💡Okay, so I need to figure out how to convert the octal number 74532 base 8 into its base ten equivalent. Hmm, I remember that octal numbers are base 8, which means each digit represents a power of 8. But I'm a bit fuzzy on the exact steps, so I should probably start by recalling how number bases work in general.First, in any base, each digit represents a value multiplied by the base raised to the power of its position, starting from 0 on the right. So, for example, in a base-10 number like 1234, it's 1*10^3 + 2*10^2 + 3*10^1 + 4*10^0. I think the same principle applies to base-8 numbers.So, applying that to 74532 base 8, I need to break it down digit by digit, starting from the rightmost digit. Let's list out the digits and their positions:From right to left, the digits are:- Position 0: 2- Position 1: 3- Position 2: 5- Position 3: 4- Position 4: 7Wait, actually, when counting positions, we start from the right, so the rightmost digit is position 0. So, in 74532, the digits are:- Position 4: 7- Position 3: 4- Position 2: 5- Position 1: 3- Position 0: 2Okay, that makes sense. So each digit is multiplied by 8 raised to the power of its position.So, the formula would be:7 * 8^4 + 4 * 8^3 + 5 * 8^2 + 3 * 8^1 + 2 * 8^0Now, I need to calculate each of these terms separately and then add them up to get the base ten equivalent.Let's start with the first term: 7 * 8^4.Calculating 8^4: 8 * 8 * 8 * 8. Let's do that step by step:8 * 8 = 6464 * 8 = 512512 * 8 = 4096So, 8^4 is 4096. Then, 7 * 4096. Let's compute that:4096 * 7. Hmm, 4000 * 7 = 28,000 and 96 * 7 = 672, so total is 28,000 + 672 = 28,672.Okay, so the first term is 28,672.Next term: 4 * 8^3.Calculating 8^3: 8 * 8 * 8. Again, step by step:8 * 8 = 6464 * 8 = 512So, 8^3 is 512. Then, 4 * 512. Let's compute that:500 * 4 = 2,000 and 12 * 4 = 48, so total is 2,000 + 48 = 2,048.So, the second term is 2,048.Third term: 5 * 8^2.Calculating 8^2: 8 * 8 = 64.Then, 5 * 64. That's straightforward: 5 * 60 = 300 and 5 * 4 = 20, so total is 300 + 20 = 320.So, the third term is 320.Fourth term: 3 * 8^1.Calculating 8^1 is just 8.Then, 3 * 8 = 24.So, the fourth term is 24.Fifth term: 2 * 8^0.Any number raised to the power of 0 is 1, so 8^0 = 1.Then, 2 * 1 = 2.So, the fifth term is 2.Now, I need to add all these terms together to get the base ten equivalent.Let's list them again:- 28,672- 2,048- 320- 24- 2Adding them step by step:First, add 28,672 and 2,048:28,672 + 2,048 = 30,720Next, add 320 to that:30,720 + 320 = 31,040Then, add 24:31,040 + 24 = 31,064Finally, add 2:31,064 + 2 = 31,066So, the total is 31,066.Wait a minute, let me double-check my calculations to make sure I didn't make any mistakes.Starting with 7 * 8^4:8^4 = 40967 * 4096 = 28,672. That seems correct.4 * 8^3:8^3 = 5124 * 512 = 2,048. Correct.5 * 8^2:8^2 = 645 * 64 = 320. Correct.3 * 8^1:8^1 = 83 * 8 = 24. Correct.2 * 8^0:8^0 = 12 * 1 = 2. Correct.Adding them up:28,672 + 2,048 = 30,72030,720 + 320 = 31,04031,040 + 24 = 31,06431,064 + 2 = 31,066Yes, that all adds up correctly. So, the base ten equivalent of 74532 base 8 is 31,066.I think that's it. I don't see any errors in my calculations, so I'm confident that 31,066 is the correct base ten equivalent.

💡Okay, so I've got this problem here with two parts, a) and b). Let me try to understand what each part is asking and how to approach them.Starting with part a): We have two sets, A and B. Set A consists of natural numbers that can be written in the form ( x^2 + 2y^2 ) where ( x ) and ( y ) are natural numbers and ( x > y ). Set B consists of natural numbers that can be written as ( frac{a^3 + b^3 + c^3}{a + b + c} ) where ( a, b, c ) are distinct natural numbers. The task is to prove that every element of A is also an element of B, meaning ( mathbb{A} subset mathbb{B} ).Hmm, so I need to show that if a number can be expressed as ( x^2 + 2y^2 ), then it can also be expressed in the form ( frac{a^3 + b^3 + c^3}{a + b + c} ) with distinct ( a, b, c ).Maybe I can find a way to represent ( x^2 + 2y^2 ) as such a fraction. Let me think about how to choose ( a, b, c ) such that when I compute ( frac{a^3 + b^3 + c^3}{a + b + c} ), I get ( x^2 + 2y^2 ).Perhaps I can express ( a, b, c ) in terms of ( x ) and ( y ). Let me try setting ( a = x - y ), ( b = x ), and ( c = x + y ). These are three numbers centered around ( x ), spaced by ( y ). Let's see if this works.First, compute ( a + b + c ):( (x - y) + x + (x + y) = 3x ).Now, compute ( a^3 + b^3 + c^3 ):( (x - y)^3 + x^3 + (x + y)^3 ).Expanding each term:- ( (x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3 )- ( x^3 = x^3 )- ( (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 )Adding them together:( (x^3 - 3x^2y + 3xy^2 - y^3) + x^3 + (x^3 + 3x^2y + 3xy^2 + y^3) )Simplify:- The ( -3x^2y ) and ( +3x^2y ) cancel out.- The ( -y^3 ) and ( +y^3 ) cancel out.- We're left with ( x^3 + x^3 + x^3 + 3xy^2 + 3xy^2 )Which is ( 3x^3 + 6xy^2 ).So, ( a^3 + b^3 + c^3 = 3x^3 + 6xy^2 ).Now, compute the fraction:( frac{3x^3 + 6xy^2}{3x} = frac{3x(x^2 + 2y^2)}{3x} = x^2 + 2y^2 ).Perfect! So, by choosing ( a = x - y ), ( b = x ), and ( c = x + y ), we've expressed ( x^2 + 2y^2 ) as ( frac{a^3 + b^3 + c^3}{a + b + c} ). Since ( x > y ), ( a = x - y ) is positive, and ( a, b, c ) are distinct because ( y ) is at least 1, making ( a < b < c ).Therefore, every element of A is indeed in B, so ( mathbb{A} subset mathbb{B} ).Moving on to part b): We need to prove that there are infinitely many positive integers ( n ) such that ( n in mathbb{B} ) but ( n notin mathbb{A} ). In other words, there are infinitely many numbers expressible as ( frac{a^3 + b^3 + c^3}{a + b + c} ) with distinct ( a, b, c ), which cannot be written as ( x^2 + 2y^2 ) with ( x > y ).To approach this, I should find a way to construct numbers in B that cannot be in A. Maybe by choosing specific forms of ( a, b, c ) that lead to numbers not representable as ( x^2 + 2y^2 ).Let me think about how to choose ( a, b, c ) such that ( frac{a^3 + b^3 + c^3}{a + b + c} ) results in a number that isn't of the form ( x^2 + 2y^2 ).Perhaps if I set two of the variables equal and let the third vary, but since ( a, b, c ) must be distinct, that won't work. Alternatively, maybe set ( c = a + b ) or some linear combination.Let me try setting ( c = a + b ). Then, compute ( frac{a^3 + b^3 + (a + b)^3}{a + b + (a + b)} ).Compute numerator:( a^3 + b^3 + (a + b)^3 = a^3 + b^3 + a^3 + 3a^2b + 3ab^2 + b^3 = 2a^3 + 2b^3 + 3a^2b + 3ab^2 ).Denominator:( a + b + a + b = 2a + 2b ).So, the fraction becomes:( frac{2a^3 + 2b^3 + 3a^2b + 3ab^2}{2a + 2b} ).Factor numerator:( 2(a^3 + b^3) + 3ab(a + b) ).Factor denominator:( 2(a + b) ).So, the fraction simplifies to:( frac{2(a^3 + b^3) + 3ab(a + b)}{2(a + b)} = frac{2(a + b)(a^2 - ab + b^2) + 3ab(a + b)}{2(a + b)} ).Factor out ( (a + b) ):( frac{(a + b)(2(a^2 - ab + b^2) + 3ab)}{2(a + b)} = frac{2(a^2 - ab + b^2) + 3ab}{2} ).Simplify numerator:( 2a^2 - 2ab + 2b^2 + 3ab = 2a^2 + ab + 2b^2 ).So, the fraction becomes:( frac{2a^2 + ab + 2b^2}{2} = a^2 + frac{ab}{2} + b^2 ).Wait, that doesn't seem right. Let me check my steps again.Wait, no, actually, when I factored ( a^3 + b^3 ), it's ( (a + b)(a^2 - ab + b^2) ). So, the numerator becomes ( 2(a + b)(a^2 - ab + b^2) + 3ab(a + b) ). Then, factoring ( (a + b) ), we have ( (a + b)(2(a^2 - ab + b^2) + 3ab) ).Expanding inside the parentheses:( 2a^2 - 2ab + 2b^2 + 3ab = 2a^2 + ab + 2b^2 ).So, the fraction is ( frac{(a + b)(2a^2 + ab + 2b^2)}{2(a + b)} = frac{2a^2 + ab + 2b^2}{2} ).Wait, that still leaves a fraction. Maybe I made a mistake in the substitution.Alternatively, perhaps I should choose specific values for ( a ) and ( b ) such that the resulting ( n ) cannot be expressed as ( x^2 + 2y^2 ).Let me try setting ( b = 1 ) and ( a ) varying. Then, ( c = a + 1 ).Compute ( n = frac{a^3 + 1 + (a + 1)^3}{a + 1 + (a + 1)} ).Compute numerator:( a^3 + 1 + (a^3 + 3a^2 + 3a + 1) = 2a^3 + 3a^2 + 3a + 2 ).Denominator:( 2a + 2 ).So, ( n = frac{2a^3 + 3a^2 + 3a + 2}{2a + 2} ).Let me perform polynomial division or factor the numerator.Factor numerator:Let me try to factor ( 2a^3 + 3a^2 + 3a + 2 ).Try rational roots: possible roots are ( pm1, pm2, pm1/2 ).Test ( a = -1 ):( 2(-1)^3 + 3(-1)^2 + 3(-1) + 2 = -2 + 3 - 3 + 2 = 0 ). So, ( a = -1 ) is a root.Thus, ( (a + 1) ) is a factor. Perform polynomial division:Divide ( 2a^3 + 3a^2 + 3a + 2 ) by ( a + 1 ).Using synthetic division:- Coefficients: 2 | 3 | 3 | 2- Root: -1Bring down 2.Multiply by -1: 2*(-1) = -2. Add to next coefficient: 3 + (-2) = 1.Multiply by -1: 1*(-1) = -1. Add to next coefficient: 3 + (-1) = 2.Multiply by -1: 2*(-1) = -2. Add to last coefficient: 2 + (-2) = 0.So, the quotient is ( 2a^2 + a + 2 ).Thus, ( 2a^3 + 3a^2 + 3a + 2 = (a + 1)(2a^2 + a + 2) ).Therefore, ( n = frac{(a + 1)(2a^2 + a + 2)}{2(a + 1)} = frac{2a^2 + a + 2}{2} ).Simplify:( n = a^2 + frac{a}{2} + 1 ).Hmm, but ( a ) is a natural number, so ( frac{a}{2} ) must be an integer, meaning ( a ) must be even. Let me set ( a = 2k ) where ( k ) is a natural number.Then, ( n = (2k)^2 + frac{2k}{2} + 1 = 4k^2 + k + 1 ).So, ( n = 4k^2 + k + 1 ).Now, I need to show that this ( n ) cannot be expressed as ( x^2 + 2y^2 ) for any ( x, y ) with ( x > y ).Assume for contradiction that ( 4k^2 + k + 1 = x^2 + 2y^2 ).Let me analyze this equation modulo some number to find a contradiction.Let me consider modulo 4.Compute ( 4k^2 + k + 1 ) mod 4:- ( 4k^2 equiv 0 mod 4 )- ( k mod 4 ) can be 0,1,2,3- So, ( k + 1 mod 4 ) can be 1,2,3,0.Thus, ( 4k^2 + k + 1 equiv k + 1 mod 4 ).Now, ( x^2 + 2y^2 mod 4 ):Squares mod 4 are 0 or 1.So, ( x^2 equiv 0 ) or 1, and ( 2y^2 equiv 0 ) or 2.Thus, possible values of ( x^2 + 2y^2 mod 4 ):- 0 + 0 = 0- 0 + 2 = 2- 1 + 0 = 1- 1 + 2 = 3So, possible residues are 0,1,2,3.But ( 4k^2 + k + 1 equiv k + 1 mod 4 ).So, depending on ( k mod 4 ), ( n mod 4 ) can be:- If ( k equiv 0 mod 4 ): ( n equiv 1 mod 4 )- If ( k equiv 1 mod 4 ): ( n equiv 2 mod 4 )- If ( k equiv 2 mod 4 ): ( n equiv 3 mod 4 )- If ( k equiv 3 mod 4 ): ( n equiv 0 mod 4 )Now, let's see if ( n ) can be expressed as ( x^2 + 2y^2 ).If ( n equiv 0 mod 4 ), then ( x^2 + 2y^2 equiv 0 mod 4 ). This requires ( x^2 equiv 0 mod 4 ) and ( 2y^2 equiv 0 mod 4 ), so ( x ) and ( y ) must be even.If ( n equiv 1 mod 4 ), then ( x^2 + 2y^2 equiv 1 mod 4 ). Possible if ( x^2 equiv 1 ) and ( 2y^2 equiv 0 ), so ( x ) is odd and ( y ) is even.If ( n equiv 2 mod 4 ), then ( x^2 + 2y^2 equiv 2 mod 4 ). This requires ( x^2 equiv 0 ) and ( 2y^2 equiv 2 ), so ( x ) even and ( y ) odd.If ( n equiv 3 mod 4 ), then ( x^2 + 2y^2 equiv 3 mod 4 ). This would require ( x^2 equiv 1 ) and ( 2y^2 equiv 2 ), so ( x ) odd and ( y ) odd.But let's look deeper. Let me consider the equation ( 4k^2 + k + 1 = x^2 + 2y^2 ).Let me rearrange it:( x^2 = 4k^2 + k + 1 - 2y^2 ).I need to see if this can hold for some integers ( x, y ).Alternatively, perhaps I can look at the equation modulo 8 to find a contradiction.Compute ( 4k^2 + k + 1 mod 8 ).Let me consider ( k ) modulo 8.But maybe a better approach is to consider specific values of ( k ) and see if ( n ) can be expressed as ( x^2 + 2y^2 ).Wait, but I need a general argument for all ( k ), or at least infinitely many ( k ).Alternatively, perhaps I can choose ( k ) such that ( n = 4k^2 + k + 1 ) is not of the form ( x^2 + 2y^2 ).Let me consider ( k equiv 3 mod 8 ). Let ( k = 8m + 3 ) for some integer ( m geq 0 ).Then, ( n = 4(8m + 3)^2 + (8m + 3) + 1 ).Compute ( n ):First, ( (8m + 3)^2 = 64m^2 + 48m + 9 ).So, ( 4(64m^2 + 48m + 9) = 256m^2 + 192m + 36 ).Add ( 8m + 3 + 1 = 8m + 4 ).Total ( n = 256m^2 + 192m + 36 + 8m + 4 = 256m^2 + 200m + 40 ).Simplify:( n = 256m^2 + 200m + 40 ).Now, let's compute ( n mod 8 ):- ( 256m^2 equiv 0 mod 8 )- ( 200m equiv 0 mod 8 ) because 200 = 25*8, so 200m ≡ 0 mod 8- ( 40 equiv 0 mod 8 )So, ( n equiv 0 mod 8 ).Now, if ( n = x^2 + 2y^2 ), then ( x^2 + 2y^2 equiv 0 mod 8 ).Let's see the possible residues of ( x^2 ) and ( 2y^2 ) modulo 8.Squares modulo 8: 0,1,4.So, ( x^2 ) can be 0,1,4 mod 8.( 2y^2 ): If ( y^2 equiv 0 mod 8 ), then ( 2y^2 equiv 0 mod 8 ).If ( y^2 equiv 1 mod 8 ), then ( 2y^2 equiv 2 mod 8 ).If ( y^2 equiv 4 mod 8 ), then ( 2y^2 equiv 0 mod 8 ) because 2*4=8≡0.Thus, possible ( x^2 + 2y^2 mod 8 ):- If ( x^2 ≡ 0 ), ( 2y^2 ≡ 0 ) or 2: total 0 or 2- If ( x^2 ≡ 1 ), ( 2y^2 ≡ 0 ) or 2: total 1 or 3- If ( x^2 ≡ 4 ), ( 2y^2 ≡ 0 ) or 2: total 4 or 6So, possible residues are 0,1,2,3,4,6.But ( n ≡ 0 mod 8 ), so ( x^2 + 2y^2 ≡ 0 mod 8 ).This requires ( x^2 ≡ 0 mod 8 ) and ( 2y^2 ≡ 0 mod 8 ).Thus, ( x ) must be even, say ( x = 2x' ), and ( y ) must be even, say ( y = 2y' ).Substitute into ( n = x^2 + 2y^2 ):( n = (2x')^2 + 2(2y')^2 = 4x'^2 + 8y'^2 = 4(x'^2 + 2y'^2) ).But ( n = 256m^2 + 200m + 40 = 4(64m^2 + 50m + 10) ).So, ( x'^2 + 2y'^2 = 64m^2 + 50m + 10 ).Now, let's compute ( 64m^2 + 50m + 10 mod 4 ):- ( 64m^2 ≡ 0 mod 4 )- ( 50m ≡ 2m mod 4 )- ( 10 ≡ 2 mod 4 )So, total ( 0 + 2m + 2 ≡ 2(m + 1) mod 4 ).Thus, ( x'^2 + 2y'^2 ≡ 2(m + 1) mod 4 ).But ( x'^2 ) can be 0 or 1 mod 4, and ( 2y'^2 ) can be 0 or 2 mod 4.So, possible residues:- If ( x'^2 ≡ 0 ), ( 2y'^2 ≡ 0 ) or 2: total 0 or 2- If ( x'^2 ≡ 1 ), ( 2y'^2 ≡ 0 ) or 2: total 1 or 3Thus, ( x'^2 + 2y'^2 ) can be 0,1,2,3 mod 4.But we have ( x'^2 + 2y'^2 ≡ 2(m + 1) mod 4 ).So, ( 2(m + 1) ) must be congruent to 0,1,2,3 mod 4.But ( 2(m + 1) ) can be 0,2 mod 4, depending on ( m ).If ( m ) is even, ( m + 1 ) is odd, so ( 2(m + 1) ≡ 2 mod 4 ).If ( m ) is odd, ( m + 1 ) is even, so ( 2(m + 1) ≡ 0 mod 4 ).Thus, ( x'^2 + 2y'^2 ≡ 0 ) or 2 mod 4.But let's look at ( x'^2 + 2y'^2 = 64m^2 + 50m + 10 ).Compute ( 64m^2 + 50m + 10 mod 4 ):- ( 64m^2 ≡ 0 mod 4 )- ( 50m ≡ 2m mod 4 )- ( 10 ≡ 2 mod 4 )So, total ( 0 + 2m + 2 ≡ 2(m + 1) mod 4 ).As before, this is 0 or 2 mod 4.But let's look at ( x'^2 + 2y'^2 ).If ( x'^2 + 2y'^2 ≡ 0 mod 4 ), then ( x' ) and ( y' ) must be even, so ( x' = 2x'' ), ( y' = 2y'' ).Then, ( x'^2 + 2y'^2 = 4x''^2 + 8y''^2 = 4(x''^2 + 2y''^2) ).Thus, ( 64m^2 + 50m + 10 = 4(x''^2 + 2y''^2) ).But ( 64m^2 + 50m + 10 ) divided by 4 is ( 16m^2 + 12.5m + 2.5 ), which is not an integer because of the fractions. Contradiction.Similarly, if ( x'^2 + 2y'^2 ≡ 2 mod 4 ), then ( x'^2 ≡ 0 mod 4 ) and ( 2y'^2 ≡ 2 mod 4 ), so ( y'^2 ≡ 1 mod 2 ), meaning ( y' ) is odd.Thus, ( x' ) is even, ( y' ) is odd.But then, ( x'^2 + 2y'^2 = 4x''^2 + 2(2y'' + 1)^2 ).Wait, maybe this is getting too complicated. Let me try a different approach.Let me consider the equation ( n = 4k^2 + k + 1 = x^2 + 2y^2 ).Rearrange to ( x^2 = 4k^2 + k + 1 - 2y^2 ).I need to see if this can hold for some integers ( x, y ).Let me consider the equation modulo 8 again.Compute ( n = 4k^2 + k + 1 mod 8 ).As before, ( 4k^2 mod 8 ) depends on ( k mod 2 ).If ( k ) is even, ( k = 2m ), then ( 4k^2 = 16m^2 ≡ 0 mod 8 ).If ( k ) is odd, ( k = 2m + 1 ), then ( 4k^2 = 4(4m^2 + 4m + 1) = 16m^2 + 16m + 4 ≡ 4 mod 8 ).So, ( n mod 8 ) is:- If ( k ) even: ( 0 + 2m + 1 mod 8 ) (since ( k = 2m ), ( k + 1 = 2m + 1 ))Wait, no, ( k ) is even, so ( k = 2m ), then ( n = 4(4m^2) + 2m + 1 = 16m^2 + 2m + 1 ).Thus, ( n ≡ 2m + 1 mod 8 ).Similarly, if ( k ) is odd, ( k = 2m + 1 ), then ( n = 4(4m^2 + 4m + 1) + (2m + 1) + 1 = 16m^2 + 16m + 4 + 2m + 2 = 16m^2 + 18m + 6 ).Thus, ( n ≡ 18m + 6 mod 8 ). Since 18 ≡ 2 mod 8, this is ( 2m + 6 mod 8 ).So, depending on ( m ), ( n mod 8 ) can be:- If ( k ) even: ( 2m + 1 mod 8 )- If ( k ) odd: ( 2m + 6 mod 8 )Now, let's see what ( x^2 + 2y^2 mod 8 ) can be.As before, possible residues are 0,1,2,3,4,6.But let's look at specific cases.Suppose ( k ) is odd, so ( n ≡ 2m + 6 mod 8 ).Let me choose ( m ) such that ( 2m + 6 ≡ 7 mod 8 ). That is, ( 2m ≡ 1 mod 8 ). But 2m ≡1 mod8 has no solution because 2 and 8 are not coprime. So, ( n ) cannot be ≡7 mod8.Wait, but ( n ) is constructed as ( 4k^2 + k + 1 ), so let's see what ( n mod 8 ) can be.Wait, when ( k ) is odd, ( n ≡ 2m + 6 mod 8 ). Let me set ( m ) such that ( 2m + 6 ≡ 7 mod 8 ). Then, ( 2m ≡1 mod8 ), which is impossible. So, ( n ) cannot be ≡7 mod8.But ( x^2 + 2y^2 ) can be 0,1,2,3,4,6 mod8, so 7 is not among them. Thus, if ( n ≡7 mod8 ), it cannot be expressed as ( x^2 + 2y^2 ).But wait, can ( n ) ever be ≡7 mod8?From earlier, when ( k ) is odd, ( n ≡2m +6 mod8 ). Let's see if ( 2m +6 ≡7 mod8 ).This would require ( 2m ≡1 mod8 ), which has no solution because 2m is always even, and 1 is odd.Thus, ( n ) cannot be ≡7 mod8 when ( k ) is odd.But wait, when ( k ) is even, ( n ≡2m +1 mod8 ). Let me see if ( n ) can be ≡7 mod8 in this case.Set ( 2m +1 ≡7 mod8 ), so ( 2m ≡6 mod8 ), which implies ( m ≡3 mod4 ).Thus, if ( k ) is even and ( m ≡3 mod4 ), then ( n ≡7 mod8 ).But in this case, ( n ≡7 mod8 ), which cannot be expressed as ( x^2 + 2y^2 ) because, as we saw, ( x^2 + 2y^2 ) cannot be ≡7 mod8.Thus, for ( k ) even and ( m ≡3 mod4 ), ( n =4k^2 +k +1 ≡7 mod8 ), which cannot be expressed as ( x^2 + 2y^2 ).Therefore, for infinitely many ( k ) (specifically, when ( k ) is even and ( m ≡3 mod4 )), ( n ) is in B but not in A.Hence, there are infinitely many positive integers ( n ) such that ( n in mathbb{B} ) and ( n notin mathbb{A} ).I think this approach works. By choosing ( a = 2k ), ( b =1 ), ( c =2k +1 ), we get ( n =4k^2 +k +1 ), and for specific ( k ), ( n ) cannot be expressed as ( x^2 + 2y^2 ), thus proving part b).

💡Okay, so I need to figure out what sin 60 degrees is. Hmm, I remember that sine has something to do with right triangles and the unit circle. Let me think about this step by step.First, I know that in a right triangle, the sine of an angle is the ratio of the length of the side opposite the angle to the length of the hypotenuse. So, if I have a right triangle with a 60-degree angle, then sin 60° would be the length of the side opposite 60 degrees divided by the hypotenuse.But wait, I also remember something about special triangles. There's a 30-60-90 triangle, right? In such a triangle, the sides are in a specific ratio. Let me recall: the side opposite the 30-degree angle is the shortest one and is usually denoted as 1, the side opposite the 60-degree angle is √3, and the hypotenuse is 2. So, the sides are in the ratio 1 : √3 : 2.If that's the case, then in a 30-60-90 triangle, the side opposite 60 degrees is √3, and the hypotenuse is 2. Therefore, sin 60° should be √3 divided by 2, which is √3/2. That seems straightforward.But I also want to think about this in terms of the unit circle because I remember that the unit circle is a way to define sine and cosine for any angle, not just those in right triangles. On the unit circle, the radius is 1, and any point on the circle can be represented as (cos θ, sin θ), where θ is the angle from the positive x-axis.So, if I consider a 60-degree angle on the unit circle, the coordinates of that point should be (cos 60°, sin 60°). I know that cos 60° is 0.5 because in the 30-60-90 triangle, the adjacent side to the 60-degree angle is 1, and the hypotenuse is 2, so cos 60° = 1/2. Therefore, sin 60° should be the y-coordinate, which, based on the triangle, is √3/2.Wait, let me make sure I'm not mixing things up. In the unit circle, the radius is 1, so the hypotenuse of the right triangle formed by the radius is 1, not 2. Does that change anything? Hmm, no, because the ratios still hold. If I scale down the 30-60-90 triangle so that the hypotenuse is 1 instead of 2, then the sides become 1/2 and √3/2. So, the side opposite 60 degrees is √3/2, and the adjacent side is 1/2. Therefore, sin 60° is indeed √3/2.I think I've got it. Both the special triangle approach and the unit circle approach lead me to the same answer. So, sin 60° is √3/2. I should also remember that this is one of the standard angles, so it's good to memorize these values for future use.Just to double-check, I can think about the graph of the sine function. I know that sin 0° is 0, sin 30° is 0.5, sin 90° is 1, and so on. So, sin 60° should be somewhere between sin 30° and sin 90°, which is between 0.5 and 1. √3/2 is approximately 0.866, which fits between 0.5 and 1. That makes sense.Another way to think about it is using the Pythagorean identity. Since sin² θ + cos² θ = 1, and I know that cos 60° is 0.5, then sin² 60° + (0.5)² = 1. So, sin² 60° + 0.25 = 1, which means sin² 60° = 0.75. Taking the square root of 0.75 gives √3/2, which is approximately 0.866. So, that confirms it again.I think I'm confident now that sin 60° is √3/2. It comes up in a lot of problems, especially in geometry and trigonometry, so it's a good value to remember. Plus, understanding both the triangle and unit circle perspectives helps solidify the concept.

💡Alright, so I've got this problem about a rectangular parallelepiped with integer coordinates and a volume of 2017. I need to find the diagonal of the parallelepiped when its edges are parallel to the coordinate axes, and then determine if there's such a shape where not all edges are parallel to the axes. Hmm, okay, let's break this down step by step.Starting with part (a). A rectangular parallelepiped is like a 3D shape with all sides meeting at right angles, so it's basically a rectangular box. The volume is given by the product of its length, width, and height. Since the volume is 2017, which is a prime number, I remember that prime numbers only have two positive divisors: 1 and themselves. So, the possible dimensions of the box must be 1, 1, and 2017 because those are the only integers that multiply to 2017.Now, to find the diagonal of this box. I recall that in a rectangular box, the space diagonal can be found using the formula derived from the Pythagorean theorem in three dimensions. The formula is:[ d = sqrt{a^2 + b^2 + c^2} ]where ( a ), ( b ), and ( c ) are the lengths of the sides. Plugging in the values we have:[ d = sqrt{1^2 + 1^2 + 2017^2} ][ d = sqrt{1 + 1 + 2017^2} ][ d = sqrt{2 + 2017^2} ]So, that's the diagonal. It seems straightforward because 2017 is prime, so the only possible integer dimensions are 1, 1, and 2017.Moving on to part (b). The question is asking if there's a rectangular parallelepiped with the same volume but where not all edges are parallel to the coordinate axes. Hmm, that means the shape might be "tilted" in some way, not aligned perfectly with the x, y, and z axes.I remember that in 3D geometry, a parallelepiped can be defined by three vectors that don't necessarily have to be orthogonal or aligned with the axes. The volume of a parallelepiped can also be calculated using the scalar triple product of these vectors. The formula is:[ V = | mathbf{a} cdot (mathbf{b} times mathbf{c}) | ]where ( mathbf{a} ), ( mathbf{b} ), and ( mathbf{c} ) are vectors defining the parallelepiped.Since we need integer coordinates, the vectors must have integer components. The volume is 2017, which is prime, so similar to part (a), we need to find three vectors whose scalar triple product is 2017.But here's the catch: the vectors don't have to be orthogonal. So, maybe we can find vectors that are not aligned with the axes but still result in the volume being 2017.I also recall that 2017 is a prime number of the form ( 4k + 1 ). Numbers of this form can be expressed as the sum of two squares. Let me check:[ 2017 = 44^2 + 9^2 ][ 44^2 = 1936 ][ 9^2 = 81 ][ 1936 + 81 = 2017 ]Yes, that works. So, 2017 can be written as the sum of two squares. This is important because it means we can have vectors in the plane (not necessarily aligned with the axes) that have integer components and whose lengths squared add up to 2017.For example, consider vectors in the xy-plane with components (44, 9, 0) and (9, -44, 0). These vectors are perpendicular to each other because their dot product is:[ (44)(9) + (9)(-44) + (0)(0) = 396 - 396 + 0 = 0 ]So, they form a right angle, which is good. The area of the base parallelogram formed by these vectors is:[ text{Area} = | mathbf{a} times mathbf{b} | = | (44, 9, 0) times (9, -44, 0) | ][ = | (0, 0, (44)(-44) - (9)(9)) | ][ = | (0, 0, -1936 - 81) | ][ = | (0, 0, -2017) | ][ = 2017 ]So, the area of the base is 2017. If we take the height along the z-axis as 1, then the volume of the parallelepiped would be:[ V = text{Base Area} times text{Height} = 2017 times 1 = 2017 ]This satisfies the volume requirement. Moreover, the vectors defining the base are not aligned with the coordinate axes, meaning the parallelepiped is "tilted" in the xy-plane.Therefore, such a parallelepiped exists where not all edges are parallel to the coordinate axes.Wait, but I need to make sure that all vertices have integer coordinates. The vectors I chose have integer components, so when you add them together, the resulting vertices will also have integer coordinates. For example, starting from the origin (0,0,0), adding vector (44,9,0) gives (44,9,0), adding vector (9,-44,0) gives (9,-44,0), and adding vector (0,0,1) gives (0,0,1). All these points have integer coordinates, so it seems to satisfy the condition.I think that covers both parts. For part (a), the diagonal is straightforward with the given dimensions, and for part (b), by expressing 2017 as a sum of squares, we can construct vectors that form a tilted parallelepiped with integer coordinates and the required volume.**Final Answer**a) boxed{sqrt{2017^2 + 2}}b) boxed{text{Yes}}

💡Okay, so I have this problem here about functions and whether they're even or not. Let me try to understand what it's asking. The function f(x) is defined on all real numbers, but it's not an even function. I need to figure out which statement among the options A to D is true.First, I remember that an even function satisfies the condition f(-x) = f(x) for all x in its domain. So, if a function is even, no matter what x you pick, plugging in -x will give you the same value as f(x). For example, f(x) = x² is even because (-x)² = x².But in this case, the function f(x) is not even. That means it doesn't satisfy f(-x) = f(x) for all x. So, the negation of that statement must be true. In logic, the negation of "for all x, P(x)" is "there exists an x such that not P(x)". So, if it's not true that f(-x) = f(x) for all x, then there must be at least one x where f(-x) ≠ f(x).Looking at the options:A: For all x in R, f(-x) ≠ f(x). This would mean that f is never equal to itself when you plug in -x. But that's too strong because a function that's not even might still have some points where f(-x) = f(x). For example, f(x) = x³ is not even, but f(0) = 0, and f(-0) = f(0). So, A is not necessarily true.B: For all x in R, f(-x) = -f(x). This would mean that f is an odd function. But just because f is not even doesn't mean it has to be odd. There are functions that are neither even nor odd. So, B is not necessarily true.C: There exists an x₀ in R such that f(-x₀) ≠ f(x₀). This aligns with what I thought earlier. Since f is not even, there must be at least one x where f(-x) ≠ f(x). So, C seems to be true.D: There exists an x₀ in R such that f(-x₀) = -f(x₀). This would mean that f is odd at that point, but again, just because f is not even doesn't mean it has to be odd somewhere. It could be neither. So, D isn't necessarily true.Therefore, the correct answer should be option C.

💡Alright, so I have this problem where a line intersects a circle at two points, and the triangle formed by these two points and the center of the circle is equilateral. I need to find the area of the circle. Hmm, okay, let's break this down step by step.First, the equation of the circle is given as ( x^2 + y^2 - 2ax - 2y + 2 = 0 ). I remember that to analyze a circle, it's helpful to rewrite its equation in the standard form by completing the square. So, let me try that.For the x-terms: ( x^2 - 2ax ). To complete the square, I take half of the coefficient of x, which is -2a, so half of that is -a, and then square it, getting ( a^2 ). So, I can write ( x^2 - 2ax = (x - a)^2 - a^2 ).For the y-terms: ( y^2 - 2y ). Similarly, half of -2 is -1, and squaring that gives 1. So, ( y^2 - 2y = (y - 1)^2 - 1 ).Putting it all back into the equation:[ (x - a)^2 - a^2 + (y - 1)^2 - 1 + 2 = 0 ]Simplify the constants:[ (x - a)^2 + (y - 1)^2 - a^2 - 1 + 2 = 0 ]Which simplifies to:[ (x - a)^2 + (y - 1)^2 = a^2 - 1 ]So, the circle has center at ( (a, 1) ) and radius ( sqrt{a^2 - 1} ). Okay, that makes sense.Next, the line given is ( y = ax ). This line intersects the circle at points A and B. So, points A and B lie on both the line and the circle. The triangle formed by points A, B, and the center C is equilateral. That means all sides are equal: AC = BC = AB.Since C is the center of the circle, AC and BC are both radii of the circle, so they are equal by definition. Therefore, AB must also be equal to the radius. So, the distance between points A and B is equal to the radius of the circle.But wait, in a circle, the length of the chord AB can be related to the radius and the distance from the center to the chord. The formula for the length of a chord is ( 2sqrt{r^2 - d^2} ), where ( d ) is the distance from the center to the chord, and ( r ) is the radius.In this case, since triangle ABC is equilateral, the distance from the center C to the chord AB must be such that the chord length AB equals the radius. So, setting up the equation:[ AB = 2sqrt{r^2 - d^2} = r ]So,[ 2sqrt{r^2 - d^2} = r ]Divide both sides by 2:[ sqrt{r^2 - d^2} = frac{r}{2} ]Square both sides:[ r^2 - d^2 = frac{r^2}{4} ]Subtract ( frac{r^2}{4} ) from both sides:[ frac{3r^2}{4} = d^2 ]So,[ d = frac{sqrt{3}r}{2} ]Okay, so the distance from the center to the chord AB is ( frac{sqrt{3}r}{2} ).Now, I need to find the distance from the center ( C(a, 1) ) to the line ( y = ax ). The formula for the distance from a point ( (x_0, y_0) ) to the line ( Ax + By + C = 0 ) is:[ text{Distance} = frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}} ]First, let's write the line ( y = ax ) in standard form:[ ax - y = 0 ]So, ( A = a ), ( B = -1 ), and ( C = 0 ).Plugging in the center ( (a, 1) ):[ text{Distance} = frac{|a cdot a + (-1) cdot 1 + 0|}{sqrt{a^2 + (-1)^2}} = frac{|a^2 - 1|}{sqrt{a^2 + 1}} ]Earlier, we found that this distance ( d ) should be ( frac{sqrt{3}r}{2} ). And since the radius ( r = sqrt{a^2 - 1} ), substituting that in:[ frac{|a^2 - 1|}{sqrt{a^2 + 1}} = frac{sqrt{3} cdot sqrt{a^2 - 1}}{2} ]Let me write that equation again:[ frac{|a^2 - 1|}{sqrt{a^2 + 1}} = frac{sqrt{3} cdot sqrt{a^2 - 1}}{2} ]Hmm, since ( a^2 - 1 ) is under a square root in the radius, it must be positive. So, ( a^2 - 1 > 0 ), which means ( |a^2 - 1| = a^2 - 1 ). So, I can drop the absolute value:[ frac{a^2 - 1}{sqrt{a^2 + 1}} = frac{sqrt{3} cdot sqrt{a^2 - 1}}{2} ]Let me square both sides to eliminate the square roots:[ left( frac{a^2 - 1}{sqrt{a^2 + 1}} right)^2 = left( frac{sqrt{3} cdot sqrt{a^2 - 1}}{2} right)^2 ]Simplify both sides:Left side:[ frac{(a^2 - 1)^2}{a^2 + 1} ]Right side:[ frac{3(a^2 - 1)}{4} ]So, the equation becomes:[ frac{(a^2 - 1)^2}{a^2 + 1} = frac{3(a^2 - 1)}{4} ]Assuming ( a^2 - 1 neq 0 ) (since the radius can't be zero), we can divide both sides by ( a^2 - 1 ):[ frac{a^2 - 1}{a^2 + 1} = frac{3}{4} ]Now, solve for ( a^2 ):Multiply both sides by ( a^2 + 1 ):[ a^2 - 1 = frac{3}{4}(a^2 + 1) ]Multiply both sides by 4 to eliminate the fraction:[ 4(a^2 - 1) = 3(a^2 + 1) ]Expand both sides:[ 4a^2 - 4 = 3a^2 + 3 ]Subtract ( 3a^2 ) from both sides:[ a^2 - 4 = 3 ]Add 4 to both sides:[ a^2 = 7 ]Okay, so ( a^2 = 7 ). Now, recall that the radius ( r = sqrt{a^2 - 1} ), so:[ r = sqrt{7 - 1} = sqrt{6} ]Therefore, the area of the circle is:[ pi r^2 = pi (sqrt{6})^2 = 6pi ]So, the area of circle ( C ) is ( 6pi ).**Final Answer**The area of circle ( C ) is boxed{6pi}.