💡Alright, so I've got this problem here where Masha wrote all the natural divisors of a number ( N ) in ascending order. The first divisor is 1, and the last one is ( N ) itself. It says that the third-to-last divisor is 21 times larger than the second divisor. I need to find the largest possible value of ( N ).Okay, let's break this down. First, let's recall that the divisors of a number ( N ) are all the numbers that divide ( N ) without leaving a remainder. They come in pairs that multiply to ( N ). So, if ( d ) is a divisor of ( N ), then so is ( frac{N}{d} ).Given that the divisors are written in ascending order, the first one is 1, and the last one is ( N ). The second divisor is the smallest prime factor of ( N ), let's call it ( p ). Then, the third divisor could be either ( p^2 ) if ( p^2 ) divides ( N ), or another prime factor ( q ) if ( N ) has more than one prime factor.Now, the problem states that the third-to-last divisor is 21 times the second divisor. The third-to-last divisor is the same as the third divisor from the end, which would be ( frac{N}{d_3} ) where ( d_3 ) is the third divisor from the start. So, we have:[frac{N}{d_3} = 21 times d_2]Given that ( d_2 = p ), this simplifies to:[frac{N}{d_3} = 21p]So, ( N = 21p times d_3 ).Now, let's consider two cases for ( d_3 ):**Case 1: ( d_3 = p^2 )**In this case, ( N = 21p times p^2 = 21p^3 ).Let's check if this works. If ( N = 21p^3 ), then the divisors would be:1, ( p ), ( p^2 ), ( p^3 ), 3, 7, ( frac{N}{7} ), ( frac{N}{3} ), ( frac{N}{p} ), ( frac{N}{p^2} ), ( frac{N}{p^3} ), ..., ( N ).Wait, but this might not necessarily be in order. Let's take a specific example. Let ( p = 2 ). Then ( N = 21 times 8 = 168 ).Divisors of 168: 1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168.So, the third-to-last divisor is 56, and the second divisor is 2. Is 56 equal to 21 times 2? 21*2=42, which is not 56. So, this doesn't work.Maybe ( p = 3 ). Then ( N = 21 times 27 = 567 ).Divisors of 567: 1, 3, 7, 9, 21, 27, 63, 81, 189, 567.Third-to-last divisor is 189, second divisor is 3. Is 189 = 21*3? 21*3=63, which is not 189. So, this doesn't work either.Hmm, maybe this case doesn't hold. Let's try the other case.**Case 2: ( d_3 = q ), where ( q ) is another prime factor**So, ( N = 21p times q ).Since ( N ) has at least two prime factors, ( p ) and ( q ), and ( p < q ).Given that ( d_2 = p ) and ( d_3 = q ), the divisors would be:1, ( p ), ( q ), ( p^2 ), ( pq ), ( q^2 ), ..., ( N ).Wait, but the order depends on the values of ( p ) and ( q ). Let's assume ( p = 3 ) and ( q = 7 ), since 21 is 3*7.So, ( N = 21 times 3 times 7 = 441 ).Let's list the divisors of 441:1, 3, 7, 9, 21, 49, 63, 147, 441.So, the third-to-last divisor is 63, and the second divisor is 3. Is 63 equal to 21*3? Yes, 21*3=63. Perfect!So, ( N = 441 ) satisfies the condition. Is this the largest possible?Let's see if we can get a larger ( N ). Suppose we take ( p = 2 ) and ( q = 21 ), but 21 isn't prime. Or ( p = 2 ) and ( q = 3 ), but then ( N = 21*2*3=126 ). Let's check the divisors:1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, 126.Third-to-last divisor is 63, second divisor is 2. Is 63=21*2? 21*2=42≠63. Doesn't work.What if ( p = 3 ) and ( q = 7 ), which gives us 441, which works. If we try ( p = 3 ) and ( q = 11 ), then ( N = 21*3*11=693 ).Divisors of 693: 1, 3, 7, 9, 11, 21, 33, 63, 77, 99, 231, 693.Third-to-last divisor is 99, second divisor is 3. Is 99=21*3? 21*3=63≠99. Doesn't work.What if ( p = 3 ) and ( q = 5 )? Then ( N = 21*3*5=315 ).Divisors of 315: 1, 3, 5, 7, 9, 15, 21, 35, 45, 63, 105, 315.Third-to-last divisor is 105, second divisor is 3. Is 105=21*3? 21*3=63≠105. Doesn't work.Hmm, seems like 441 is the only one that works so far. Let's try ( p = 5 ) and ( q = 7 ). Then ( N = 21*5*7=735 ).Divisors of 735: 1, 3, 5, 7, 15, 21, 35, 49, 105, 147, 245, 735.Third-to-last divisor is 245, second divisor is 3. Is 245=21*3? 21*3=63≠245. Doesn't work.What if ( p = 7 ) and ( q = 3 )? But ( p ) should be the smallest prime, so ( p = 3 ) is better.Wait, maybe ( N ) has more than two prime factors? Let's see.Suppose ( N = 21pqr ), but that might complicate things. Let's try ( N = 21p^2 ), but earlier that didn't work.Alternatively, maybe ( N ) has multiple exponents. For example, ( N = p^k times q^m times ... ).Wait, in the case of ( N = 441 = 3^2 times 7^2 ), the divisors are 1, 3, 7, 9, 21, 49, 63, 147, 441. So, the third divisor is 7, and the third-to-last is 63, which is 21*3. Perfect.If I try to increase the exponents, say ( N = 3^3 times 7^2 = 3^3 times 49 = 27 times 49 = 1323 ).Divisors of 1323: 1, 3, 7, 9, 21, 27, 49, 63, 147, 189, 343, 441, 1029, 1323.Third-to-last divisor is 441, second divisor is 3. Is 441=21*3? 21*3=63≠441. Doesn't work.Hmm, so increasing the exponents beyond 2 for 3 and 7 doesn't help because the third-to-last divisor becomes too large.What if I try to include another prime factor? Let's say ( N = 3^2 times 7 times 11 = 9 times 7 times 11 = 693 ). Wait, we tried that earlier, and it didn't work.Alternatively, ( N = 3^2 times 7^2 times 2 = 441 times 2 = 882 ).Divisors of 882: 1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 49, 63, 98, 126, 147, 294, 441, 882.Third-to-last divisor is 441, second divisor is 2. Is 441=21*2? 21*2=42≠441. Doesn't work.So, adding another prime factor seems to mess up the condition.Wait, maybe if I take ( p = 3 ) and ( q = 7 ), and have ( N = 3^2 times 7^2 = 441 ), which works. If I try to make ( N ) larger by adding another factor, say ( 3^2 times 7^2 times 2 = 882 ), but as we saw, it doesn't satisfy the condition.Alternatively, maybe ( N = 3^2 times 7^2 times 5 = 441 times 5 = 2205 ).Divisors of 2205: 1, 3, 5, 7, 9, 15, 21, 35, 45, 49, 63, 105, 147, 245, 315, 441, 735, 2205.Third-to-last divisor is 735, second divisor is 3. Is 735=21*3? 21*3=63≠735. Doesn't work.Hmm, seems like adding more prime factors or increasing exponents beyond 2 for 3 and 7 doesn't help because the third-to-last divisor becomes too large or doesn't satisfy the 21 times condition.Therefore, the largest ( N ) that satisfies the condition is 441.

💡First, I need to determine how much gasoline was used. The car used 0.8333333333333334 of a full 10-gallon tank. Multiplying these values gives the total gallons used.Next, I'll calculate the distance traveled using the car's fuel efficiency. Since the car consumes 1 gallon every 30 miles, multiplying the gallons used by 30 miles per gallon will provide the total distance.Finally, to find out how long the car traveled, I'll divide the total distance by the car's speed of 50 miles per hour. This will give the time in hours.

💡First, I recognize that the expression ( x^{2} - 4x + 5 ) is a quadratic function, and its graph is a parabola opening upwards because the coefficient of ( x^{2} ) is positive. This means the function has a minimum value at its vertex.To find the vertex of the parabola, I can use the formula for the x-coordinate of the vertex, which is ( x = -frac{b}{2a} ). In this case, ( a = 1 ) and ( b = -4 ), so:[x = -frac{-4}{2 times 1} = frac{4}{2} = 2]Now that I have the x-coordinate of the vertex, I can substitute ( x = 2 ) back into the original expression to find the minimum value:[(2)^{2} - 4(2) + 5 = 4 - 8 + 5 = 1]Therefore, the minimum value of the expression is 1.

💡First, I need to determine how many apples the fruit seller sold. He originally had 9999.9980000004 apples and was left with 5000 apples after selling some.To find the number of apples sold, I subtract the remaining apples from the original number:9999.9980000004 - 5000 = 4999.9980000004 apples sold.Next, to find the percentage of apples sold, I divide the number of apples sold by the original number of apples and then multiply by 100:(4999.9980000004 / 9999.9980000004) * 100.Simplifying this calculation:4999.998 / 9999.998 ≈ 0.5,so 0.5 * 100 = 50%.Therefore, the fruit seller sold approximately 50% of his apples.

💡Okay, so I have this problem about temperatures in Chicago and Denver. Let me try to understand it step by step. At 10:00 A.M., the temperature in Chicago is P degrees warmer than in Denver. So, if I let the temperature in Denver be D degrees, then the temperature in Chicago would be D + P degrees. That makes sense because Chicago is P degrees warmer.Then, by 2:00 P.M., which is four hours later, the temperature in Chicago decreases by 4 degrees. So, the temperature in Chicago at 2:00 P.M. would be (D + P) - 4. On the other hand, the temperature in Denver increases by 2 degrees, so the temperature in Denver at 2:00 P.M. would be D + 2.Now, the problem states that by 2:00 P.M., the temperatures in the two cities differ by 6 degrees. So, the absolute difference between the temperature in Chicago and Denver at 2:00 P.M. is 6 degrees. Let me write that down as an equation:| (Temperature in Chicago at 2:00 P.M.) - (Temperature in Denver at 2:00 P.M.) | = 6Substituting the expressions I have:| (D + P - 4) - (D + 2) | = 6Let me simplify the expression inside the absolute value:(D + P - 4) - (D + 2) = D + P - 4 - D - 2 = P - 6So, the equation becomes:| P - 6 | = 6Now, to solve this absolute value equation, I know that |x| = a means x = a or x = -a. So, applying that here:P - 6 = 6 or P - 6 = -6Solving the first equation:P - 6 = 6 Add 6 to both sides: P = 12Solving the second equation:P - 6 = -6 Add 6 to both sides: P = 0So, the possible values of P are 12 and 0. The problem asks for the product of all possible values of P. So, I need to multiply these two values together:12 * 0 = 0Therefore, the product of all possible values of P is 0.Wait, let me double-check to make sure I didn't make a mistake. Starting from the beginning: At 10:00 A.M., Chicago is P degrees warmer than Denver. So, if Denver is D, Chicago is D + P. At 2:00 P.M., Chicago is D + P - 4, and Denver is D + 2. The difference is 6 degrees. So, |(D + P - 4) - (D + 2)| = |P - 6| = 6. That leads to P = 12 or P = 0. Multiplying them gives 0. Seems correct.I think that's solid. The answer should be 0.

💡Okay, so I'm trying to figure out whether these two fractions are always equivalent in each case. Let's start with Case 1: (frac{x + a}{x + c}) versus (frac{a}{c}). Hmm, I need to see if these two fractions are the same for all values of x, given that c and x are not zero.First, I'll write down both fractions:1. (frac{x + a}{x + c})2. (frac{a}{c})I wonder if these are equal. Maybe I can cross-multiply to check. If (frac{x + a}{x + c} = frac{a}{c}), then cross-multiplying should give me:((x + a) cdot c = (x + c) cdot a)Let me expand both sides:Left side: (x cdot c + a cdot c = cx + ac)Right side: (x cdot a + c cdot a = ax + ac)So, comparing both sides:(cx + ac = ax + ac)If I subtract (ac) from both sides, I get:(cx = ax)Now, if x is not zero, I can divide both sides by x:(c = a)Wait, so this means that (frac{x + a}{x + c}) is only equal to (frac{a}{c}) if (a = c). But the problem doesn't state that (a) and (c) are equal. It just says they are nonzero. So, unless (a) and (c) are the same, these fractions aren't equivalent. Therefore, Case 1 isn't always equivalent.Now, let's look at Case 2: (frac{b}{d}) versus (frac{b}{d}), with (b) and (d) not equal to zero. Well, these are exactly the same fractions. The numerator is the same, and the denominator is the same. So, of course, they are equal. It's like saying (frac{5}{3}) is equal to (frac{5}{3}), which is obviously true.So, in summary, Case 1 isn't always equivalent because it depends on (a) and (c) being equal, which isn't guaranteed. Case 2 is always equivalent because both fractions are identical. Therefore, the answer should be Case 2.