💡Okay, so I've got this problem here about four natural numbers that add up to 243. The problem says that if we do some operations on each of these numbers—adding 8 to the first, subtracting 8 from the second, multiplying the third by 8, and dividing the fourth by 8—all four results will be equal. Then, I need to find the product of the largest and smallest of the original four numbers.Hmm, let's break this down step by step. First, let me define the four numbers as a, b, c, and d. So, according to the problem, a + b + c + d = 243. Got that part.Now, the tricky part is the transformations. If I add 8 to the first number, subtract 8 from the second, multiply the third by 8, and divide the fourth by 8, all these results are equal. Let's call this common result x. So, I can write four equations:1. a + 8 = x2. b - 8 = x3. c * 8 = x4. d / 8 = xAlright, so from these equations, I can express a, b, c, and d in terms of x:- a = x - 8- b = x + 8- c = x / 8- d = 8xNow, since I know that a + b + c + d = 243, I can substitute these expressions into that equation. Let's do that:(x - 8) + (x + 8) + (x / 8) + 8x = 243Let me simplify this step by step. First, combine like terms:x - 8 + x + 8 + x/8 + 8xLooking at the constants, -8 and +8 cancel each other out. So, I'm left with:x + x + x/8 + 8xNow, let's combine the x terms. I have x + x, which is 2x, plus 8x, which is 10x. So, altogether, it's 10x + x/8.So, the equation becomes:10x + (x / 8) = 243Hmm, to solve for x, I need to combine these terms. Let me find a common denominator. The common denominator for 10x and x/8 is 8. So, I'll convert 10x to 80x/8:(80x / 8) + (x / 8) = 243Now, combine the fractions:(80x + x) / 8 = 24381x / 8 = 243To solve for x, multiply both sides by 8:81x = 243 * 8Let me calculate 243 * 8. 243 * 8 is 1944.So, 81x = 1944Now, divide both sides by 81:x = 1944 / 81Let me compute that. 81 goes into 1944 how many times? 81 * 24 is 1944 because 80*24=1920 and 1*24=24, so 1920+24=1944. So, x = 24.Alright, so x is 24. Now, let's find the original numbers a, b, c, and d.Starting with a:a = x - 8 = 24 - 8 = 16Then, b:b = x + 8 = 24 + 8 = 32Next, c:c = x / 8 = 24 / 8 = 3And finally, d:d = 8x = 8 * 24 = 192So, the four original numbers are 16, 32, 3, and 192.Now, the problem asks for the product of the largest and smallest of these original four numbers. Let's identify the largest and smallest.Looking at the numbers: 3, 16, 32, 192. The smallest is 3, and the largest is 192.So, the product is 3 * 192. Let me calculate that. 3 * 192 is 576.Therefore, the product of the largest and smallest original numbers is 576.**Final Answer**The product of the largest and smallest original numbers is boxed{576}.

💡Okay, so I have this problem here: If ( c ) and ( d ) are integers with ( c > d ), what is the smallest possible positive value of ( frac{c+2d}{c-d} + frac{c-d}{c+2d} )? Hmm, interesting. Let me try to figure this out step by step.First, I notice that the expression is the sum of two fractions: ( frac{c+2d}{c-d} ) and ( frac{c-d}{c+2d} ). It might be helpful to simplify this expression or find a way to make it easier to work with. Maybe I can set one of these fractions equal to a variable to make things simpler.Let me let ( y = frac{c+2d}{c-d} ). Then, what is ( frac{c-d}{c+2d} )? Well, that's just the reciprocal of ( y ), so ( frac{c-d}{c+2d} = frac{1}{y} ). So, the original expression becomes ( y + frac{1}{y} ). That simplifies things a bit.So now, I need to find the smallest possible positive value of ( y + frac{1}{y} ). Hmm, I remember that for positive real numbers, the expression ( y + frac{1}{y} ) has a minimum value when ( y = 1 ), which gives ( 2 ). But wait, ( y ) isn't just any real number here; it's defined in terms of integers ( c ) and ( d ) with ( c > d ). So, I need to check if ( y = 1 ) is achievable with integer values of ( c ) and ( d ).If ( y = 1 ), then ( frac{c+2d}{c-d} = 1 ). Let's solve this equation for ( c ) and ( d ). Multiplying both sides by ( c - d ), we get ( c + 2d = c - d ). Subtracting ( c ) from both sides, we have ( 2d = -d ), which simplifies to ( 3d = 0 ). Therefore, ( d = 0 ).So, if ( d = 0 ), then ( c ) can be any integer greater than ( d ), which is 0. Let's pick ( c = 1 ) for simplicity. Then, substituting back into the original expression, we have ( frac{1 + 2(0)}{1 - 0} + frac{1 - 0}{1 + 2(0)} = frac{1}{1} + frac{1}{1} = 1 + 1 = 2 ). So, that works! It gives us the value 2.But wait, is 2 the smallest possible positive value? Let me think. If I can find another pair of integers ( c ) and ( d ) where ( y + frac{1}{y} ) is less than 2, then 2 wouldn't be the minimum. But from what I remember, ( y + frac{1}{y} ) is minimized at ( y = 1 ), giving 2, and for any other positive ( y ), the value is larger. So, unless there's a way to make ( y ) negative, which might complicate things, 2 should be the minimum.But hold on, ( c ) and ( d ) are integers with ( c > d ). If ( d ) is negative, could that make ( y ) negative? Let's explore that. Suppose ( d ) is negative. Let me pick some values.Let's say ( c = 2 ) and ( d = -1 ). Then, ( c - d = 2 - (-1) = 3 ), and ( c + 2d = 2 + 2(-1) = 0 ). Oh, wait, that would make the first fraction ( frac{0}{3} = 0 ), and the second fraction ( frac{3}{0} ), which is undefined. So, that doesn't work.How about ( c = 3 ) and ( d = -1 )? Then, ( c - d = 3 - (-1) = 4 ), and ( c + 2d = 3 + 2(-1) = 1 ). So, the first fraction is ( frac{1}{4} ), and the second fraction is ( frac{4}{1} = 4 ). Adding them together, we get ( frac{1}{4} + 4 = frac{17}{4} = 4.25 ), which is larger than 2.What if ( c = 2 ) and ( d = -2 )? Then, ( c - d = 2 - (-2) = 4 ), and ( c + 2d = 2 + 2(-2) = -2 ). So, the first fraction is ( frac{-2}{4} = -0.5 ), and the second fraction is ( frac{4}{-2} = -2 ). Adding them together, we get ( -0.5 + (-2) = -2.5 ). But the problem asks for the smallest possible positive value, so negative results don't count.Hmm, maybe trying ( d ) negative isn't helpful because either the fractions become undefined or the sum is negative or larger than 2. Let me try another approach.Since ( y = frac{c+2d}{c-d} ), maybe I can express this in terms of another variable. Let me let ( k = c - d ). Then, ( c = k + d ). Substituting back into ( y ), we get:( y = frac{(k + d) + 2d}{k} = frac{k + 3d}{k} = 1 + frac{3d}{k} ).So, ( y = 1 + frac{3d}{k} ). Therefore, ( frac{1}{y} = frac{1}{1 + frac{3d}{k}} ).Hmm, not sure if that helps directly. Maybe I can think about the expression ( y + frac{1}{y} ) in terms of ( k ) and ( d ).Alternatively, let's consider that ( c ) and ( d ) are integers, so ( c - d ) and ( c + 2d ) must also be integers. Let me denote ( a = c - d ) and ( b = c + 2d ). Then, our expression becomes ( frac{b}{a} + frac{a}{b} ).So, we need to find integers ( a ) and ( b ) such that ( a ) and ( b ) are positive (since ( c > d ), ( a = c - d > 0 ), and ( b = c + 2d ) must also be positive because otherwise, the fractions could be negative or undefined). Wait, is ( b ) necessarily positive?Let me check. If ( c + 2d ) is positive, then both ( a ) and ( b ) are positive. If ( c + 2d ) is negative, then ( frac{b}{a} ) is negative, and ( frac{a}{b} ) is also negative, so their sum would be negative. But we're looking for the smallest positive value, so we can focus on cases where both ( a ) and ( b ) are positive.Therefore, ( a = c - d > 0 ) and ( b = c + 2d > 0 ). So, both ( a ) and ( b ) are positive integers.So, our problem reduces to finding positive integers ( a ) and ( b ) such that ( frac{b}{a} + frac{a}{b} ) is minimized.Wait, but ( a ) and ( b ) are related through ( c ) and ( d ). Specifically, ( a = c - d ) and ( b = c + 2d ). Let me see if I can express ( a ) and ( b ) in terms of each other.From ( a = c - d ) and ( b = c + 2d ), I can solve for ( c ) and ( d ) in terms of ( a ) and ( b ).Let me subtract the first equation from the second:( b - a = (c + 2d) - (c - d) = 3d ).So, ( d = frac{b - a}{3} ).Similarly, adding ( 2a ) to ( b ):( b + 2a = (c + 2d) + 2(c - d) = 3c ).So, ( c = frac{b + 2a}{3} ).Since ( c ) and ( d ) must be integers, ( b - a ) must be divisible by 3, and ( b + 2a ) must also be divisible by 3.Therefore, ( b equiv a mod 3 ), because ( b - a ) is divisible by 3, so ( b equiv a mod 3 ).Similarly, ( b + 2a equiv 0 mod 3 ). Since ( b equiv a mod 3 ), substituting, we get ( a + 2a = 3a equiv 0 mod 3 ), which is always true. So, the only condition is that ( b equiv a mod 3 ).Therefore, ( a ) and ( b ) must be congruent modulo 3.So, our problem now is to find positive integers ( a ) and ( b ) such that ( b equiv a mod 3 ), and ( frac{b}{a} + frac{a}{b} ) is minimized.Hmm, okay. Let's denote ( k = frac{b}{a} ). Then, ( k ) is a positive rational number, and our expression becomes ( k + frac{1}{k} ).We know that for real numbers, the minimum of ( k + frac{1}{k} ) is 2 when ( k = 1 ). But since ( k ) must be a rational number such that ( b ) and ( a ) are integers with ( b equiv a mod 3 ), we need to see if ( k = 1 ) is achievable.If ( k = 1 ), then ( b = a ). So, ( b = a ). Then, ( b equiv a mod 3 ) is automatically satisfied because ( b = a ). So, that works.Therefore, the minimal value of ( k + frac{1}{k} ) is 2, achieved when ( k = 1 ), which corresponds to ( b = a ).So, going back, ( b = a ) implies ( c + 2d = c - d ), which simplifies to ( 3d = 0 ), so ( d = 0 ). Then, ( c ) can be any integer greater than ( d ), which is 0. So, ( c ) is a positive integer.Therefore, substituting ( d = 0 ) into the original expression, we get ( frac{c + 0}{c - 0} + frac{c - 0}{c + 0} = 1 + 1 = 2 ).So, the minimal positive value is 2.But wait, let me check if there are any other possible values of ( a ) and ( b ) that could give a smaller positive value. Since ( k + frac{1}{k} ) is minimized at 2, and any other value of ( k ) would give a larger result, I don't think there's a way to get a smaller positive value.Just to be thorough, let me try some small integer values for ( a ) and ( b ) that satisfy ( b equiv a mod 3 ) and see what the expression evaluates to.Let's try ( a = 1 ), ( b = 1 ). Then, the expression is ( 1 + 1 = 2 ).How about ( a = 2 ), ( b = 2 ). Then, ( 2 + frac{1}{2} = 2.5 ), which is larger than 2.What if ( a = 1 ), ( b = 4 )? Wait, does ( 4 equiv 1 mod 3 )? Yes, because ( 4 - 1 = 3 ), which is divisible by 3. So, ( a = 1 ), ( b = 4 ). Then, the expression is ( 4 + frac{1}{4} = 4.25 ), which is larger.How about ( a = 2 ), ( b = 5 )? ( 5 equiv 2 mod 3 ) because ( 5 - 2 = 3 ). So, ( frac{5}{2} + frac{2}{5} = 2.5 + 0.4 = 2.9 ), which is still larger than 2.What if ( a = 3 ), ( b = 3 )? Then, ( 3 + frac{1}{3} approx 3.333 ), which is larger.Alternatively, ( a = 3 ), ( b = 6 ). Then, ( frac{6}{3} + frac{3}{6} = 2 + 0.5 = 2.5 ), still larger.Wait, but in all these cases, the minimal value seems to be 2. So, I think it's safe to conclude that the smallest possible positive value is indeed 2.Just to double-check, let me pick another pair where ( d ) is not zero. Let's say ( c = 2 ), ( d = 1 ). Then, ( c - d = 1 ), ( c + 2d = 4 ). So, the expression is ( frac{4}{1} + frac{1}{4} = 4 + 0.25 = 4.25 ), which is larger than 2.Another example: ( c = 3 ), ( d = 1 ). Then, ( c - d = 2 ), ( c + 2d = 5 ). So, ( frac{5}{2} + frac{2}{5} = 2.5 + 0.4 = 2.9 ), still larger.What about ( c = 4 ), ( d = 1 )? ( c - d = 3 ), ( c + 2d = 6 ). So, ( frac{6}{3} + frac{3}{6} = 2 + 0.5 = 2.5 ), again larger.It seems consistent that 2 is the minimal value, achieved when ( d = 0 ) and ( c ) is any positive integer.Therefore, after all this exploration and checking, I'm confident that the smallest possible positive value of the given expression is 2.

💡Okay, so Petya has these 12 identical multicolored train cars, and he thinks he can make more different 12-car trains than 11-car trains. Hmm, that's interesting. I need to figure out if he's right or not.First, let me make sure I understand the problem correctly. We have 12 train cars, and some of them might be the same color. The exact number of cars of each color isn't specified, so it could be any combination. Petya believes that the number of unique 12-car trains he can make is greater than the number of unique 11-car trains. I need to determine if this is true.Alright, so when we talk about trains being different, it's based on the sequence of colors. If two trains have the same colors in the same order, they're considered the same. So, for example, if you have two red cars followed by a blue car, that's different from a blue car followed by two red cars.Now, Petya is comparing 12-car trains to 11-car trains. Intuitively, I might think that having more cars would allow for more combinations, but maybe that's not the case here because we're dealing with sequences where the order matters.Let me break it down. If I have 12 cars, the number of possible sequences is based on the number of ways to arrange these cars considering their colors. Since some cars might be the same color, the number of unique sequences would be less than if all cars were different colors.Similarly, for 11-car trains, we're looking at sequences of 11 cars, again considering the colors. The question is, does the number of unique 12-car sequences exceed the number of unique 11-car sequences?Maybe I can think about it in terms of permutations. If all cars were different, the number of 12-car sequences would be 12 factorial, which is a huge number. The number of 11-car sequences would be 12 choose 11 times 11 factorial, which is also a large number but slightly less than 12 factorial. So in that case, 12-car sequences would be more.But in our case, cars can be the same color, so the number of unique sequences is reduced because swapping identical cars doesn't create a new sequence. So the exact number depends on how many cars of each color there are.Wait, but Petya doesn't know how many cars of each color there are. He just knows there are 12 cars, some may be the same color. So we need to consider the worst-case scenario or maybe the general case.Let me think about an example. Suppose all 12 cars are the same color. Then, there's only one unique 12-car train and only one unique 11-car train. So in this case, the number of 12-car trains is equal to the number of 11-car trains.On the other hand, if all 12 cars are different colors, then the number of 12-car trains is 12 factorial, and the number of 11-car trains is 12 choose 11 times 11 factorial, which is 12 times 11 factorial. Since 12 factorial is 12 times 11 factorial, they are equal in this case as well.Hmm, interesting. So in both extremes—when all cars are the same color and when all cars are different colors—the number of 12-car trains equals the number of 11-car trains.What about a case where there are some repeated colors but not all? Let's say we have two colors: 6 red and 6 blue cars. How many unique 12-car trains can we make? It would be the number of distinct permutations of 6 red and 6 blue cars, which is 12 choose 6, equal to 924.For 11-car trains, we would need to choose 11 cars out of 12, which could be either 5 red and 6 blue or 6 red and 5 blue. The number of unique 11-car trains would be the sum of the permutations for each case. So, for 5 red and 6 blue, it's 11 choose 5, which is 462, and for 6 red and 5 blue, it's also 462. So total unique 11-car trains would be 462 + 462 = 924.Wait, that's the same as the number of 12-car trains. So again, they are equal.Is this a coincidence? Let me try another example. Suppose we have 3 colors: 4 red, 4 blue, and 4 green cars. The number of unique 12-car trains would be 12! divided by (4!4!4!), which is a large number.For 11-car trains, we would need to remove one car of any color. So, we would have either 3 red, 4 blue, and 4 green; 4 red, 3 blue, and 4 green; or 4 red, 4 blue, and 3 green. The number of unique 11-car trains would be the sum of the permutations for each case:- For 3 red, 4 blue, 4 green: 11! / (3!4!4!) = 330,330- For 4 red, 3 blue, 4 green: 11! / (4!3!4!) = 330,330- For 4 red, 4 blue, 3 green: 11! / (4!4!3!) = 330,330Total unique 11-car trains: 330,330 * 3 = 990,990Now, the number of unique 12-car trains is 12! / (4!4!4!) = 34,650Wait, that's way less than 990,990. So in this case, the number of 11-car trains is actually greater than the number of 12-car trains.Hmm, that's different from the previous examples. So in this case, Petya would be incorrect because there are more 11-car trains than 12-car trains.But wait, in the previous examples, when all cars were the same color or all different, the numbers were equal. When there was a mix, sometimes 12-car trains were equal or less.So maybe Petya's belief depends on the distribution of colors among the cars.But the problem states that Petya has 12 identical multicolored train cars, but it's unknown how many cars of each color there are. So we need to consider the general case.From the examples, it seems that depending on the color distribution, the number of 12-car trains can be equal to or less than the number of 11-car trains, but never more.Wait, in the first example with 6 red and 6 blue, the numbers were equal. In the second example with 4 red, 4 blue, and 4 green, the 11-car trains were more.Is there a case where 12-car trains are more than 11-car trains?Let me think. Suppose we have 11 cars of one color and 1 car of another color. So, 11 red and 1 blue.Number of unique 12-car trains: Since there are 11 red and 1 blue, the number of unique sequences is 12 (the position of the blue car).For 11-car trains, we need to remove one car. We can either remove a red car or the blue car.- If we remove a red car: We have 10 red and 1 blue. The number of unique sequences is 11 (the position of the blue car).- If we remove the blue car: We have 11 red cars, which is only 1 unique sequence.So total unique 11-car trains: 11 + 1 = 12.So, number of 12-car trains is 12, and number of 11-car trains is also 12. They are equal.Another example: 10 red, 1 blue, 1 green.Number of unique 12-car trains: 12! / (10!1!1!) = 66.For 11-car trains, we can remove one red, one blue, or one green.- Remove one red: 9 red, 1 blue, 1 green. Number of sequences: 11! / (9!1!1!) = 55.- Remove one blue: 10 red, 0 blue, 1 green. Number of sequences: 11 (position of green).- Remove one green: 10 red, 1 blue, 0 green. Number of sequences: 11 (position of blue).Total unique 11-car trains: 55 + 11 + 11 = 77.So, 12-car trains: 66, 11-car trains: 77. So again, 11-car trains are more.Wait, so in all these cases, 11-car trains are either equal or more than 12-car trains. Is there any scenario where 12-car trains are more?Let me think differently. Suppose we have 12 different colors, so all cars are unique. Then, number of 12-car trains is 12!, and number of 11-car trains is 12 * 11! = 12!.So they are equal.If we have 11 unique colors and 1 duplicate, say 2 red and 10 unique colors.Number of 12-car trains: 12! / 2! = 667,920.Number of 11-car trains: We can remove one red or one unique color.- Remove one red: 11 cars with 1 red and 10 unique. Number of sequences: 11! / 1! = 39,916,800.- Remove one unique color: 12 cars minus one unique, so 11 cars with 2 red and 9 unique. Number of sequences: 11! / 2! = 19,958,400.Total unique 11-car trains: 39,916,800 + 19,958,400 = 59,875,200.Compare to 12-car trains: 667,920.So, 11-car trains are way more.Wait, but in this case, 12-car trains are much less than 11-car trains.So, in all the cases I've considered, 12-car trains are either equal or less than 11-car trains.Is there any case where 12-car trains are more?Let me try with more duplicates.Suppose we have 6 pairs of colors: 2 red, 2 blue, 2 green, 2 yellow, 2 black, 2 white.Number of 12-car trains: 12! / (2!^6) = 12! / 64 = 199,584,000 / 64 = 3,118,500.For 11-car trains, we need to remove one car. Since all colors have 2 cars, removing one will leave us with 1 car of one color and 2 of the others.Number of unique 11-car trains: For each color, we can remove one, so 6 possibilities. For each, the number of sequences is 11! / (2!^5 * 1!) = 11! / (32) = 39,916,800 / 32 = 1,247,400.Total unique 11-car trains: 6 * 1,247,400 = 7,484,400.Compare to 12-car trains: 3,118,500.So, 11-car trains are more.Hmm, again, 11-car trains are more.Wait, is there a mathematical reason behind this?Let me think about the general case.Suppose we have n cars, and we want to compare the number of unique n-car trains to (n-1)-car trains.If we denote the number of unique n-car trains as T(n), and the number of unique (n-1)-car trains as T(n-1), we want to compare T(n) and T(n-1).From the examples, it seems that T(n) ≤ T(n-1).Is this always true?Let me think about how T(n) relates to T(n-1).To form a unique n-car train, you can think of it as adding one car to an (n-1)-car train. However, because of possible duplicates, adding a car might not always lead to a new unique train.But in terms of counting, the number of unique n-car trains is equal to the number of unique (n-1)-car trains multiplied by the number of possible ways to add a car, divided by the symmetries due to identical cars.Wait, maybe it's better to think in terms of generating functions or multinomial coefficients.The number of unique n-car trains is given by the multinomial coefficient:T(n) = n! / (k1! * k2! * ... * km!),where ki is the number of cars of each color.For T(n-1), we have:T(n-1) = (n-1)! / ( (k1 - a1)! * (k2 - a2)! * ... * (km - am)! ),where ai is 0 or 1, indicating whether we removed a car of color i.But this seems complicated.Alternatively, consider that T(n-1) can be obtained by removing one car from T(n). Since removing a car can be done in n ways, but some removals might lead to the same (n-1)-car train.Therefore, T(n-1) ≥ T(n) / n.But since T(n) is an integer, T(n-1) ≥ ceil(T(n) / n).But this doesn't directly help.Wait, maybe it's better to think about the ratio T(n) / T(n-1).If T(n) > T(n-1), then the ratio is greater than 1.But from the examples, it seems that T(n) ≤ T(n-1).Is there a general proof for this?Let me consider that T(n) is the number of distinct permutations of n cars with given color counts.T(n-1) is the number of distinct permutations of n-1 cars, which can be obtained by removing one car from T(n).However, each T(n-1) can be obtained in multiple ways by removing different cars from T(n).Therefore, T(n-1) ≥ T(n) / n.But since T(n) and T(n-1) are integers, T(n-1) ≥ ceil(T(n) / n).But this doesn't necessarily mean T(n-1) ≥ T(n). It just gives a lower bound.Wait, maybe I need to think about it differently.Suppose we have a set of n cars with color counts k1, k2, ..., km.Then, the number of unique n-car trains is T(n) = n! / (k1! k2! ... km!).The number of unique (n-1)-car trains is the sum over each color of T(n) * (ki / n).Because for each color, the probability of removing a car of that color is ki / n, and the number of unique trains after removing one car of color i is T(n) * (ki / n).Wait, that might not be accurate.Alternatively, the number of unique (n-1)-car trains is equal to the sum over each color of the number of unique trains obtained by removing one car of that color.For each color i, the number of unique (n-1)-car trains obtained by removing one car of color i is T(n) * (ki / n).Therefore, T(n-1) = sum_{i=1 to m} [ T(n) * (ki / n) ].But since sum_{i=1 to m} ki = n, we have T(n-1) = T(n) * sum_{i=1 to m} (ki / n) = T(n) * (n / n) = T(n).Wait, that suggests T(n-1) = T(n), which contradicts our earlier examples.Hmm, maybe this approach is flawed.Let me think again.Actually, when you remove a car of color i, the number of unique (n-1)-car trains is T(n) * (ki / n) only if all the cars are distinguishable, which they are not because of duplicates.So, this approach might not work.Perhaps a better way is to consider that T(n-1) is at least T(n) because each T(n) can generate multiple T(n-1) by removing different cars, but due to duplicates, some removals might lead to the same T(n-1).Therefore, T(n-1) ≥ T(n).Wait, but in our earlier examples, sometimes T(n-1) was equal to T(n), and sometimes it was greater.So, in general, T(n-1) ≥ T(n).Therefore, Petya's belief that T(12) > T(11) is incorrect.Hence, Petya is wrong.

💡Okay, so I have this problem here: When the greatest common divisor (gcd) and least common multiple (lcm) of two integers are multiplied, the product is 240. I need to find how many different values could be the greatest common divisor of the two integers. Hmm, that sounds interesting. Let me try to figure this out step by step.First, I remember that there's a relationship between gcd and lcm of two numbers. Specifically, the product of the gcd and lcm of two numbers equals the product of the numbers themselves. So, if we have two integers a and b, then:gcd(a, b) * lcm(a, b) = a * bSo, in this problem, it's given that gcd(a, b) * lcm(a, b) = 240. That means a * b = 240. Okay, so the product of the two numbers is 240. Got it.Now, I need to find how many different values could be the gcd of a and b. So, essentially, I need to find all possible values of gcd(a, b) where a and b are positive integers such that a * b = 240.To approach this, I think I should factorize 240 into its prime factors. Let me do that:240 can be divided by 2: 240 / 2 = 120120 divided by 2 is 6060 divided by 2 is 3030 divided by 2 is 1515 divided by 3 is 55 divided by 5 is 1So, the prime factorization of 240 is 2^4 * 3^1 * 5^1. Okay, so 240 = 2^4 * 3 * 5.Now, since a and b are two numbers whose product is 240, I can represent them in terms of their prime factors. Let me denote:a = 2^p * 3^q * 5^rb = 2^s * 3^t * 5^uWhere p, q, r, s, t, u are non-negative integers. Then, the product a * b would be:a * b = (2^p * 3^q * 5^r) * (2^s * 3^t * 5^u) = 2^(p+s) * 3^(q+t) * 5^(r+u)But we know that a * b = 240 = 2^4 * 3^1 * 5^1. Therefore, we can set up the following equations:p + s = 4q + t = 1r + u = 1Okay, so now I have these equations for the exponents. Now, the gcd of a and b is given by taking the minimum exponents for each prime factor. That is:gcd(a, b) = 2^min(p, s) * 3^min(q, t) * 5^min(r, u)Similarly, the lcm is given by taking the maximum exponents:lcm(a, b) = 2^max(p, s) * 3^max(q, t) * 5^max(r, u)But since we already know that gcd(a, b) * lcm(a, b) = 240, which is equal to a * b, maybe I can focus on finding all possible gcds given the exponents.So, let me first consider the exponents for each prime factor separately.Starting with the prime 2: p + s = 4. The possible pairs (p, s) are (0,4), (1,3), (2,2), (3,1), (4,0). For each of these pairs, the minimum of p and s will be the exponent for 2 in the gcd. So, let's list the min(p, s) for each pair:- (0,4): min(0,4) = 0- (1,3): min(1,3) = 1- (2,2): min(2,2) = 2- (3,1): min(3,1) = 1- (4,0): min(4,0) = 0So, the possible exponents for 2 in the gcd are 0, 1, and 2.Next, for the prime 3: q + t = 1. The possible pairs (q, t) are (0,1) and (1,0). The minimum of q and t will be the exponent for 3 in the gcd:- (0,1): min(0,1) = 0- (1,0): min(1,0) = 0So, the only possible exponent for 3 in the gcd is 0.Similarly, for the prime 5: r + u = 1. The possible pairs (r, u) are (0,1) and (1,0). The minimum of r and u will be the exponent for 5 in the gcd:- (0,1): min(0,1) = 0- (1,0): min(1,0) = 0So, the only possible exponent for 5 in the gcd is 0.Putting this all together, the gcd(a, b) can be expressed as:gcd(a, b) = 2^k * 3^0 * 5^0 = 2^kWhere k can be 0, 1, or 2. Therefore, the possible gcds are:- 2^0 = 1- 2^1 = 2- 2^2 = 4Wait, that gives me three possible gcds: 1, 2, and 4. But hold on, I think I might have missed something because the problem is asking for how many different values could be the gcd, and I only have three here. But I recall that sometimes when dealing with multiple primes, the gcd can have combinations of different primes, but in this case, since the exponents for 3 and 5 in the gcd are fixed at 0, the gcd can only be a power of 2.But let me double-check. The exponents for 3 and 5 in the gcd are both 0 because in both cases, the minimum of (q, t) and (r, u) is 0. Therefore, the gcd cannot have any factors of 3 or 5. So, indeed, the gcd must be a power of 2, specifically 2^k where k is 0, 1, or 2.Therefore, the possible gcds are 1, 2, and 4. So, there are 3 different possible values for the gcd.Wait, but in the initial problem statement, it's mentioned that the product of gcd and lcm is 240. So, if the gcd is 1, then the lcm would be 240. If the gcd is 2, then the lcm would be 120. If the gcd is 4, then the lcm would be 60. Let me verify if these are valid.For gcd = 1: Then a and b are coprime, so their lcm is 240. So, possible pairs (a, b) could be (1, 240), (3, 80), (5, 48), etc., as long as they multiply to 240 and are coprime. That seems valid.For gcd = 2: Then a = 2 * m and b = 2 * n, where m and n are coprime. Then, a * b = 4 * m * n = 240, so m * n = 60. The lcm(a, b) would be 2 * lcm(m, n) = 2 * m * n = 120. So, yes, that works.For gcd = 4: Then a = 4 * m and b = 4 * n, where m and n are coprime. Then, a * b = 16 * m * n = 240, so m * n = 15. The lcm(a, b) would be 4 * lcm(m, n) = 4 * m * n = 60. That also works.So, all three gcd values are possible. Therefore, the number of different possible values for the gcd is 3.Wait, but earlier, when I thought about the exponents, I considered the exponents for 2, 3, and 5 separately. But since the exponents for 3 and 5 in the gcd are fixed at 0, the gcd can only vary based on the exponent of 2. So, that gives us 3 different values.But hold on, let me think again. Is there a possibility that the gcd could include other primes? For example, could the gcd be 3 or 5? Let me check.If the gcd were 3, then both a and b would have to be multiples of 3. So, a = 3 * m and b = 3 * n, where m and n are integers. Then, a * b = 9 * m * n = 240. So, m * n = 240 / 9 ≈ 26.666, which is not an integer. Therefore, m and n cannot be integers in this case. So, gcd cannot be 3.Similarly, if the gcd were 5, then a = 5 * m and b = 5 * n, so a * b = 25 * m * n = 240. Then, m * n = 240 / 25 = 9.6, which is also not an integer. Therefore, gcd cannot be 5 either.What about gcd = 6? That would mean both a and b are multiples of 6. So, a = 6 * m and b = 6 * n, with m and n coprime. Then, a * b = 36 * m * n = 240. So, m * n = 240 / 36 ≈ 6.666, which is not an integer. Therefore, gcd cannot be 6.Similarly, gcd = 10: a = 10 * m, b = 10 * n, so a * b = 100 * m * n = 240. Then, m * n = 2.4, which is not an integer. So, gcd cannot be 10.Gcd = 12: a = 12 * m, b = 12 * n, so a * b = 144 * m * n = 240. Then, m * n = 240 / 144 ≈ 1.666, not an integer. So, gcd cannot be 12.Gcd = 20: a = 20 * m, b = 20 * n, so a * b = 400 * m * n = 240. Then, m * n = 0.6, which is not possible since m and n are positive integers. So, gcd cannot be 20.Wait, so actually, the only possible gcds are 1, 2, and 4. So, that would mean there are 3 different possible values for the gcd.But earlier, when I considered the exponents, I thought that the gcd could be 1, 2, 4, 3, 6, 12, 5, 10, 20, which would give 9 different values. But now, when I check, those higher gcds like 3, 5, 6, etc., don't actually work because they result in non-integer products for m * n. So, perhaps my initial approach was wrong.Wait, why did I think there were 9 possible gcds? Let me go back to that.I had considered that the exponents for 2 in the gcd could be 0, 1, or 2, and for 3 and 5, they could be 0 or 1. So, in total, 3 * 2 * 2 = 12 possibilities. But then, I realized that for 3 and 5, the exponents in the gcd have to be 0 because the minimum of (q, t) and (r, u) is 0. So, actually, the exponents for 3 and 5 in the gcd must be 0, leaving only the exponent for 2 to vary. Therefore, the gcd can only be 2^0, 2^1, or 2^2, which are 1, 2, and 4.So, that would mean there are only 3 possible gcds. But in my initial thought process, I thought there were 9, but that was incorrect because I didn't consider that the exponents for 3 and 5 in the gcd must be 0.Wait, but let me think again. Maybe I was overcomplicating it. Since the exponents for 3 and 5 in the gcd are fixed at 0, the gcd can only be a power of 2. Therefore, the possible gcds are 1, 2, and 4, which are 3 different values.But hold on, let me check with specific examples. For gcd = 1, possible pairs are (1, 240), (3, 80), (5, 48), (15, 16), etc. For gcd = 2, possible pairs are (2, 120), (4, 60), (6, 40), (10, 24), (12, 20), etc. For gcd = 4, possible pairs are (4, 60), (8, 30), (12, 20), etc. Wait, but some of these pairs might overlap or have different gcds.Wait, for example, take the pair (12, 20). The gcd of 12 and 20 is 4, and their lcm is 60. So, 4 * 60 = 240, which works. Similarly, take (6, 40). The gcd is 2, and the lcm is 120, so 2 * 120 = 240. That works too.But if I try to find a pair with gcd = 3, like (3, 80). The gcd is 1, not 3, because 3 and 80 are coprime. Similarly, (5, 48) has gcd 1. So, actually, there are no pairs with gcd = 3 or 5 because when you factor out the gcd, the remaining factors must multiply to 240 divided by gcd squared, which isn't an integer for gcd = 3 or 5.Therefore, the only possible gcds are 1, 2, and 4. So, the number of different possible values for the gcd is 3.Wait, but in my initial thought process, I thought there were 9 possible gcds, but that was incorrect because I didn't properly consider the constraints on the exponents for 3 and 5. So, the correct answer should be 3.But let me double-check with another approach. Since a * b = 240, and gcd(a, b) = d, then we can write a = d * m and b = d * n, where m and n are coprime integers. Then, a * b = d^2 * m * n = 240. Therefore, d^2 must divide 240.So, let's find all d such that d^2 divides 240.First, factorize 240: 240 = 2^4 * 3^1 * 5^1.For d^2 to divide 240, the exponents in d^2 must be less than or equal to those in 240.Let d = 2^x * 3^y * 5^z.Then, d^2 = 2^(2x) * 3^(2y) * 5^(2z).For d^2 to divide 240, we must have:2x ≤ 4 ⇒ x ≤ 22y ≤ 1 ⇒ y ≤ 0.5 ⇒ y = 02z ≤ 1 ⇒ z ≤ 0.5 ⇒ z = 0Therefore, x can be 0, 1, or 2; y and z must be 0.Thus, d can be 2^0 = 1, 2^1 = 2, or 2^2 = 4.So, the possible values of d are 1, 2, and 4. Therefore, there are 3 different possible values for the gcd.This confirms my earlier conclusion. So, the number of different possible values for the gcd is 3.**Final Answer**The number of different possible values for the greatest common divisor is boxed{3}.

💡Alright, so I've got this problem here: I need to find the value of ( x ) given by the expression ( x = left(log_{frac{1}{2}} frac{1}{3}right)^{-2} + left(log_{frac{1}{3}} frac{1}{3}right)^{-1} ), and then determine which interval ( x ) falls into. The options are A. (-2, -1), B. (1, 2), C. (-3, -2), and D. (2, 3). Hmm, okay, let's break this down step by step.First, I remember that logarithms can sometimes be tricky, especially when dealing with bases that are fractions. I think it might help to recall some logarithm properties. One key property is that ( log_b a = frac{ln a}{ln b} ), which could be useful here. Also, I remember that ( log_{frac{1}{b}} a = -log_b a ) because ( frac{1}{b} ) is the reciprocal of ( b ), and that flips the sign of the logarithm.Let me tackle each term separately. Starting with the first term: ( left(log_{frac{1}{2}} frac{1}{3}right)^{-2} ). Using the property I just mentioned, ( log_{frac{1}{2}} frac{1}{3} = -log_2 frac{1}{3} ). Wait, but ( log_2 frac{1}{3} ) is the same as ( -log_2 3 ) because ( frac{1}{3} = 3^{-1} ), and logarithm of a reciprocal is the negative logarithm. So putting it all together, ( log_{frac{1}{2}} frac{1}{3} = -(-log_2 3) = log_2 3 ).But hold on, I think I made a mistake there. Let me double-check. If ( log_{frac{1}{2}} frac{1}{3} = -log_2 frac{1}{3} ), and ( log_2 frac{1}{3} = -log_2 3 ), then substituting back, it's ( -(-log_2 3) = log_2 3 ). Yeah, that seems right. So ( log_{frac{1}{2}} frac{1}{3} = log_2 3 ).Now, since we have ( left(log_{frac{1}{2}} frac{1}{3}right)^{-2} ), which is ( (log_2 3)^{-2} ). That's the same as ( frac{1}{(log_2 3)^2} ). Okay, so that's the first term.Moving on to the second term: ( left(log_{frac{1}{3}} frac{1}{3}right)^{-1} ). Again, using the same property, ( log_{frac{1}{3}} frac{1}{3} = -log_3 frac{1}{3} ). But ( log_3 frac{1}{3} = log_3 3^{-1} = -1 ). So, ( log_{frac{1}{3}} frac{1}{3} = -(-1) = 1 ). Therefore, ( left(log_{frac{1}{3}} frac{1}{3}right)^{-1} = 1^{-1} = 1 ).So now, putting it all together, ( x = frac{1}{(log_2 3)^2} + 1 ). Hmm, okay. I need to figure out the value of ( log_2 3 ) to approximate ( x ).I recall that ( log_2 3 ) is approximately 1.58496. Let me verify that. Since ( 2^1 = 2 ) and ( 2^{1.58496} approx 3 ). Yeah, that's correct. So ( log_2 3 approx 1.58496 ).Therefore, ( (log_2 3)^2 approx (1.58496)^2 approx 2.51189 ). So, ( frac{1}{(log_2 3)^2} approx frac{1}{2.51189} approx 0.398 ).Adding 1 to that, ( x approx 0.398 + 1 = 1.398 ). So, approximately 1.4.Looking back at the options: A. (-2, -1), B. (1, 2), C. (-3, -2), D. (2, 3). So, 1.4 falls into interval B. (1, 2).Wait, but let me make sure I didn't make any mistakes in my calculations. Let's go through it again.First term: ( log_{frac{1}{2}} frac{1}{3} ). As I did before, this is ( log_2 3 ). So, ( (log_2 3)^{-2} ) is ( 1/(log_2 3)^2 ). That's approximately ( 1/(1.58496)^2 approx 1/2.51189 approx 0.398 ).Second term: ( log_{frac{1}{3}} frac{1}{3} ). That's 1, as I found earlier, so ( 1^{-1} = 1 ).Adding them together: 0.398 + 1 = 1.398, which is approximately 1.4. So, yes, that's between 1 and 2.But just to be thorough, let me consider if there's another way to approach this without approximating ( log_2 3 ). Maybe using inequalities.I know that ( 2^1 = 2 ) and ( 2^{1.58496} = 3 ), as before. So, ( log_2 3 ) is between 1 and 2, specifically between 1 and 1.6.Therefore, ( (log_2 3)^2 ) is between ( 1^2 = 1 ) and ( 1.6^2 = 2.56 ). So, ( (log_2 3)^2 ) is between 1 and 2.56.Thus, ( 1/(log_2 3)^2 ) is between ( 1/2.56 approx 0.3906 ) and ( 1/1 = 1 ).So, ( x = 1/(log_2 3)^2 + 1 ) is between ( 0.3906 + 1 = 1.3906 ) and ( 1 + 1 = 2 ).Therefore, ( x ) is between approximately 1.39 and 2. So, it's definitely in the interval (1, 2), which is option B.Wait, but let me think again about the exact value of ( log_2 3 ). I know that ( 2^{1.58496} = 3 ), so ( log_2 3 approx 1.58496 ). Squaring that gives approximately 2.51189, as before. So, ( 1/2.51189 approx 0.398 ). Adding 1 gives approximately 1.398, which is about 1.4.So, 1.4 is definitely between 1 and 2, so option B is correct.But just to make sure, let me think if there's any chance I could have messed up the negative exponents or the logarithm properties.Starting again: ( x = left(log_{frac{1}{2}} frac{1}{3}right)^{-2} + left(log_{frac{1}{3}} frac{1}{3}right)^{-1} ).First term: ( log_{frac{1}{2}} frac{1}{3} ). Since the base is less than 1, the logarithm is negative. Specifically, ( log_{frac{1}{2}} frac{1}{3} = frac{ln frac{1}{3}}{ln frac{1}{2}} = frac{-ln 3}{-ln 2} = frac{ln 3}{ln 2} = log_2 3 ). So, that's positive, approximately 1.58496.Therefore, ( left(log_{frac{1}{2}} frac{1}{3}right)^{-2} = (log_2 3)^{-2} = 1/(log_2 3)^2 approx 1/(1.58496)^2 approx 0.398 ).Second term: ( log_{frac{1}{3}} frac{1}{3} ). Again, using change of base: ( frac{ln frac{1}{3}}{ln frac{1}{3}} = 1 ). So, ( left(log_{frac{1}{3}} frac{1}{3}right)^{-1} = 1^{-1} = 1 ).Adding them together: 0.398 + 1 = 1.398, which is approximately 1.4.So, yes, 1.4 is in the interval (1, 2), which is option B.I think I've double-checked enough. I don't see any mistakes in my reasoning. The key was recognizing how to handle the logarithms with fractional bases and correctly applying the properties to simplify the expressions. Once I had the simplified terms, approximating ( log_2 3 ) helped me find the numerical value of ( x ), which clearly falls into interval B.

💡To find the equation of the line ( l_1 ) that is symmetric to the line ( y = 3x - 2 ) with respect to the y-axis, I need to understand how symmetry about the y-axis affects the equation of a line.First, symmetry about the y-axis means that for every point ( (x, y) ) on the original line, there is a corresponding point ( (-x, y) ) on the symmetric line. This implies that the x-coordinates of the points are negated, while the y-coordinates remain the same.Given the original line ( y = 3x - 2 ), I can determine the slope and y-intercept. The slope is 3, and the y-intercept is -2.When reflecting a line about the y-axis, the slope changes sign because the direction of the line reverses horizontally. Therefore, the slope of the symmetric line ( l_1 ) will be -3.The y-intercept remains unchanged because the reflection does not affect the vertical position of the line. Thus, the y-intercept of ( l_1 ) is still -2.Putting it all together, the equation of the symmetric line ( l_1 ) is ( y = -3x - 2 ).