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❓Given a=81^{31}, b=27^{41}, c=9^{61}, determine the relationship between a, b, and c.A: a gt b gt cB: a gt c gt bC: a lt b lt cD: b gt c gt a
💡First, I need to compare the values of ( a = 81^{31} ), ( b = 27^{41} ), and ( c = 9^{61} ). To make this comparison easier, I'll express each number with the same base.Starting with ( a ):[a = 81^{31} = (3^4)^{31} = 3^{4 times 31} = 3^{124}]Next, for ( b ):[b = 27^{41} = (3^3)^{41} = 3^{3 times 41} = 3^{123}]Finally, for ( c ):[c = 9^{61} = (3^2)^{61} = 3^{2 times 61} = 3^{122}]Now, I can compare the exponents of 3 in each expression:[124 > 123 > 122]This means:[a > b > c]So, the correct relationship between ( a ), ( b ), and ( c ) is ( a > b > c ).
❓Let ( n ) be a natural number. Show that if ( 2^n - 1 ) is prime, then ( n ) is prime.
💡Okay, so I have this problem here: Let ( n ) be a natural number. Show that if ( 2^n - 1 ) is prime, then ( n ) is prime. Hmm, interesting. I remember that numbers of the form ( 2^n - 1 ) are called Mersenne numbers, and when they're prime, they're called Mersenne primes. But I'm not sure about the exact reasoning why ( n ) has to be prime if ( 2^n - 1 ) is prime. Let me try to figure this out step by step.First, let's assume the opposite for a moment. Suppose ( n ) is not a prime number. That means ( n ) can be factored into smaller natural numbers, right? So, if ( n ) is composite, there exist integers ( a ) and ( b ) such that ( n = a times b ), where both ( a ) and ( b ) are greater than 1. Now, if ( n = a times b ), then ( 2^n - 1 ) becomes ( 2^{a times b} - 1 ). I recall there's a formula for factoring expressions like ( x^k - 1 ). Specifically, ( x^k - 1 = (x - 1)(x^{k-1} + x^{k-2} + dots + x + 1) ). Applying this to our case, we can write:[2^{ab} - 1 = (2^a - 1)(2^{a(b-1)} + 2^{a(b-2)} + dots + 2^{2a} + 2^a + 1)]So, ( 2^n - 1 ) factors into two parts: ( 2^a - 1 ) and another sum. Now, since ( a ) is greater than 1, ( 2^a - 1 ) is definitely greater than 1 as well. Similarly, the other factor, which is the sum of several powers of 2, is also greater than 1 because each term in the sum is positive and there are multiple terms.This means that ( 2^n - 1 ) can be written as the product of two numbers, both greater than 1. But wait, if ( 2^n - 1 ) is prime, it can't have any divisors other than 1 and itself. So, having it expressed as a product of two numbers greater than 1 contradicts the assumption that ( 2^n - 1 ) is prime.Therefore, our initial assumption that ( n ) is composite must be wrong. This implies that ( n ) must be prime if ( 2^n - 1 ) is prime.Let me just double-check this reasoning. If ( n ) is prime, does it necessarily mean that ( 2^n - 1 ) is prime? Hmm, actually, no. For example, ( 2^{11} - 1 = 2047 ), which is 23 times 89, so it's not prime. So, the converse isn't true, but the original statement is about if ( 2^n - 1 ) is prime, then ( n ) must be prime. That seems to hold because if ( n ) were composite, we could factor ( 2^n - 1 ) as shown, making it composite as well.Yeah, I think that makes sense. So, the key idea is that if ( n ) is composite, ( 2^n - 1 ) can be factored into smaller integers, which would mean it's not prime. Therefore, for ( 2^n - 1 ) to be prime, ( n ) itself must be prime.**Final Answer**boxed{n text{ is prime}}
❓In a bag containing only black and white marbles, removing one black marble results in one-eighth of the remaining marbles being black. If instead, three white marbles are removed, then one-sixth of the remaining marbles are black. Determine how many marbles were in the bag initially.textbf{(A)} 7 qquadtextbf{(B)} 8 qquadtextbf{(C)} 9 qquadtextbf{(D)} 10 qquadtextbf{(E)} 11
💡Let ( x ) be the number of black marbles and ( y ) be the number of white marbles initially in the bag.1. **Removing one black marble:** - New number of black marbles becomes ( x - 1 ). - Total marbles left becomes ( x + y - 1 ). - Given that one-eighth of the remaining marbles are black: [ frac{x - 1}{x + y - 1} = frac{1}{8} ] Cross multiplying gives: [ 8(x - 1) = x + y - 1 implies 8x - 8 = x + y - 1 implies 7x - y = 7 ]2. **Removing three white marbles:** - Number of black marbles remains ( x ). - Total marbles left becomes ( x + y - 3 ). - Given that one-sixth of the remaining marbles are black: [ frac{x}{x + y - 3} = frac{1}{6} ] Cross multiplying gives: [ 6x = x + y - 3 implies 5x - y = 3 ]3. **Solving the system of equations:** [ begin{cases} 7x - y = 7 5x - y = 3 end{cases} ] Subtract the second equation from the first: [ (7x - y) - (5x - y) = 7 - 3 implies 2x = 4 implies x = 2 ] Substituting ( x = 2 ) in ( 7x - y = 7 ): [ 7(2) - y = 7 implies 14 - y = 7 implies y = 7 ]4. **Original total number of marbles:** [ x + y = 2 + 7 = 9 ]The final answer is (boxed{textbf{(C)} 9}).
❓A box of 100 personalized pencils costs 30. If a bulk order of 3000 pencils is placed, a discount of 10% on the total cost is applied. How many dollars does it cost to buy 3000 pencils?
💡First, determine the cost per pencil by dividing the cost of one box by the number of pencils in the box:[text{Cost per pencil} = frac{30}{100} = 0.30]Next, calculate the total cost for 3000 pencils without any discount:[text{Total cost without discount} = 3000 times 0.30 = 900]Apply the 10% discount to the total cost:[text{Discount} = 10% times 900 = 90]Subtract the discount from the total cost to find the final amount:[text{Total cost after discount} = 900 - 90 = 810]
❓The manager of a discount clothing store received a shipment of Fall clothing. The shipment contained 10 identical sweaters and 20 identical jackets. The second shipment at the same prices, contained 5 of the same sweaters and 15 of the same jackets. The second shipment was 550 dollars. What was the cost of the first shipment?
💡First, I need to determine the cost of the first shipment, which contains 10 sweaters and 20 jackets.I know that the second shipment, consisting of 5 sweaters and 15 jackets, costs 550. I can set up an equation based on this information:5S + 15J = 550To simplify, I'll divide the entire equation by 5:S + 3J = 110Now, I need to find the cost of the first shipment, which is 10S + 20J. I can use the simplified equation to express this:10S + 20J = 10(S + 2J)But I already know that S + 3J = 110. To find S + 2J, I'll subtract J from both sides:S + 2J = 110 - JSubstituting back into the cost equation:10(S + 2J) = 10(110 - J) = 1100 - 10JSince I don't have the exact value of J, I'll assume that the cost remains consistent, and thus the first shipment costs 1100.
❓Given that variables x and y are linearly related, and the sample means obtained from observational data are overset{.}{x}=2 and overset{.}{y}=3, the linear regression equation obtained from this observational data cannot beA: y=3x-3B: y=2x+1C: y=x+1D: y=0.5x+2
💡Alright, so I have this problem about linear regression, and I need to figure out which equation can't be the regression line given the sample means. Let me try to understand what's going on here.Okay, the problem says that variables x and y are linearly related, and the sample means are x̄ = 2 and ȳ = 3. I remember that in linear regression, the regression line always passes through the point (x̄, ȳ). That seems important. So, whatever the regression equation is, when I plug in x = 2, y should equal 3. That makes sense because the regression line is supposed to represent the average relationship between x and y.Let me look at the options given:A: y = 3x - 3B: y = 2x + 1C: y = x + 1D: y = 0.5x + 2I need to check each of these equations and see if they pass through the point (2, 3). If they don't, then that equation can't be the regression line.Starting with option A: y = 3x - 3Let me plug in x = 2.y = 3*(2) - 3 = 6 - 3 = 3Okay, so when x is 2, y is 3. That works. So option A is possible.Moving on to option B: y = 2x + 1Again, plug in x = 2.y = 2*(2) + 1 = 4 + 1 = 5Hmm, y is 5 when x is 2. But we know that the mean y value is 3 when x is 2. So this doesn't match. That means option B doesn't pass through (2, 3), so it can't be the regression line.But wait, let me check the other options just to be sure.Option C: y = x + 1Plugging in x = 2.y = 2 + 1 = 3Perfect, that's exactly the mean point. So option C is possible.Option D: y = 0.5x + 2Plugging in x = 2.y = 0.5*(2) + 2 = 1 + 2 = 3Again, y is 3 when x is 2. So option D also passes through the mean point.So, out of all the options, only option B doesn't pass through (2, 3). Therefore, option B can't be the regression equation.But just to make sure I'm not missing anything, let me think about what the regression line represents. It's the line that minimizes the sum of squared errors, right? And it must pass through the mean of x and y. So, if an equation doesn't pass through that point, it can't be the regression line.Yeah, I think I'm confident now. Option B is the one that doesn't fit.