, pairs:- 1 + 1 = 2 mod 4- 1 + 3 = 0 mod 4So, in this case, we have pairs that sum to 0 modulo 4, so we're done.It seems that for ( n = 4 ), ( k = 5 ) works. So, maybe for even ( n ), ( k = n + 1 ) is sufficient.Putting this together, it seems that the minimal ( k ) is ( 2n ) when ( n ) is odd and ( n + 1 ) when ( n ) is even.Let me try to formalize this.**For odd ( n ):**We need ( k = 2n ). Suppose we have ( 2n ) numbers. Consider their partial sums modulo ( n ). There are ( 2n ) partial sums (including the empty sum, which is 0). By the Pigeonhole Principle, two of these partial sums must be congruent modulo ( n ). The difference between these two sums gives a subset whose sum is 0 modulo ( n ). However, the size of this subset could be odd or even. To ensure it's even, we might need to adjust our approach.Wait, actually, if we consider the partial sums of even lengths, we can apply the Pigeonhole Principle specifically to even-sized subsets. Since there are ( n ) possible residues and ( n ) even-sized partial sums (from 2 to ( 2n )), if none of them are 0 modulo ( n ), then by the Pigeonhole Principle, two must be congruent, and their difference would give an even-sized subset summing to 0 modulo ( n ).**For even ( n ):**We need ( k = n + 1 ). Suppose we have ( n + 1 ) numbers. If any two sum to 0 modulo ( n ), we're done. If not, then each pair has a sum that is non-zero modulo ( n ). Since there are ( binom{n+1}{2} ) pairs and ( n - 1 ) non-zero residues, by the Pigeonhole Principle, at least ( frac{binom{n+1}{2}}{n - 1} ) pairs share the same residue. From these, we can find four numbers whose total sum is 0 modulo ( n ).Wait, this is a bit hand-wavy. Maybe a better approach is to use the Erdős–Ginzburg–Ziv theorem, which states that for any ( 2n - 1 ) integers, there exists a subset of ( n ) integers whose sum is divisible by ( n ). But in our case, we need an even number of integers. So, if ( n ) is even, ( n ) itself is even, so the theorem directly gives us that ( k = 2n - 1 ) suffices. But wait, earlier I thought ( k = n + 1 ) suffices for even ( n ). There's a discrepancy here.Wait, no. The Erdős–Ginzburg–Ziv theorem requires ( 2n - 1 ) integers to guarantee a subset of size ( n ) with sum divisible by ( n ). But in our problem, we need an even number of integers, not necessarily ( n ) of them. So, perhaps for even ( n ), ( k = n + 1 ) is sufficient because we can apply a similar principle but for even-sized subsets.Let me think again. For even ( n ), say ( n = 2m ). If we have ( 2m + 1 ) numbers, can we always find an even number of them summing to 0 modulo ( 2m )?Yes, because if we consider all possible even-sized subsets, their sums modulo ( 2m ) must cover all residues, and by the Pigeonhole Principle, some subset must sum to 0 modulo ( 2m ).Alternatively, using the theorem, since ( n ) is even, ( n = 2m ), and we have ( n + 1 = 2m + 1 ) numbers. By the theorem, there exists a subset of ( m ) numbers whose sum is divisible by ( 2m ). But ( m ) could be odd or even. Wait, no, the theorem says a subset of size ( n ), which is ( 2m ). So, actually, the theorem would say that among ( 4m - 1 ) numbers, there's a subset of ( 2m ) numbers summing to 0 modulo ( 2m ). But that's not directly helpful.Wait, maybe I'm overcomplicating. Let me go back to the earlier approach.For even ( n ), if we have ( n + 1 ) numbers, we can pair them up. If any pair sums to 0 modulo ( n ), we're done. If not, then each pair has a sum that is non-zero modulo ( n ). Since there are ( frac{n + 1}{2} ) pairs (if ( n + 1 ) is odd), but actually, ( n + 1 ) is odd, so we can't pair all of them. Hmm, maybe this isn't the right way.Alternatively, consider the residues of the numbers modulo ( n ). If any residue appears at least twice, their difference is 0 modulo ( n ), but we need an even number of terms. Wait, no, the difference would be the sum of those two numbers, which is 0 modulo ( n ). So, if any two numbers have the same residue, their sum is 0 modulo ( n ), and since two is even, we're done.If all residues are distinct, then since there are ( n + 1 ) numbers and ( n ) residues, by the Pigeonhole Principle, at least two numbers share the same residue. Therefore, their sum is 0 modulo ( n ), and we're done.Wait, that seems too simple. So, for even ( n ), ( k = n + 1 ) works because either two numbers share the same residue (sum to 0 modulo ( n )) or all residues are distinct, which is impossible because there are ( n + 1 ) numbers and only ( n ) residues. Therefore, at least two must share a residue, giving us an even subset (two numbers) summing to 0 modulo ( n ).But wait, what if ( n ) is even and the numbers are arranged such that no two share the same residue, but their sums can still be 0 modulo ( n ) with an even number of terms? Actually, the Pigeonhole Principle ensures that at least two numbers share a residue, so their sum is 0 modulo ( n ). Therefore, ( k = n + 1 ) suffices for even ( n ).For odd ( n ), the situation is different because the Pigeonhole Principle doesn't directly give us two numbers with the same residue unless we have ( n + 1 ) numbers. But in that case, we might only get a single number with a unique residue, but we need an even number of terms. Therefore, we need a different approach, possibly involving more numbers to ensure that an even-sized subset sums to 0 modulo ( n ).In summary, after considering various cases and applying the Pigeonhole Principle, it seems that:- If ( n ) is odd, the minimal ( k ) is ( 2n ).- If ( n ) is even, the minimal ( k ) is ( n + 1 ).This aligns with the Erdős–Ginzburg–Ziv theorem for the even case, where ( k = n + 1 ) suffices, and for the odd case, we need ( 2n ) to ensure an even-sized subset with the desired property.

💡Alright, so I've got this math problem here with four options, and I need to figure out which one is correct. Let's take it step by step because I want to make sure I understand each part properly.First, let's look at option A: √2 + √3 = √5. Hmm, okay. I remember that when you add square roots, you can't just combine them like that unless they're the same. For example, √2 + √2 would be 2√2, but √2 and √3 are different. So, adding them together shouldn't give me another square root like √5. To check, maybe I can square both sides to see if it holds true. If I square the left side, (√2 + √3)², that would be (√2)² + 2*√2*√3 + (√3)², which is 2 + 2√6 + 3, totaling 5 + 2√6. On the right side, (√5)² is just 5. So, 5 + 2√6 is definitely not equal to 5. Therefore, option A is incorrect.Moving on to option B: (2√3)² = 6. Let's calculate that. Squaring 2√3 should be (2)²*(√3)², which is 4*3, resulting in 12. But the option says it's equal to 6, which is half of 12. So, that doesn't add up. Option B is also incorrect.Now, option C: -√3 + 4√3 = 4. Okay, combining like terms here. If I have -√3 and add 4√3, that's like having -1√3 + 4√3, which should be 3√3. But the option says it equals 4, which is a whole number, not involving √3. That doesn't seem right. So, option C is incorrect as well.Finally, option D: √32 ÷ √2 = 4. Let's see. I know that dividing square roots can be simplified by dividing the numbers under the roots. So, √32 ÷ √2 is the same as √(32/2), which is √16. And √16 is 4. That matches the option exactly. So, option D is correct.Wait, let me double-check everything to make sure I didn't make any mistakes. For option A, squaring both sides definitely didn't work out. Option B, squaring 2√3 gives 12, not 6. Option C, combining the terms gives 3√3, not 4. And option D, simplifying the division of square roots gives 4, which is correct. Yeah, I think I got it right.