💡Okay, so I need to figure out how many values of ( n leq 30 ) satisfy ( g_{50}(n) = 243 ). Hmm, let's start by understanding what ( g_j(n) ) means.First, ( g_1(n) ) is defined as three times the square of the number of positive integer divisors of ( n ). So, if I denote ( d(n) ) as the number of divisors of ( n ), then ( g_1(n) = 3 times (d(n))^2 ). For example, if ( n = 9 ), the divisors are 1, 3, and 9, so ( d(9) = 3 ) and ( g_1(9) = 3 times 3^2 = 27 ).For ( j geq 2 ), ( g_j(n) ) is defined recursively as ( g_1(g_{j-1}(n)) ). So, each subsequent ( g_j(n) ) is just applying ( g_1 ) to the previous result. Therefore, ( g_{50}(n) ) is just applying ( g_1 ) fifty times starting from ( n ).Our goal is to find all ( n leq 30 ) such that after applying ( g_1 ) fifty times, we get 243. That is, ( g_{50}(n) = 243 ).Let me think about what 243 is. 243 is ( 3^5 ). So, 243 is a power of 3. Maybe that can help us understand the behavior of ( g_1 ).First, let's understand ( g_1(n) ). Since ( g_1(n) = 3 times (d(n))^2 ), it's always a multiple of 3. So, ( g_1(n) ) is divisible by 3. Therefore, ( g_j(n) ) for any ( j geq 1 ) will also be a multiple of 3. Since 243 is a multiple of 3, that's consistent.Now, let's think about what ( g_1(n) ) can be. Since ( d(n) ) is the number of divisors of ( n ), which is always at least 1 (for ( n = 1 )) and increases as ( n ) has more factors. For ( n leq 30 ), the maximum number of divisors is for 24, which has 8 divisors. So, ( d(n) ) can range from 1 to 8 for ( n leq 30 ).Therefore, ( g_1(n) ) can be ( 3 times 1^2 = 3 ), ( 3 times 2^2 = 12 ), ( 3 times 3^2 = 27 ), ( 3 times 4^2 = 48 ), ( 3 times 5^2 = 75 ), ( 3 times 6^2 = 108 ), ( 3 times 7^2 = 147 ), or ( 3 times 8^2 = 192 ).So, ( g_1(n) ) can be one of these values: 3, 12, 27, 48, 75, 108, 147, 192.Now, let's see what ( g_2(n) = g_1(g_1(n)) ) would be. For each possible ( g_1(n) ), we need to compute ( g_1 ) of that value.Let's compute ( g_1 ) for each possible ( g_1(n) ):1. If ( g_1(n) = 3 ): - ( d(3) = 2 ) (divisors: 1, 3) - ( g_1(3) = 3 times 2^2 = 12 )2. If ( g_1(n) = 12 ): - ( d(12) = 6 ) (divisors: 1, 2, 3, 4, 6, 12) - ( g_1(12) = 3 times 6^2 = 108 )3. If ( g_1(n) = 27 ): - ( d(27) = 4 ) (divisors: 1, 3, 9, 27) - ( g_1(27) = 3 times 4^2 = 48 )4. If ( g_1(n) = 48 ): - ( d(48) = 10 ) (divisors: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48) - ( g_1(48) = 3 times 10^2 = 300 )5. If ( g_1(n) = 75 ): - ( d(75) = 6 ) (divisors: 1, 3, 5, 15, 25, 75) - ( g_1(75) = 3 times 6^2 = 108 )6. If ( g_1(n) = 108 ): - ( d(108) = 12 ) (divisors: 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108) - ( g_1(108) = 3 times 12^2 = 432 )7. If ( g_1(n) = 147 ): - ( d(147) = 6 ) (divisors: 1, 3, 7, 21, 49, 147) - ( g_1(147) = 3 times 6^2 = 108 )8. If ( g_1(n) = 192 ): - ( d(192) = 14 ) (divisors: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 192) - ( g_1(192) = 3 times 14^2 = 588 )So, ( g_2(n) ) can be 12, 108, 48, 300, 108, 432, 108, or 588 depending on ( g_1(n) ).Wait, but 243 is not among these results. So, does that mean that ( g_2(n) ) can never be 243? Or maybe it can be reached in more steps?But we need ( g_{50}(n) = 243 ). So, perhaps we need to see if 243 is part of a cycle or a fixed point in the function ( g_1 ).Let's check if 243 is a fixed point. Compute ( g_1(243) ):- ( d(243) ): 243 is ( 3^5 ), so the number of divisors is 6 (since exponents are 5, so 5+1=6).- ( g_1(243) = 3 times 6^2 = 108 ).So, ( g_1(243) = 108 ). Then ( g_2(243) = g_1(108) = 432 ), as computed earlier.Then ( g_3(243) = g_1(432) ). Let's compute ( g_1(432) ):- ( d(432) ): 432 factors into ( 2^4 times 3^3 ), so number of divisors is (4+1)(3+1) = 20.- ( g_1(432) = 3 times 20^2 = 1200 ).Continuing, ( g_4(243) = g_1(1200) ):- ( d(1200) ): 1200 factors into ( 2^4 times 3^1 times 5^2 ), so number of divisors is (4+1)(1+1)(2+1) = 5*2*3=30.- ( g_1(1200) = 3 times 30^2 = 2700 ).Hmm, this seems to be increasing. Let's see if we can find a cycle or a fixed point.Wait, let's check if 108 leads to a cycle:- ( g_1(108) = 432 )- ( g_1(432) = 1200 )- ( g_1(1200) = 2700 )- ( g_1(2700) ): 2700 factors into ( 2^2 times 3^3 times 5^2 ), so number of divisors is (2+1)(3+1)(2+1)=3*4*3=36.- ( g_1(2700) = 3 times 36^2 = 3888 )This is increasing each time. So, it seems like once we get to 108, it's increasing each time we apply ( g_1 ). Therefore, 243 is not part of a cycle, and once we go beyond a certain point, the values keep increasing.But wait, let's check if 243 can be reached from some other number.Earlier, we saw that ( g_1(27) = 48 ), ( g_1(48) = 300 ), ( g_1(300) ): let's compute that.- ( d(300) ): 300 factors into ( 2^2 times 3^1 times 5^2 ), so number of divisors is (2+1)(1+1)(2+1)=3*2*3=18.- ( g_1(300) = 3 times 18^2 = 972 ).Then ( g_1(972) ): 972 factors into ( 2^2 times 3^5 ), so number of divisors is (2+1)(5+1)=3*6=18.- ( g_1(972) = 3 times 18^2 = 972 ).Oh, so 972 is a fixed point because ( g_1(972) = 972 ). So, once we reach 972, it stays there.But we need to reach 243. Let's see if 243 can be part of a cycle or if it can be reached from some number.Wait, let's go back. Maybe I should consider the reverse. If ( g_{50}(n) = 243 ), then ( g_{49}(n) ) must be a number ( m ) such that ( g_1(m) = 243 ). So, we need to find ( m ) such that ( 3 times (d(m))^2 = 243 ). Solving for ( d(m) ):( 3 times (d(m))^2 = 243 )( (d(m))^2 = 81 )( d(m) = 9 )So, ( m ) must have exactly 9 divisors. Therefore, ( g_{49}(n) ) must be a number with exactly 9 divisors.Similarly, ( g_{48}(n) ) must be a number ( k ) such that ( g_1(k) = m ), where ( m ) has 9 divisors. So, ( 3 times (d(k))^2 = m ), and ( m ) has 9 divisors.Wait, this is getting complicated. Maybe it's better to work backwards from 243.Let me try to find all numbers ( m ) such that ( g_1(m) = 243 ). As above, ( d(m) = 9 ). So, ( m ) must have exactly 9 divisors.Numbers with exactly 9 divisors are numbers of the form ( p^8 ) or ( p^2 q^2 ) where ( p ) and ( q ) are distinct primes. Because the number of divisors is given by multiplying one more than each exponent in the prime factorization. So, for 9 divisors, the exponents plus one must multiply to 9. The possible factorizations are 9 = 9*1 or 3*3. Therefore, the number is either ( p^8 ) or ( p^2 q^2 ).Now, let's find all such numbers ( m ) with exactly 9 divisors, and then see if ( g_1(m) = 243 ).But wait, ( g_1(m) = 3 times (d(m))^2 = 3 times 81 = 243 ). So, any ( m ) with exactly 9 divisors will satisfy ( g_1(m) = 243 ).Therefore, ( g_{49}(n) ) must be a number with exactly 9 divisors. So, we need to find ( n leq 30 ) such that after 49 applications of ( g_1 ), we get a number with exactly 9 divisors.But this seems too vague. Maybe we can find cycles or fixed points.Earlier, we saw that 972 is a fixed point because ( g_1(972) = 972 ). So, if we ever reach 972, we stay there. But 972 is much larger than 30, so ( n leq 30 ) can't reach 972 in 50 steps? Wait, no, because ( g_j(n) ) can increase beyond 30.But we need ( g_{50}(n) = 243 ). So, perhaps we need to see if starting from some ( n leq 30 ), after 50 iterations, we reach 243.But 50 iterations is a lot. Maybe the sequence stabilizes or enters a cycle before that.Alternatively, perhaps we can find numbers ( n ) such that ( g_j(n) ) eventually becomes 243 and stays there. But earlier, we saw that ( g_1(243) = 108 ), and then ( g_1(108) = 432 ), which leads to increasing values. So, 243 is not a fixed point, but rather leads to increasing values.Wait, but if we have ( g_{50}(n) = 243 ), that means that at step 50, it's 243, but step 51 would be ( g_1(243) = 108 ), and step 52 would be ( g_1(108) = 432 ), etc. So, unless 243 is part of a cycle, but we saw that 243 leads to 108, which leads to 432, which leads to 1200, which leads to 2700, which leads to 3888, which is a fixed point? Wait, no, 3888: let's compute ( g_1(3888) ).3888 factors into ( 2^4 times 3^5 ), so number of divisors is (4+1)(5+1)=5*6=30.So, ( g_1(3888) = 3 times 30^2 = 2700 ).Wait, so ( g_1(3888) = 2700 ), which was previously computed as ( g_1(2700) = 3888 ). So, 2700 and 3888 form a cycle of length 2.So, if we ever reach 2700 or 3888, we cycle between them.But 243 is not part of this cycle. So, if we have ( g_{50}(n) = 243 ), then ( g_{51}(n) = 108 ), ( g_{52}(n) = 432 ), ( g_{53}(n) = 1200 ), ( g_{54}(n) = 2700 ), ( g_{55}(n) = 3888 ), ( g_{56}(n) = 2700 ), and so on, cycling between 2700 and 3888.Therefore, 243 is only at step 50, but then it goes to 108, which is not part of the cycle. So, unless the sequence can reach 243 at step 50 and then proceed to the cycle, but since 243 is not part of the cycle, it's just a transient state.But we need ( g_{50}(n) = 243 ). So, we need to find ( n leq 30 ) such that after 50 iterations, it's 243. But considering that 50 iterations is a lot, and the function tends to increase beyond 30, it's unlikely that any ( n leq 30 ) can reach 243 at step 50.Wait, but maybe some numbers ( n leq 30 ) can reach 243 in fewer steps, and then stay there? But earlier, we saw that ( g_1(243) = 108 ), so 243 doesn't stay; it goes to 108.Alternatively, maybe some numbers ( n leq 30 ) can reach 243 in exactly 50 steps, but I think that's too vague.Alternatively, perhaps we can find numbers ( n leq 30 ) such that ( g_j(n) ) eventually reaches 243 and then cycles or something. But since 243 leads to 108, which leads to 432, which leads to 1200, which leads to 2700, which leads to 3888, which cycles between 2700 and 3888, I don't see how 243 can be part of a cycle.Therefore, perhaps there are no such ( n leq 30 ) that ( g_{50}(n) = 243 ).Wait, but let's check some specific numbers ( n leq 30 ) and see what happens.Let's take ( n = 27 ). ( d(27) = 4 ), so ( g_1(27) = 3*16=48 ). Then ( g_2(27) = g_1(48) = 3*10^2=300 ). Then ( g_3(27) = g_1(300) = 3*18^2=972 ). Then ( g_4(27) = g_1(972) = 972 ). So, it gets stuck at 972.Similarly, take ( n = 16 ). ( d(16)=5 ), so ( g_1(16)=3*25=75 ). Then ( g_2(16)=g_1(75)=3*6^2=108 ). Then ( g_3(16)=g_1(108)=432 ). Then ( g_4(16)=g_1(432)=1200 ). Then ( g_5(16)=2700 ). Then ( g_6(16)=3888 ). Then ( g_7(16)=2700 ), and it cycles.Another example: ( n = 12 ). ( d(12)=6 ), so ( g_1(12)=108 ). Then ( g_2(12)=432 ), and so on, same as above.Another example: ( n = 1 ). ( d(1)=1 ), so ( g_1(1)=3 ). Then ( g_2(1)=g_1(3)=12 ). Then ( g_3(1)=g_1(12)=108 ), and then same as above.Another example: ( n = 2 ). ( d(2)=2 ), so ( g_1(2)=12 ). Then ( g_2(2)=108 ), same as above.Another example: ( n = 3 ). ( d(3)=2 ), so ( g_1(3)=12 ). Then ( g_2(3)=108 ), same as above.Another example: ( n = 4 ). ( d(4)=3 ), so ( g_1(4)=27 ). Then ( g_2(4)=g_1(27)=48 ). Then ( g_3(4)=g_1(48)=300 ). Then ( g_4(4)=972 ), and it stays.Another example: ( n = 5 ). ( d(5)=2 ), so ( g_1(5)=12 ). Then ( g_2(5)=108 ), same as above.Another example: ( n = 6 ). ( d(6)=4 ), so ( g_1(6)=48 ). Then ( g_2(6)=300 ). Then ( g_3(6)=972 ), and it stays.Another example: ( n = 7 ). ( d(7)=2 ), so ( g_1(7)=12 ). Then ( g_2(7)=108 ), same as above.Another example: ( n = 8 ). ( d(8)=4 ), so ( g_1(8)=48 ). Then ( g_2(8)=300 ). Then ( g_3(8)=972 ), and it stays.Another example: ( n = 9 ). ( d(9)=3 ), so ( g_1(9)=27 ). Then ( g_2(9)=48 ). Then ( g_3(9)=300 ). Then ( g_4(9)=972 ), and it stays.Another example: ( n = 10 ). ( d(10)=4 ), so ( g_1(10)=48 ). Then ( g_2(10)=300 ). Then ( g_3(10)=972 ), and it stays.Another example: ( n = 11 ). ( d(11)=2 ), so ( g_1(11)=12 ). Then ( g_2(11)=108 ), same as above.Another example: ( n = 12 ). Already did that.Another example: ( n = 13 ). ( d(13)=2 ), so ( g_1(13)=12 ). Then ( g_2(13)=108 ), same as above.Another example: ( n = 14 ). ( d(14)=4 ), so ( g_1(14)=48 ). Then ( g_2(14)=300 ). Then ( g_3(14)=972 ), and it stays.Another example: ( n = 15 ). ( d(15)=4 ), so ( g_1(15)=48 ). Then ( g_2(15)=300 ). Then ( g_3(15)=972 ), and it stays.Another example: ( n = 16 ). Already did that.Another example: ( n = 17 ). ( d(17)=2 ), so ( g_1(17)=12 ). Then ( g_2(17)=108 ), same as above.Another example: ( n = 18 ). ( d(18)=6 ), so ( g_1(18)=108 ). Then ( g_2(18)=432 ), and so on.Another example: ( n = 19 ). ( d(19)=2 ), so ( g_1(19)=12 ). Then ( g_2(19)=108 ), same as above.Another example: ( n = 20 ). ( d(20)=6 ), so ( g_1(20)=108 ). Then ( g_2(20)=432 ), and so on.Another example: ( n = 21 ). ( d(21)=4 ), so ( g_1(21)=48 ). Then ( g_2(21)=300 ). Then ( g_3(21)=972 ), and it stays.Another example: ( n = 22 ). ( d(22)=4 ), so ( g_1(22)=48 ). Then ( g_2(22)=300 ). Then ( g_3(22)=972 ), and it stays.Another example: ( n = 23 ). ( d(23)=2 ), so ( g_1(23)=12 ). Then ( g_2(23)=108 ), same as above.Another example: ( n = 24 ). ( d(24)=8 ), so ( g_1(24)=3*64=192 ). Then ( g_2(24)=g_1(192)=588 ). Then ( g_3(24)=g_1(588) ). Let's compute ( d(588) ).588 factors into ( 2^2 times 3^1 times 7^2 ), so number of divisors is (2+1)(1+1)(2+1)=3*2*3=18. So, ( g_1(588)=3*18^2=972 ). Then ( g_4(24)=972 ), and it stays.Another example: ( n = 25 ). ( d(25)=3 ), so ( g_1(25)=27 ). Then ( g_2(25)=48 ). Then ( g_3(25)=300 ). Then ( g_4(25)=972 ), and it stays.Another example: ( n = 26 ). ( d(26)=4 ), so ( g_1(26)=48 ). Then ( g_2(26)=300 ). Then ( g_3(26)=972 ), and it stays.Another example: ( n = 27 ). Already did that.Another example: ( n = 28 ). ( d(28)=6 ), so ( g_1(28)=108 ). Then ( g_2(28)=432 ), and so on.Another example: ( n = 29 ). ( d(29)=2 ), so ( g_1(29)=12 ). Then ( g_2(29)=108 ), same as above.Another example: ( n = 30 ). ( d(30)=8 ), so ( g_1(30)=192 ). Then ( g_2(30)=588 ). Then ( g_3(30)=972 ), and it stays.So, from all these examples, it seems that starting from any ( n leq 30 ), the sequence either goes to 972 and stays, or goes to the cycle between 2700 and 3888. But 243 is not part of any cycle, and once you reach 243, you go to 108, then to 432, then to 1200, then to 2700, and then cycles.Therefore, for ( g_{50}(n) = 243 ), we would need that at step 50, it's 243, but from the above, it seems that all ( n leq 30 ) either reach 972 or enter the 2700-3888 cycle, and 243 is not part of that.Moreover, since 50 is a large number, and the function tends to increase, it's unlikely that any ( n leq 30 ) can reach 243 at step 50.Therefore, the answer is 0.**Final Answer**boxed{A}

💡Alright, so I have this problem here about vectors and bases. Let me try to understand it step by step. We have two non-collinear vectors, (overrightarrow{e_{1}}) and (overrightarrow{e_{2}}). That means they're not pointing in the same direction or opposite directions; they're linearly independent, right? So, they can form a basis for the plane on their own. Now, we're given two other vectors, (overrightarrow{a}) and (overrightarrow{b}), defined as:[overrightarrow{a} = 2overrightarrow{e_{1}} + overrightarrow{e_{2}}]and[overrightarrow{b} = koverrightarrow{e_{1}} + 3overrightarrow{e_{2}}]where (k) is a real number. The question is asking for the range of values (k) can take such that (overrightarrow{a}) and (overrightarrow{b}) can serve as a basis for plane vectors. Okay, so if (overrightarrow{a}) and (overrightarrow{b}) are to form a basis, they must also be non-collinear. That makes sense because a basis requires vectors to be linearly independent. If they were collinear, one would just be a scalar multiple of the other, and they wouldn't span the entire plane. So, the key here is to ensure that (overrightarrow{a}) and (overrightarrow{b}) are not collinear. If they were collinear, there would exist some scalar (lambda) such that:[overrightarrow{a} = lambda overrightarrow{b}]Let me write that out:[2overrightarrow{e_{1}} + overrightarrow{e_{2}} = lambda (koverrightarrow{e_{1}} + 3overrightarrow{e_{2}})]Expanding the right side:[2overrightarrow{e_{1}} + overrightarrow{e_{2}} = lambda k overrightarrow{e_{1}} + 3lambda overrightarrow{e_{2}}]Since (overrightarrow{e_{1}}) and (overrightarrow{e_{2}}) are non-collinear, their coefficients must be equal on both sides of the equation. So, setting the coefficients equal gives us a system of equations:[2 = lambda k]and[1 = 3lambda]From the second equation, I can solve for (lambda):[1 = 3lambda implies lambda = frac{1}{3}]Now, plugging this value of (lambda) into the first equation:[2 = frac{1}{3}k implies k = 2 times 3 = 6]So, if (k = 6), then (overrightarrow{a}) and (overrightarrow{b}) are collinear. But we don't want them to be collinear because we need them to form a basis. Therefore, (k) must not be equal to 6. Since (k) is a real number, the only restriction is that (k neq 6). There are no other restrictions because any other real number would ensure that (overrightarrow{a}) and (overrightarrow{b}) are non-collinear, hence forming a valid basis for the plane. Let me just double-check my reasoning. If (k = 6), then (overrightarrow{b}) becomes (6overrightarrow{e_{1}} + 3overrightarrow{e_{2}}), which is exactly three times (overrightarrow{a}) (since (overrightarrow{a} = 2overrightarrow{e_{1}} + overrightarrow{e_{2}}), multiplying by 3 gives (6overrightarrow{e_{1}} + 3overrightarrow{e_{2}})). So, yes, they are scalar multiples of each other when (k = 6), making them collinear. Therefore, to ensure they are not collinear, (k) must be any real number except 6. I think that's solid. I don't see any mistakes in my calculations or reasoning. It all checks out.**Final Answer**The range of values for (k) is (boxed{k neq 6}).

💡First, I need to calculate the perimeter of the rectangular field. The formula for the perimeter of a rectangle is ( 2 times (text{length} + text{width}) ).Given the length is 75 meters and the width is 15 meters, I can substitute these values into the formula:[text{Perimeter} = 2 times (75 , text{m} + 15 , text{m}) = 2 times 90 , text{m} = 180 , text{m}]Next, since Sofie ran around the track 3 times, I need to multiply the perimeter by the number of laps:[text{Total Distance} = 180 , text{m} times 3 = 540 , text{m}]

💡Okay, so I have this problem about an arithmetic sequence. Let me try to figure it out step by step. First, I remember that in an arithmetic sequence, each term is obtained by adding a common difference to the previous term. So, if the first term is ( a_1 ) and the common difference is ( d ), then the nth term can be written as ( a_n = a_1 + (n-1)d ).The problem says that ( a_2 + a_3 = a_6 ). Hmm, let me write down what each of these terms is in terms of ( a_1 ) and ( d ). - ( a_2 = a_1 + d )- ( a_3 = a_1 + 2d )- ( a_6 = a_1 + 5d )So, substituting these into the equation ( a_2 + a_3 = a_6 ), we get:( (a_1 + d) + (a_1 + 2d) = a_1 + 5d )Let me simplify the left side first:( a_1 + d + a_1 + 2d = 2a_1 + 3d )So, the equation becomes:( 2a_1 + 3d = a_1 + 5d )Now, let me solve for ( a_1 ). Subtract ( a_1 ) from both sides:( a_1 + 3d = 5d )Then, subtract ( 3d ) from both sides:( a_1 = 2d )Okay, so the first term ( a_1 ) is equal to twice the common difference ( d ). That's useful information.Now, the problem asks for the value of ( frac{a_1 + a_2}{a_3 + a_4 + a_5} ). Let me write down each of these terms again using ( a_1 = 2d ):- ( a_1 = 2d )- ( a_2 = a_1 + d = 2d + d = 3d )- ( a_3 = a_1 + 2d = 2d + 2d = 4d )- ( a_4 = a_1 + 3d = 2d + 3d = 5d )- ( a_5 = a_1 + 4d = 2d + 4d = 6d )So, substituting these into the numerator and denominator:Numerator: ( a_1 + a_2 = 2d + 3d = 5d )Denominator: ( a_3 + a_4 + a_5 = 4d + 5d + 6d = 15d )Therefore, the fraction becomes:( frac{5d}{15d} )I can simplify this by dividing numerator and denominator by ( 5d ) (since ( d ) is non-zero, as given in the problem):( frac{5d}{15d} = frac{1}{3} )So, the value of the fraction is ( frac{1}{3} ).Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the beginning, I expressed each term correctly in terms of ( a_1 ) and ( d ). Then, I substituted into the given equation and solved for ( a_1 ) correctly, getting ( a_1 = 2d ). Then, I found each term again using this value and substituted into the numerator and denominator. The arithmetic seems correct: 2d + 3d is 5d, and 4d + 5d + 6d is 15d. Dividing 5d by 15d gives 1/3. I think that's correct. So, the value should be ( frac{1}{3} ).**Final Answer**The value is boxed{dfrac{1}{3}}.

💡First, I need to determine the total number of parts in the given ratio. The ratio is 3:3:4:5:6, so adding these together gives 3 + 3 + 4 + 5 + 6 = 21 parts.Next, I'll calculate the value of each part by dividing the total profit of 42,000 by the total number of parts, which is 21. This results in 2,000 per part.Finally, to find the largest share, I'll multiply the highest ratio value (6) by the value of each part (2,000), resulting in 12,000.

💡Alright, so I've got this problem about "sum-product equal number pairs." It sounds a bit complicated, but let me try to break it down. The definition says that two real numbers are called "sum-product equal number pairs" if their sum equals their product. In other words, if I have two numbers, say a and b, then a + b should be equal to a * b.Okay, part (1) asks me to determine whether two specific pairs are "sum-product equal number pairs." The pairs are (-2, 4) and (√2 + 2, √2). I need to check for each pair whether the sum equals the product.Let's start with the first pair: (-2, 4). I'll calculate the sum and the product separately and see if they're equal.Sum: -2 + 4. Hmm, that's straightforward. -2 plus 4 is 2.Product: -2 * 4. That's -8.Wait, 2 is not equal to -8. So, that means (-2, 4) is not a "sum-product equal number pair." Okay, that seems clear.Now, onto the second pair: (√2 + 2, √2). Again, I'll calculate the sum and the product.Sum: (√2 + 2) + √2. Let me simplify that. Combining like terms, √2 + √2 is 2√2, and then we have the +2. So, the sum is 2√2 + 2.Product: (√2 + 2) * √2. Let me distribute that. √2 * √2 is 2, and 2 * √2 is 2√2. So, adding those together, we get 2 + 2√2.Wait a minute, that's the same as the sum: 2√2 + 2. So, the sum and the product are equal. Therefore, (√2 + 2, √2) is a "sum-product equal number pair."Alright, part (1) seems done. Now, part (2) is a bit more involved. It says that if (m, n) is a "sum-product equal number pair" where m and n are not equal to 1, then express m as an algebraic expression containing n.So, given that m + n = m * n, and m, n ≠ 1, I need to solve for m in terms of n.Let me write down the equation: m + n = m * n.I want to isolate m. Let's see. Subtract m from both sides: n = m * n - m.Factor out m on the right side: n = m(n - 1).Now, to solve for m, divide both sides by (n - 1): m = n / (n - 1).So, m is equal to n divided by (n - 1). That should be the expression.Wait, let me double-check. If I substitute m = n / (n - 1) back into the original equation, does it hold?Let's see: m + n = (n / (n - 1)) + n.To add these, I need a common denominator. So, that's (n / (n - 1)) + (n(n - 1) / (n - 1)).Simplify the numerator: n + n(n - 1) = n + n^2 - n = n^2.So, the sum is n^2 / (n - 1).Now, the product: m * n = (n / (n - 1)) * n = n^2 / (n - 1).Yes, both the sum and the product are equal to n^2 / (n - 1). So, the expression m = n / (n - 1) is correct.I think that's it. So, m is equal to n divided by (n - 1).**Final Answer**(1) left(-2,4right) is not a "sum-product equal number pair" and (sqrt{2}+2,sqrt{2}) is a "sum-product equal number pair". (2) m = boxed{dfrac{n}{n - 1}}