answer：Okay, so I have this problem where I need to find the value of ( m ) such that the lines ( 13x + 11y = 700 ) and ( y = mx - 1 ) intersect at a point with integer coordinates. The options given are 4, 5, 6, 7, or one of those. Hmm, let me try to figure this out step by step.First, I know that if two lines intersect, their point of intersection satisfies both equations. So, I can substitute ( y ) from the second equation into the first one. That should give me an equation in terms of ( x ) only, which I can solve.So, substituting ( y = mx - 1 ) into ( 13x + 11y = 700 ), we get:( 13x + 11(mx - 1) = 700 )Let me expand that:( 13x + 11mx - 11 = 700 )Combine like terms:( (13 + 11m)x - 11 = 700 )Now, add 11 to both sides:( (13 + 11m)x = 711 )So, ( x = frac{711}{13 + 11m} )Since the coordinates must be integers, both ( x ) and ( y ) have to be integers. So, ( x ) must be an integer, which means that ( 13 + 11m ) must divide 711 exactly.Therefore, ( 13 + 11m ) is a divisor of 711.So, I need to find all positive integers ( m ) such that ( 13 + 11m ) divides 711.First, let me factorize 711 to find its divisors.711 divided by 3 is 237, because 7 + 1 + 1 = 9, which is divisible by 3. Then, 237 divided by 3 is 79. 79 is a prime number. So, the prime factorization of 711 is ( 3 times 3 times 79 ), or ( 3^2 times 79 ).Therefore, the positive divisors of 711 are:1, 3, 9, 79, 237, and 711.So, ( 13 + 11m ) must be one of these divisors. Let me write that down:( 13 + 11m = d ), where ( d ) is a positive divisor of 711.So, ( 13 + 11m in {1, 3, 9, 79, 237, 711} ).Now, let's solve for ( m ) in each case:1. If ( 13 + 11m = 1 ): ( 11m = 1 - 13 = -12 ) ( m = -12 / 11 ) But ( m ) must be a positive integer, so discard this case.2. If ( 13 + 11m = 3 ): ( 11m = 3 - 13 = -10 ) ( m = -10 / 11 ) Again, negative and not integer, discard.3. If ( 13 + 11m = 9 ): ( 11m = 9 - 13 = -4 ) ( m = -4 / 11 ) Still negative, discard.4. If ( 13 + 11m = 79 ): ( 11m = 79 - 13 = 66 ) ( m = 66 / 11 = 6 ) Okay, ( m = 6 ) is a positive integer. Let's keep this.5. If ( 13 + 11m = 237 ): ( 11m = 237 - 13 = 224 ) ( m = 224 / 11 ) Let me compute that: 11*20=220, so 224-220=4, so ( m = 20 + 4/11 ) Not an integer, so discard.6. If ( 13 + 11m = 711 ): ( 11m = 711 - 13 = 698 ) ( m = 698 / 11 ) Let's compute: 11*63=693, so 698-693=5, so ( m = 63 + 5/11 ) Not an integer, discard.So, the only possible value of ( m ) is 6. Hmm, but wait, the options include 4,5,6,7, or one of them. So, is 6 the only possible value? But let me double-check.Wait, perhaps I missed something. Let me think again.I considered only the positive divisors, but maybe I should also consider negative divisors? Because ( 13 + 11m ) could be negative, but since ( m ) is a positive integer, ( 13 + 11m ) must be positive. So, negative divisors won't give us positive ( m ). So, I think I was correct to only consider positive divisors.But wait, let me check if 711 has any other divisors. Wait, 711 is 3^2 * 79, so the only divisors are 1, 3, 9, 79, 237, 711. So, I think I covered all.But wait, let me check if 13 + 11m can be 79, which gives m=6, which is acceptable. But are there any other possibilities?Wait, maybe I made a mistake in the substitution. Let me go back.We had:( 13x + 11y = 700 )( y = mx - 1 )Substituting:( 13x + 11(mx - 1) = 700 )Which is:( 13x + 11mx - 11 = 700 )Combine terms:( (13 + 11m)x = 711 )So, ( x = 711 / (13 + 11m) )So, yes, that's correct.So, ( 13 + 11m ) must divide 711, which is 3^2 * 79.So, the possible values of ( 13 + 11m ) are 1,3,9,79,237,711.But as we saw, only 79 gives an integer ( m = 6 ). So, m=6 is the only possible value.But wait, the options are (A)4, (B)5, (C)6, (D)7, (E) One of 4,5,6,7.So, according to my calculation, only 6 is possible, so answer is (C). But wait, let me make sure.Wait, perhaps I missed that 711 is also divisible by other numbers? Let me check.Wait, 711 divided by 79 is 9, which is correct. 711 divided by 9 is 79, 711 divided by 3 is 237, 711 divided by 1 is 711. So, no, I think I have all the divisors.Wait, but let me think differently. Maybe I can express 711 as 13 + 11m times x, where x is an integer. So, 711 must be divisible by 13 + 11m, but perhaps I can find m such that 13 + 11m divides 711.Alternatively, maybe I can write 711 as 13 + 11m times x, so 711 = (13 + 11m) * x.So, x must be an integer, so 13 + 11m must be a divisor of 711.But I think I did that already.Wait, but maybe I can find m such that 13 + 11m divides 711, but perhaps I can also consider that x must be positive, since m is positive, and y = mx -1, so x must be such that y is also integer.Wait, but x could be positive or negative, but since 13x + 11y = 700, which is positive, x and y must be such that 13x +11y is positive. So, x and y can't be too negative.Wait, but maybe x is positive. Let me check.If x is positive, then 13 + 11m must be a positive divisor of 711, which is what I considered.But perhaps I can also consider that 13 + 11m must be less than or equal to 711, which it is in all cases.Wait, but let me check for m=4,5,6,7, which are the options.Let me compute 13 + 11m for each m:- m=4: 13 + 44 = 57- m=5: 13 +55=68- m=6:13+66=79- m=7:13+77=90Now, check if 57,68,79,90 divide 711.711 divided by 57: 57*12=684, 711-684=27, so 57*12 +27=711, so 57 doesn't divide 711.711 divided by 68: 68*10=680, 711-680=31, so 68 doesn't divide 711.711 divided by 79: 79*9=711, so yes, 79 divides 711.711 divided by 90: 90*7=630, 711-630=81, which is less than 90, so 90 doesn't divide 711.So, only m=6 gives 13 +11m=79, which divides 711.Therefore, the only possible value is m=6, which is option (C).Wait, but the options include (E) One of 4,5,6,7. So, why is that? Maybe I made a mistake.Wait, let me check again.Wait, perhaps I should consider that x could be negative, but in that case, 13 +11m would have to be a negative divisor of 711, but since m is positive, 13 +11m is positive, so x must be positive as well.Wait, but let me check if x is positive.From x = 711 / (13 +11m). Since 13 +11m is positive, x is positive.So, x must be positive integer.So, only m=6 is possible.Wait, but let me check if m=5 gives x=711/68, which is not integer, as I saw earlier.Similarly, m=4 gives x=711/57, which is 12.473..., not integer.m=7 gives x=711/90=7.9, not integer.So, only m=6 gives x=9, which is integer.Then, y = mx -1 = 6*9 -1=54-1=53, which is integer.So, the point is (9,53), which is integer coordinates.Therefore, m=6 is the only possible value.So, the answer should be (C)6.But wait, the options include (E) One of 4,5,6,7. So, perhaps I missed something.Wait, maybe I should consider that 13 +11m could be a negative divisor, but since m is positive, 13 +11m is positive, so negative divisors would give negative m, which is not allowed.Alternatively, perhaps I made a mistake in the substitution.Wait, let me check substitution again.We have:13x +11y=700y=mx -1Substitute y into first equation:13x +11(mx -1)=70013x +11mx -11=700(13 +11m)x=711So, x=711/(13 +11m)Yes, that's correct.So, 13 +11m must divide 711.Divisors of 711 are 1,3,9,79,237,711.So, 13 +11m must be one of these.So, solving for m:13 +11m=1: m negative, discard.13 +11m=3: m negative, discard.13 +11m=9: m negative, discard.13 +11m=79: m=6.13 +11m=237: m=(237-13)/11=224/11=20.363..., not integer.13 +11m=711: m=(711-13)/11=698/11≈63.454..., not integer.So, only m=6 is possible.Therefore, the answer is (C)6.But wait, the options include (E) One of 4,5,6,7. So, perhaps the problem is designed such that multiple m's are possible, but according to my calculations, only m=6 works.Alternatively, maybe I made a mistake in the substitution.Wait, let me try m=5.If m=5, then y=5x -1.Substitute into 13x +11y=700:13x +11*(5x -1)=70013x +55x -11=70068x=711x=711/68≈10.456, which is not integer.Similarly, m=4:y=4x -113x +11*(4x -1)=70013x +44x -11=70057x=711x=711/57=12.473..., not integer.m=7:y=7x -113x +11*(7x -1)=70013x +77x -11=70090x=711x=711/90=7.9, not integer.So, indeed, only m=6 gives integer x=9, y=53.Therefore, the answer is (C)6.But wait, the options include (E) as well, so maybe I need to check if there's another way.Alternatively, perhaps I can consider that 13x +11y=700, and y=mx -1.So, substituting, 13x +11(mx -1)=700Which is 13x +11mx -11=700(13 +11m)x=711So, x=711/(13 +11m)So, x must be integer, so 13 +11m must divide 711.As before, divisors of 711 are 1,3,9,79,237,711.So, 13 +11m must be one of these.So, solving for m:13 +11m=79: m=613 +11m=237: m=(237-13)/11=224/11≈20.36, not integer.13 +11m=711: m=(711-13)/11=698/11≈63.45, not integer.So, only m=6.Therefore, the answer is (C)6.I think that's correct. So, I'll go with option (C)6.

answer：Okay, so I have this problem where I need to form four-digit numbers using the numbers 1 to 6. The catch is I have to select two even numbers and two odd numbers, and I can't repeat any digits. There are three questions to answer, so I'll take them one by one.Starting with the first question: How many different four-digit numbers can be formed?Alright, let's break this down. First, I need to figure out how many ways I can choose two even numbers and two odd numbers from the set 1 to 6.Looking at the numbers 1 to 6, the even numbers are 2, 4, 6. So that's three even numbers. The odd numbers are 1, 3, 5. That's also three odd numbers.So, I need to choose two even numbers from three. The number of ways to do that is given by the combination formula: C(n, k) = n! / (k! (n - k)!).So, for the even numbers: C(3, 2) = 3! / (2! * 1!) = (6)/(2 * 1) = 3.Similarly, for the odd numbers, I also need to choose two from three: C(3, 2) = 3.So, the number of ways to choose the numbers is 3 (for evens) multiplied by 3 (for odds), which is 9.But wait, that's just the selection part. Once I've selected the four numbers, I need to arrange them into a four-digit number. Since all digits are distinct, the number of permutations is 4! = 24.So, putting it all together, the total number of four-digit numbers is 9 * 24.Calculating that: 9 * 24 = 216.Hmm, let me double-check. So, 3 ways to choose evens, 3 ways to choose odds, 4! ways to arrange them. Yep, 3*3=9, 9*24=216. That seems right.So, the answer to the first question is 216.Moving on to the second question: Among these four-digit numbers, how many have two even numbers placed together?Hmm, okay. So, I need to count how many four-digit numbers have the two even digits adjacent to each other.I think one way to approach this is to treat the two even numbers as a single entity or "block." Then, along with the two odd numbers, we have three "items" to arrange: the block and the two odd digits.But wait, actually, the two even numbers can be arranged within the block in two different ways, right? Since they are distinct, swapping them gives a different number.So, let's formalize this.First, the number of ways to choose the two even numbers is still C(3, 2) = 3, and the number of ways to choose the two odd numbers is also C(3, 2) = 3. So, 3*3=9 ways to choose the numbers.Now, for each such selection, we need to arrange the four digits with the two even numbers together.As I thought earlier, we can consider the two even digits as a single block. So, instead of four separate digits, we have three items: the block and the two odd digits.The number of ways to arrange these three items is 3! = 6.But within the block, the two even digits can be arranged in 2! = 2 ways.Therefore, for each selection of numbers, the number of arrangements where the two even digits are together is 3! * 2! = 6 * 2 = 12.So, for each of the 9 selections, we have 12 arrangements. Therefore, the total number is 9 * 12 = 108.Wait, but hold on a second. Let me make sure I'm not overcounting or undercounting.Another way to think about this is: in the four-digit number, how many ways can the two even digits be adjacent?In a four-digit number, the two even digits can be in positions 1-2, 2-3, or 3-4. So, there are three possible positions for the block of two even digits.For each of these positions, we can arrange the two even digits in 2! ways, and then arrange the two odd digits in the remaining two positions in 2! ways.So, for each selection of numbers, the number of such arrangements is 3 (positions) * 2! (even arrangements) * 2! (odd arrangements) = 3 * 2 * 2 = 12.Which is the same as before. So, 12 arrangements per selection, 9 selections, 108 total.Therefore, the answer to the second question is 108.Now, the third question: How many four-digit numbers have two even numbers not adjacent to each other?Hmm, okay. So, this is the complement of the second question. Instead of counting the numbers where the two even digits are together, we need to count where they are not.From the first question, we know the total number of four-digit numbers is 216. From the second question, we know that 108 of them have the two even digits together. Therefore, the number of four-digit numbers where the two even digits are not adjacent is 216 - 108 = 108.Wait, that seems too straightforward. Is that correct?Alternatively, let me think about it another way. Maybe using permutations.We have four positions: 1, 2, 3, 4.We need to place two even digits such that they are not adjacent.First, let's count the number of ways to choose positions for the two even digits such that they are not adjacent.In four positions, the number of ways to choose two non-adjacent positions is equal to the total number of ways to choose two positions minus the number of adjacent pairs.Total number of ways to choose two positions: C(4, 2) = 6.Number of adjacent pairs: as before, positions 1-2, 2-3, 3-4. So, 3 adjacent pairs.Therefore, the number of non-adjacent pairs is 6 - 3 = 3.So, for each selection of numbers, the number of ways to arrange the two even digits in non-adjacent positions is 3.But wait, for each such pair of positions, the two even digits can be arranged in 2! ways, and the two odd digits can be arranged in the remaining two positions in 2! ways.Therefore, for each selection of numbers, the number of arrangements is 3 (non-adjacent position pairs) * 2! (even arrangements) * 2! (odd arrangements) = 3 * 2 * 2 = 12.Wait, that's the same as before. So, 12 arrangements per selection, 9 selections, 108 total.So, that corroborates the previous result.But wait, hold on. If I use the total number of arrangements (216) minus the number of arrangements where evens are together (108), I get 108. But when I calculated using the non-adjacent positions, I also got 108. So, that seems consistent.But wait, is that correct? Because in the second question, we had 108 numbers where the two even digits are together, and here, 108 where they are not. So, 108 + 108 = 216, which is the total. That seems correct.Wait, but let me think again. Is there a different way to compute this? Maybe using inclusion-exclusion or something else.Alternatively, think of arranging the two odd digits first, then placing the even digits in the gaps.So, arranging the two odd digits: since they are distinct, the number of ways is 2! = 2.Once the two odd digits are arranged, they create three possible gaps where the even digits can be placed: before the first odd, between the two odds, and after the second odd.So, for example, if the odd digits are O1 and O2, the gaps are: _ O1 _ O2 _.We need to choose two gaps out of these three to place the even digits. The number of ways is C(3, 2) = 3.For each such choice, the two even digits can be arranged in 2! = 2 ways.Therefore, the number of ways is 2 (arrangements of odds) * 3 (choices of gaps) * 2 (arrangements of evens) = 12.Again, same result. So, 12 per selection, 9 selections, 108 total.Therefore, regardless of the method, the number is 108.So, the answer to the third question is 108.Wait, but hold on. Let me just think if there is any mistake here.Wait, in the first question, we have 216 total numbers. Then, in the second question, 108 have the two even digits together, and the third question, 108 have them not together. So, 108 + 108 = 216, which is correct.But sometimes, in combinatorics, when you have two overlapping conditions, you have to be careful. But in this case, the two cases are mutually exclusive and cover all possibilities: either the two even digits are together or they are not. So, their counts should add up to the total.Therefore, I think 108 is correct for both the second and third questions.Wait, but hold on, in the second question, the two even digits are placed together, and in the third, they are not. So, both are 108. That seems a bit surprising, but considering the symmetry, maybe it's correct.Alternatively, let's think about the number of ways where the two even digits are together versus not. Since the total is 216, and the two cases are equal, each is 108.Alternatively, maybe I can think of the probability. The probability that two even digits are together is 108/216 = 1/2. Similarly, the probability they are not is also 1/2. That seems plausible.Alternatively, let's think of arranging the four digits. The number of ways where the two even digits are adjacent is equal to the number of ways where they are not adjacent, which is 108 each.So, yeah, I think that's correct.Therefore, summarizing:1) Total four-digit numbers: 2162) Numbers with two even digits together: 1083) Numbers with two even digits not adjacent: 108So, all three answers are 216, 108, and 108.**Final Answer**1) boxed{216}2) boxed{108}3) boxed{108}

answer：To determine the probability that Alex sings on Saturday, let's denote the following probabilities:- ( P(Sat) ) = Probability that Alex sings on Saturday- ( P(Sun) ) = Probability that Alex sings on Sunday- ( P(Sun | neg Sat) ) = Probability that Alex sings on Sunday given that she does not sing on SaturdayFrom the problem, we know:- ( P(Sun | neg Sat) = 0.7 )- ( P(Sun) = 0.5 )- Alex never sings on both days, so ( P(Sun | Sat) = 0 )Using the law of total probability:[P(Sun) = P(Sun | Sat) cdot P(Sat) + P(Sun | neg Sat) cdot P(neg Sat)]Substituting the known values:[0.5 = 0 cdot P(Sat) + 0.7 cdot (1 - P(Sat))]Simplifying:[0.5 = 0.7 - 0.7 P(Sat)][0.7 P(Sat) = 0.2][P(Sat) = frac{0.2}{0.7} = frac{2}{7}]Therefore, the probability that Alex sings on Saturday is ( frac{2}{7} ).

answer：Okay, so I have this algebra problem here: In the expansion of ((x + frac{1}{sqrt{x}})^6), the constant term equals (boxed{text{answer}}). Hmm, I need to find the constant term in this expansion. Let me think about how to approach this.First, I remember that when you have a binomial expansion like ((a + b)^n), each term can be found using the binomial theorem. The binomial theorem states that:[(a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n - k} b^{k}]So, in this case, (a) is (x), (b) is (frac{1}{sqrt{x}}), and (n) is 6. That means each term in the expansion will be:[binom{6}{k} x^{6 - k} left(frac{1}{sqrt{x}}right)^k]I need to simplify this term to find when it becomes a constant term. A constant term is one where the exponent of (x) is zero because (x^0 = 1), which is a constant.Let me write out the general term again:[binom{6}{k} x^{6 - k} left(frac{1}{sqrt{x}}right)^k]Simplify the (left(frac{1}{sqrt{x}}right)^k) part. Since (frac{1}{sqrt{x}} = x^{-1/2}), raising that to the (k) power gives:[left(x^{-1/2}right)^k = x^{-k/2}]So, substituting back into the term:[binom{6}{k} x^{6 - k} cdot x^{-k/2}]When multiplying exponents with the same base, we add the exponents. So:[x^{6 - k - k/2} = x^{6 - (3k/2)}]We want the exponent to be zero for the term to be constant. So set up the equation:[6 - frac{3k}{2} = 0]Let me solve for (k):[6 = frac{3k}{2}]Multiply both sides by 2:[12 = 3k]Divide both sides by 3:[k = 4]So, the constant term occurs when (k = 4). Now, let's find the value of the term when (k = 4).First, compute the binomial coefficient (binom{6}{4}). I remember that (binom{n}{k} = frac{n!}{k!(n - k)!}).So,[binom{6}{4} = frac{6!}{4! cdot (6 - 4)!} = frac{6!}{4! cdot 2!}]Calculating the factorials:- (6! = 720)- (4! = 24)- (2! = 2)So,[binom{6}{4} = frac{720}{24 cdot 2} = frac{720}{48} = 15]Okay, so the binomial coefficient is 15.Now, let's compute the rest of the term:[x^{6 - 4} cdot left(frac{1}{sqrt{x}}right)^4]Simplify each part:- (x^{6 - 4} = x^2)- (left(frac{1}{sqrt{x}}right)^4 = left(x^{-1/2}right)^4 = x^{-2})Multiplying these together:[x^2 cdot x^{-2} = x^{0} = 1]So, the term when (k = 4) is (15 times 1 = 15). Therefore, the constant term is 15.Wait, let me double-check my steps to make sure I didn't make a mistake.1. I used the binomial theorem correctly.2. I simplified the exponents correctly: (6 - k - k/2 = 6 - 3k/2).3. Solved for (k) when exponent is zero: (6 - 3k/2 = 0) leading to (k = 4). That seems right.4. Calculated the binomial coefficient: (binom{6}{4} = 15). That's correct because (binom{6}{4} = binom{6}{2} = 15).5. Simplified the term: (x^2 times x^{-2} = 1). Yep, that's correct.So, all steps check out. I think 15 is the correct constant term.Just to be thorough, maybe I can write out all the terms of the expansion and see if the 4th term is indeed 15.Wait, actually, in the expansion, the terms are from (k = 0) to (k = 6). So, the term when (k = 4) is the 5th term (since we start counting from (k = 0)). Let me compute a few terms to confirm.- (k = 0): (binom{6}{0} x^6 times left(frac{1}{sqrt{x}}right)^0 = 1 times x^6 times 1 = x^6)- (k = 1): (binom{6}{1} x^5 times left(frac{1}{sqrt{x}}right)^1 = 6 x^5 times x^{-1/2} = 6 x^{9/2})- (k = 2): (binom{6}{2} x^4 times left(frac{1}{sqrt{x}}right)^2 = 15 x^4 times x^{-1} = 15 x^{3})- (k = 3): (binom{6}{3} x^3 times left(frac{1}{sqrt{x}}right)^3 = 20 x^3 times x^{-3/2} = 20 x^{3/2})- (k = 4): (binom{6}{4} x^2 times left(frac{1}{sqrt{x}}right)^4 = 15 x^2 times x^{-2} = 15 x^{0} = 15)- (k = 5): (binom{6}{5} x^1 times left(frac{1}{sqrt{x}}right)^5 = 6 x times x^{-5/2} = 6 x^{-3/2})- (k = 6): (binom{6}{6} x^0 times left(frac{1}{sqrt{x}}right)^6 = 1 times 1 times x^{-3} = x^{-3})Looking at these terms, the 5th term (when (k = 4)) is indeed 15, which is the constant term. All other terms have exponents of (x) that are not zero, so they are not constants.Therefore, I'm confident that the constant term is 15.**Final Answer**The constant term is boxed{15}.

answer：Okay, so I have this problem here: If (1 + √2)^5 equals a + √2 b, where a and b are rational numbers, then I need to find the value of a - b. Hmm, okay. Let me think about how to approach this.First, I remember that when you have expressions like (1 + √2) raised to a power, you can expand them using the binomial theorem. The binomial theorem says that (a + b)^n is the sum from k=0 to n of (n choose k) a^(n - k) b^k. So in this case, a is 1, b is √2, and n is 5. So maybe I can expand (1 + √2)^5 using the binomial theorem.Let me write that out. So, (1 + √2)^5 is equal to the sum from k=0 to 5 of (5 choose k) * 1^(5 - k) * (√2)^k. Simplifying that, since 1 to any power is just 1, so it becomes the sum from k=0 to 5 of (5 choose k) * (√2)^k.Now, let me compute each term individually. I'll list them out for k = 0 to 5.For k = 0:(5 choose 0) * (√2)^0 = 1 * 1 = 1For k = 1:(5 choose 1) * (√2)^1 = 5 * √2For k = 2:(5 choose 2) * (√2)^2 = 10 * (2) = 20For k = 3:(5 choose 3) * (√2)^3 = 10 * (2√2) = 20√2For k = 4:(5 choose 4) * (√2)^4 = 5 * (4) = 20For k = 5:(5 choose 5) * (√2)^5 = 1 * (4√2) = 4√2Now, let me add all these terms together.First, the constant terms (without √2):1 (from k=0) + 20 (from k=2) + 20 (from k=4) = 1 + 20 + 20 = 41Next, the terms with √2:5√2 (from k=1) + 20√2 (from k=3) + 4√2 (from k=5) = (5 + 20 + 4)√2 = 29√2So putting it all together, (1 + √2)^5 = 41 + 29√2. Therefore, a is 41 and b is 29.But wait, the question asks for a - b. So that would be 41 - 29. Let me compute that: 41 - 29 is 12.Hmm, that seems straightforward, but let me double-check my calculations to make sure I didn't make any mistakes.First, checking the binomial coefficients:(5 choose 0) = 1(5 choose 1) = 5(5 choose 2) = 10(5 choose 3) = 10(5 choose 4) = 5(5 choose 5) = 1Yes, those are correct.Now, computing each term:k=0: 1 * 1 = 1k=1: 5 * √2 = 5√2k=2: 10 * (√2)^2 = 10 * 2 = 20k=3: 10 * (√2)^3 = 10 * (2√2) = 20√2k=4: 5 * (√2)^4 = 5 * (4) = 20k=5: 1 * (√2)^5 = 1 * (4√2) = 4√2Adding constants: 1 + 20 + 20 = 41. That's correct.Adding √2 terms: 5√2 + 20√2 + 4√2 = (5 + 20 + 4)√2 = 29√2. That's correct too.So, a = 41, b = 29. Therefore, a - b = 41 - 29 = 12. Hmm, seems right.Alternatively, maybe I can use another method to verify. Perhaps using recursion or another expansion technique.Wait, another thought: sometimes, expressions like (1 + √2)^n can be related to solutions of Pell's equation or something similar, but I don't know if that's necessary here. Since the binomial expansion worked, maybe it's not needed, but just to be thorough, let me see.Alternatively, I can compute (1 + √2)^5 step by step, multiplying it out step by step instead of using the binomial theorem.Let me try that.First, compute (1 + √2)^2:(1 + √2)^2 = 1^2 + 2*1*√2 + (√2)^2 = 1 + 2√2 + 2 = 3 + 2√2.Then, compute (1 + √2)^3 by multiplying (1 + √2)^2 by (1 + √2):(3 + 2√2)(1 + √2) = 3*1 + 3*√2 + 2√2*1 + 2√2*√2= 3 + 3√2 + 2√2 + 4= (3 + 4) + (3√2 + 2√2)= 7 + 5√2.Next, compute (1 + √2)^4 by squaring (1 + √2)^2:(3 + 2√2)^2 = 3^2 + 2*3*2√2 + (2√2)^2= 9 + 12√2 + 8= 17 + 12√2.Then, compute (1 + √2)^5 by multiplying (1 + √2)^4 by (1 + √2):(17 + 12√2)(1 + √2) = 17*1 + 17*√2 + 12√2*1 + 12√2*√2= 17 + 17√2 + 12√2 + 24= (17 + 24) + (17√2 + 12√2)= 41 + 29√2.So, same result as before: (1 + √2)^5 = 41 + 29√2. So a = 41, b = 29, so a - b = 12.Hmm, so both methods give me the same answer. That gives me more confidence that it's correct.Alternatively, maybe another approach: using the conjugate. Since (1 + √2) and (1 - √2) are conjugates, and their product is 1 - (√2)^2 = 1 - 2 = -1.So, maybe I can use that to find a relation.Let me denote x = 1 + √2. Then, x^5 = a + √2 b.Also, the conjugate of x is y = 1 - √2. So, y^5 = a - √2 b.If I compute x^5 + y^5, that would be (a + √2 b) + (a - √2 b) = 2a.Similarly, x^5 - y^5 = (a + √2 b) - (a - √2 b) = 2√2 b.So, if I can compute x^5 + y^5 and x^5 - y^5, I can find a and b.But since I already have x^5 = 41 + 29√2, and y^5 would be 41 - 29√2, so x^5 + y^5 = 82, which is 2a, so a = 41. Similarly, x^5 - y^5 = 58√2, which is 2√2 b, so b = 29. So, same result.Alternatively, maybe I can compute x^5 + y^5 and x^5 - y^5 without expanding, using recurrence relations.But perhaps that's overcomplicating. Since both methods give me the same a and b, I think it's safe to say that a - b is 12.Wait, just to make sure, let me compute (1 + √2)^5 numerically and see if it's approximately equal to 41 + 29√2, which is approximately 41 + 29*1.4142.Compute 29*1.4142: 29*1.4142 ≈ 29*1.414 ≈ 29*1.4 = 40.6, plus 29*0.014 ≈ 0.406, so total ≈ 40.6 + 0.406 ≈ 41.006.So, 41 + 41.006 ≈ 82.006.Now, compute (1 + √2)^5 numerically. Let's compute step by step:1 + √2 ≈ 1 + 1.4142 ≈ 2.4142.Compute (2.4142)^2: 2.4142 * 2.4142 ≈ 5.8284.Then, (2.4142)^3 = (2.4142)^2 * 2.4142 ≈ 5.8284 * 2.4142 ≈ let's compute 5 * 2.4142 = 12.071, 0.8284 * 2.4142 ≈ approx 2. So total ≈ 12.071 + 2 ≈ 14.071.Wait, but actually, 5.8284 * 2.4142: let me compute it more accurately.5.8284 * 2 = 11.65685.8284 * 0.4142 ≈ 5.8284 * 0.4 = 2.33136, 5.8284 * 0.0142 ≈ 0.0828So total ≈ 2.33136 + 0.0828 ≈ 2.41416So total of 11.6568 + 2.41416 ≈ 14.07096.So, (2.4142)^3 ≈ 14.07096.Then, (2.4142)^4 = (2.4142)^3 * 2.4142 ≈ 14.07096 * 2.4142.Compute 14 * 2.4142 ≈ 33.80.07096 * 2.4142 ≈ approx 0.1715So total ≈ 33.8 + 0.1715 ≈ 33.9715.Wait, but let's compute it more accurately:14.07096 * 2.4142First, 14 * 2.4142 = 33.80.07096 * 2.4142 ≈ 0.07096 * 2 = 0.14192, 0.07096 * 0.4142 ≈ approx 0.0293So total ≈ 0.14192 + 0.0293 ≈ 0.1712Thus, total ≈ 33.8 + 0.1712 ≈ 33.9712.So, (2.4142)^4 ≈ 33.9712.Then, (2.4142)^5 = (2.4142)^4 * 2.4142 ≈ 33.9712 * 2.4142.Compute 33 * 2.4142 = 79.66860.9712 * 2.4142 ≈ approx 2.346So total ≈ 79.6686 + 2.346 ≈ 82.0146.So, (1 + √2)^5 ≈ 82.0146.Now, 41 + 29√2 ≈ 41 + 29*1.4142 ≈ 41 + 41.0118 ≈ 82.0118.So, 82.0118 vs 82.0146. These are very close, with a slight difference due to rounding errors in the decimal approximations. So, that's a good sign.Therefore, I can be confident that a = 41 and b = 29, so a - b = 12.I think that's solid. I used two different methods: binomial expansion and step-by-step multiplication, both leading to the same result. Additionally, I verified numerically that the approximate value matches, which gives me more confidence.So, the answer should be 12.**Final Answer**The value of (a - b) is boxed{12}.

answer：Okay, so I have this problem here about quadratic graphs. Let me read it again: "On the same set of axes are drawn the graph of ( y = ax^2 + bx + c ) and the graph of the equation obtained by replacing ( x ) by ( -x ) in the given equation. If ( b neq 0 ) and ( c neq 0 ) these two graphs intersect: (A) in two points, one on the x-axis and one on the y-axis; (B) in one point located neither on axis; (C) only at the origin; (D) in one point on the x-axis; (E) in one point on the y-axis."Hmm, okay. So, we have two quadratic functions: the original one ( y = ax^2 + bx + c ) and another one where ( x ) is replaced by ( -x ), which would be ( y = a(-x)^2 + b(-x) + c ). Let me write that out:Original: ( y = ax^2 + bx + c )Reflected: ( y = a(-x)^2 + b(-x) + c = ax^2 - bx + c )So, the second equation is ( y = ax^2 - bx + c ). So, both are quadratic functions, opening upwards or downwards depending on the sign of ( a ). Since ( a ) isn't mentioned, I guess it's just a general case.Now, the question is about their intersection points. So, I need to find the points where ( ax^2 + bx + c = ax^2 - bx + c ). Let me set them equal:( ax^2 + bx + c = ax^2 - bx + c )Hmm, okay, let's subtract ( ax^2 ) and ( c ) from both sides to simplify:( ax^2 + bx + c - ax^2 - c = ax^2 - bx + c - ax^2 - c )Simplifying both sides:Left side: ( bx )Right side: ( -bx )So, ( bx = -bx )Bring all terms to one side:( bx + bx = 0 )Which is ( 2bx = 0 )Since ( b neq 0 ), we can divide both sides by ( 2b ):( x = 0 )So, the only solution is ( x = 0 ). Therefore, the two parabolas intersect only at ( x = 0 ). Now, let's find the corresponding ( y )-value when ( x = 0 ).Plugging ( x = 0 ) into either equation:( y = a(0)^2 + b(0) + c = c )So, the point of intersection is ( (0, c) ). Now, since ( c neq 0 ), this point is on the y-axis but not at the origin. So, it's a single point on the y-axis.Wait, but let me make sure. Is there any other point where they might intersect? Because sometimes when you have quadratics, they can intersect at two points. But in this case, when we set them equal, we only got one solution, ( x = 0 ). So, that suggests only one point of intersection.But let me think again. Maybe I made a mistake in simplifying. Let me go through the steps again.Set ( ax^2 + bx + c = ax^2 - bx + c )Subtract ( ax^2 ) and ( c ) from both sides:( bx = -bx )Which leads to ( 2bx = 0 ), so ( x = 0 ). So, yeah, only one solution. So, only one point of intersection at ( x = 0 ), which is ( (0, c) ). Since ( c neq 0 ), it's not the origin, but on the y-axis.So, looking at the options:(A) two points, one on x-axis, one on y-axis.(B) one point, neither on axis.(C) only at origin.(D) one point on x-axis.(E) one point on y-axis.So, the correct answer seems to be (E), one point on the y-axis.Wait, but let me think again. Is there a possibility that they could intersect somewhere else? Maybe if ( a ) is zero? But no, ( a ) is the coefficient of ( x^2 ), so if ( a ) were zero, it wouldn't be a quadratic anymore. So, ( a ) is non-zero because it's a quadratic.Also, ( b neq 0 ) and ( c neq 0 ). So, we can't have ( b = 0 ), which would make the two equations identical, but since ( b neq 0 ), they are different.So, yeah, I think (E) is correct.Wait, but hold on. Let me visualize this. The original parabola is ( y = ax^2 + bx + c ), and the other one is its reflection over the y-axis, since replacing ( x ) with ( -x ) reflects over the y-axis.So, two parabolas, one is the reflection of the other over the y-axis. So, they both open in the same direction because ( a ) is the same. So, their vertices are symmetric with respect to the y-axis.So, if you have two parabolas opening upwards, symmetric over the y-axis, they should intersect at the vertex point on the y-axis, which is ( (0, c) ). But wait, is that the only intersection?Wait, actually, if you have two parabolas that are symmetric over the y-axis, they might intersect at two points: one on the y-axis and another somewhere else. Hmm, but according to my algebra earlier, they only intersect at one point.Wait, maybe I need to think about this more carefully. Let me consider specific examples.Take ( a = 1 ), ( b = 1 ), ( c = 1 ). So, the original equation is ( y = x^2 + x + 1 ), and the reflected one is ( y = x^2 - x + 1 ).Set them equal:( x^2 + x + 1 = x^2 - x + 1 )Simplify: ( x = -x ), so ( 2x = 0 ), so ( x = 0 ). Then, ( y = 1 ). So, only one intersection at (0,1). So, that's on the y-axis.Another example: ( a = 1 ), ( b = 2 ), ( c = 3 ). Original: ( y = x^2 + 2x + 3 ); reflected: ( y = x^2 - 2x + 3 ).Set equal: ( x^2 + 2x + 3 = x^2 - 2x + 3 )Simplify: ( 4x = 0 ), so ( x = 0 ), ( y = 3 ). Again, only one intersection on the y-axis.Wait, maybe if ( c = 0 ), then the intersection would be at the origin. But in the problem, ( c neq 0 ), so the intersection is at ( (0, c) ), which is on the y-axis but not the origin.So, in all these cases, only one point of intersection, on the y-axis.Therefore, the answer should be (E).But wait, let me think again. Maybe if the parabolas are wider or narrower, but since they are symmetric, they can only intersect at one point on the y-axis. Because if you have two parabolas symmetric over the y-axis, their only common point is on the y-axis.But wait, is that always the case? Suppose one is shifted up and the other is shifted down, but in this case, both have the same ( c ). So, they both have the same y-intercept. So, they both pass through ( (0, c) ). But since they are symmetric, they don't cross each other elsewhere.Wait, but if they are symmetric, wouldn't they cross at two points? Hmm, maybe not. Because if you have two symmetric parabolas, they only meet at the axis of symmetry, which is the y-axis here.Wait, let me think of another example. Suppose ( a = 1 ), ( b = 1 ), ( c = 0 ). Then, original is ( y = x^2 + x ), reflected is ( y = x^2 - x ). Setting equal: ( x^2 + x = x^2 - x ), so ( 2x = 0 ), so ( x = 0 ). So, intersection at (0,0). So, in this case, it's the origin. But in the problem, ( c neq 0 ), so it's not the origin, but on the y-axis.So, yeah, in all cases where ( c neq 0 ), the intersection is at ( (0, c) ), which is on the y-axis.Therefore, the answer is (E).Wait, but the options are:(A) two points, one on x-axis and one on y-axis.(B) one point located neither on axis.(C) only at the origin.(D) one point on the x-axis.(E) one point on the y-axis.So, (E) is correct.But wait, hold on. Let me think about another case where ( a ) is negative. Suppose ( a = -1 ), ( b = 1 ), ( c = 1 ). So, original: ( y = -x^2 + x + 1 ); reflected: ( y = -x^2 - x + 1 ).Set equal: ( -x^2 + x + 1 = -x^2 - x + 1 )Simplify: ( 2x = 0 ), so ( x = 0 ), ( y = 1 ). So, again, only one point on the y-axis.So, regardless of the value of ( a ), as long as ( b neq 0 ) and ( c neq 0 ), the two parabolas only intersect at one point on the y-axis.Therefore, the answer is (E).**Final Answer**boxed{E}