answer：Alright, so I have this problem here: I need to find the largest natural number ( x ) such that for every natural number ( y ), ( x ) divides ( 7^y + 12y - 1 ). Hmm, okay. Let me try to unpack this step by step.First off, I need to understand what the problem is asking. It wants the biggest number ( x ) that can divide the expression ( 7^y + 12y - 1 ) no matter what natural number ( y ) I plug in. So, ( x ) has to be a common divisor of all these expressions for every ( y ). That means ( x ) has to work for ( y = 1, 2, 3, ) and so on.Maybe I should start by plugging in some small values of ( y ) to see what ( 7^y + 12y - 1 ) looks like and see if I can spot a pattern or figure out what ( x ) could be.Let's try ( y = 1 ):[ 7^1 + 12(1) - 1 = 7 + 12 - 1 = 18 ]Okay, so when ( y = 1 ), the expression equals 18. That means ( x ) has to divide 18. So, ( x ) is a divisor of 18. The divisors of 18 are 1, 2, 3, 6, 9, and 18. So, ( x ) must be one of these numbers. But we need the largest such ( x ), so ideally, it would be 18, but I need to check if 18 works for all ( y ).Let me try ( y = 2 ):[ 7^2 + 12(2) - 1 = 49 + 24 - 1 = 72 ]72 divided by 18 is 4, so 18 divides 72. Good.Next, ( y = 3 ):[ 7^3 + 12(3) - 1 = 343 + 36 - 1 = 378 ]378 divided by 18 is 21, so 18 divides 378. Still good.Let's try ( y = 4 ):[ 7^4 + 12(4) - 1 = 2401 + 48 - 1 = 2448 ]2448 divided by 18 is 136, so that's also divisible. Hmm, seems like 18 is working so far.But just because it works for the first few values doesn't mean it works for all ( y ). I need a more general approach to confirm that 18 divides ( 7^y + 12y - 1 ) for every natural number ( y ).Maybe I can use mathematical induction. Induction is a common technique to prove statements for all natural numbers.Let's recall how induction works. First, I prove the base case, usually ( y = 1 ), which we've already done—it works. Then, I assume it works for some arbitrary ( y = k ), and then I prove it works for ( y = k + 1 ).So, let's set up the induction.**Base Case (( y = 1 ))**:We've already calculated that when ( y = 1 ), the expression is 18, which is divisible by 18. So, the base case holds.**Inductive Step**:Assume that for some natural number ( k ), ( 7^k + 12k - 1 ) is divisible by 18. That is, there exists some integer ( m ) such that:[ 7^k + 12k - 1 = 18m ]Now, we need to show that ( 7^{k+1} + 12(k + 1) - 1 ) is also divisible by 18.Let's compute ( 7^{k+1} + 12(k + 1) - 1 ):[ 7^{k+1} + 12(k + 1) - 1 = 7 cdot 7^k + 12k + 12 - 1 ]Simplify:[ = 7 cdot 7^k + 12k + 11 ]Now, from our inductive hypothesis, we know that ( 7^k + 12k - 1 = 18m ). Let's see if we can express ( 7^{k+1} + 12(k + 1) - 1 ) in terms of this.Let me try to manipulate the expression:[ 7 cdot 7^k + 12k + 11 = 7(7^k) + 12k + 11 ]From the inductive hypothesis, ( 7^k = 18m - 12k + 1 ). Let's substitute that in:[ 7(18m - 12k + 1) + 12k + 11 ]Expand this:[ 126m - 84k + 7 + 12k + 11 ]Combine like terms:[ 126m - 72k + 18 ]Now, factor out 18:[ 18(7m - 4k + 1) ]Since ( 7m - 4k + 1 ) is an integer, the entire expression is divisible by 18. Therefore, ( 7^{k+1} + 12(k + 1) - 1 ) is divisible by 18.Thus, by the principle of mathematical induction, ( 7^y + 12y - 1 ) is divisible by 18 for all natural numbers ( y ).But wait, just to be thorough, let me check a few more values of ( y ) to make sure I didn't miss anything.For ( y = 5 ):[ 7^5 + 12(5) - 1 = 16807 + 60 - 1 = 16866 ]Divided by 18: ( 16866 ÷ 18 = 937 ). So, yes, divisible.For ( y = 6 ):[ 7^6 + 12(6) - 1 = 117649 + 72 - 1 = 117720 ]Divided by 18: ( 117720 ÷ 18 = 6540 ). Divisible again.Okay, so it seems consistent. But just to be extra cautious, let me think if there's any other way this could fail. Maybe if ( y ) is very large? But since we've proven it via induction, it should hold for all ( y ).Another thought: perhaps 18 isn't the maximum? Maybe a larger multiple? But 18 is already the maximum divisor of 18, so it can't be larger. Unless there's a mistake in my induction step.Wait, let me double-check the induction step.Starting from:[ 7^{k+1} + 12(k + 1) - 1 = 7 cdot 7^k + 12k + 12 - 1 = 7 cdot 7^k + 12k + 11 ]Then, using the inductive hypothesis ( 7^k + 12k - 1 = 18m ), so ( 7^k = 18m - 12k + 1 ).Substituting:[ 7(18m - 12k + 1) + 12k + 11 = 126m - 84k + 7 + 12k + 11 = 126m - 72k + 18 ]Which is:[ 18(7m - 4k + 1) ]Yes, that looks correct. So, the expression is indeed divisible by 18 for ( y = k + 1 ) if it's divisible for ( y = k ). Therefore, the induction holds.So, after all this, I think 18 is indeed the largest natural number that divides ( 7^y + 12y - 1 ) for every natural number ( y ).Just to recap:1. Tested small values of ( y ) and found that 18 divides the expression.2. Noted that 18 is a divisor of the expression for ( y = 1 ), and since we need ( x ) to divide for all ( y ), ( x ) must be a divisor of 18.3. Used mathematical induction to prove that 18 divides the expression for all ( y ).4. Double-checked with additional values of ( y ) and confirmed the result.5. Verified the induction step to ensure there were no errors in the logic.Therefore, I'm confident that the largest natural number ( x ) satisfying the condition is 18.**Final Answer**The largest natural number ( x ) is boxed{18}.

answer：First, I need to understand the possible outcomes of the chess game between Players A and B. There are three possible outcomes: Player A wins, Player B wins, or the game ends in a draw.I know the probability of a draw is 1/2, and the probability of Player B winning is 1/3. Since these are the only possible outcomes, their probabilities should add up to 1.To find the probability of Player B not losing, I need to consider both the probability of Player B winning and the probability of the game ending in a draw. This is because "not losing" means either winning or drawing.So, I add the probability of Player B winning (1/3) to the probability of a draw (1/2). To add these fractions, I need a common denominator, which is 6.Converting 1/3 to 2/6 and 1/2 to 3/6, I can now add them together: 2/6 + 3/6 equals 5/6.Therefore, the probability that Player B does not lose is 5/6.

answer：First, I need to determine how much money Anna spent. She originally had 32 and now has 24 left. By subtracting the remaining amount from the original amount, I can find the amount spent.Next, to find the fraction of her money that she spent, I will divide the amount spent by the original amount. This will give me the fraction of her original money that Anna has spent.Finally, I will present the fraction in its simplest form to provide a clear answer to the problem.

answer：Okay, so I have this problem where I'm given real numbers ( a, b, c ) such that ( a^2 + b^2 + c^2 = 1 ). I need to prove that ( -frac{1}{2} leq ab + bc + ca leq 1 ). Hmm, okay, let's try to break this down step by step.First, I remember that expressions involving sums of products like ( ab + bc + ca ) often relate to the square of sums or differences. Maybe I can use some algebraic identities here. Let me think about the square of ( (a + b + c) ). That would be ( a^2 + b^2 + c^2 + 2(ab + bc + ca) ). Since I know ( a^2 + b^2 + c^2 = 1 ), maybe I can plug that in.So, ( (a + b + c)^2 = 1 + 2(ab + bc + ca) ). Hmm, since squares are always non-negative, this tells me that ( 1 + 2(ab + bc + ca) geq 0 ). If I solve for ( ab + bc + ca ), I get ( ab + bc + ca geq -frac{1}{2} ). Okay, that gives me the lower bound. That seems straightforward.Now, for the upper bound. I need to show that ( ab + bc + ca leq 1 ). Maybe I can use another identity or inequality. I recall that ( (a - b)^2 + (b - c)^2 + (c - a)^2 ) is always non-negative. Let me expand that.Expanding each term:- ( (a - b)^2 = a^2 - 2ab + b^2 )- ( (b - c)^2 = b^2 - 2bc + c^2 )- ( (c - a)^2 = c^2 - 2ca + a^2 )Adding them all together:( (a - b)^2 + (b - c)^2 + (c - a)^2 = 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca )Factor out the 2:( 2(a^2 + b^2 + c^2) - 2(ab + bc + ca) )Since ( a^2 + b^2 + c^2 = 1 ), substitute that in:( 2(1) - 2(ab + bc + ca) = 2 - 2(ab + bc + ca) )Since the sum of squares is non-negative, ( 2 - 2(ab + bc + ca) geq 0 ). Solving for ( ab + bc + ca ), we get:( -2(ab + bc + ca) geq -2 )Divide both sides by -2 (remembering to reverse the inequality):( ab + bc + ca leq 1 )Okay, so that gives me the upper bound. Putting it all together, I have shown that ( -frac{1}{2} leq ab + bc + ca leq 1 ). That seems to cover both bounds. Let me just double-check my steps to make sure I didn't make any mistakes.For the lower bound, I used the square of the sum, which is always non-negative, leading to ( ab + bc + ca geq -frac{1}{2} ). For the upper bound, I used the sum of squared differences, which is also non-negative, leading to ( ab + bc + ca leq 1 ). Both steps seem solid. I think I've got it!

answer：First, I need to determine the original purchase price of the product. Given that the original selling price is approximately 550 and the profit percentage is unknown, I'll denote the original purchase price as ( P ) and the original profit percentage as ( x % ).Using the original selling price formula:[550 = P cdot (1 + frac{x}{100})]I can express ( P ) as:[P = frac{550}{1 + frac{x}{100}} = frac{55000}{100 + x}]Next, if Bill had purchased the product for 10% less, the new purchase price would be:[P_{text{new}} = P cdot 0.9]If he sold it at a 30% profit, the new selling price would be:[SP_{text{new}} = P_{text{new}} cdot 1.3 = P cdot 1.17]According to the problem, this new selling price is 35 more than the original selling price:[P cdot 1.17 = 585]Solving for ( P ):[P = frac{585}{1.17} = 500]Now, substituting ( P = 500 ) back into the equation for the original purchase price:[frac{55000}{100 + x} = 500]Solving for ( x ):[55000 = 500 cdot (100 + x) 55000 = 50000 + 500x 500x = 5000 x = 10]Therefore, Bill's original profit percentage was 10%.

answer：Okay, so I have this expression to simplify: (2x(4x^2 - 3x + 1) - 7(2x^2 - 3x + 4)). Hmm, I remember from algebra that I need to use the distributive property, which is like multiplying each term inside the parentheses by the term outside. Let me try to break this down step by step.First, I'll tackle the first part: (2x(4x^2 - 3x + 1)). I need to distribute the 2x to each term inside the parentheses. So, multiplying 2x by 4x² should give me... let's see, 2 times 4 is 8, and x times x² is x³, so that's 8x³. Next, 2x multiplied by -3x. 2 times -3 is -6, and x times x is x², so that's -6x². Then, 2x times 1 is just 2x. So, putting that all together, the first part simplifies to (8x³ - 6x² + 2x). Okay, that seems right.Now, moving on to the second part: (-7(2x² - 3x + 4)). Again, I need to distribute the -7 across each term inside the parentheses. Starting with -7 times 2x², that should be -14x². Then, -7 times -3x. Hmm, multiplying two negatives should give a positive, so that's +21x. Finally, -7 times 4 is -28. So, the second part simplifies to (-14x² + 21x - 28). Got that down.Now, I need to combine these two results: (8x³ - 6x² + 2x) and (-14x² + 21x - 28). Let me write them together: (8x³ - 6x² + 2x - 14x² + 21x - 28). Now, I should combine like terms. Like terms are terms with the same variable raised to the same power.Looking at the (x³) terms, there's only one: (8x³). So that stays as is. Next, the (x²) terms: I have (-6x²) and (-14x²). Adding those together: (-6 - 14) is (-20), so that gives me (-20x²). Now, the (x) terms: I have (2x) and (21x). Adding those together: (2 + 21) is (23), so that's (23x). Finally, the constant term is just (-28), and there are no other constants to combine with.Putting it all together, the simplified expression is (8x³ - 20x² + 23x - 28). Let me double-check my work to make sure I didn't make any mistakes. Starting with the distribution:- (2x * 4x² = 8x³)- (2x * -3x = -6x²)- (2x * 1 = 2x)- (-7 * 2x² = -14x²)- (-7 * -3x = 21x)- (-7 * 4 = -28)Combining like terms:- (8x³) remains- (-6x² -14x² = -20x²)- (2x + 21x = 23x)- (-28) remainsYes, that all checks out. So, I think I did it correctly. The final simplified expression is (8x³ - 20x² + 23x - 28).