divides a*b. Then either c divides a or c divides b.\nFirst, we tried a bunch of examples, like when c=3 and ab=27, so a =3 and b=9, and you see that c divides a and c divides b.  Or another one is when c=7 and ab=56, so a=7 and b=8, and we know that 7 divides 56 and also 7 divides 7. Also, if c=12 and a=5 and b=24, then ab=120, and you see that c divides 120, and slo c divides b, which is all we need because the conjecture is an "or" statement so we only need one of the conditions to be true.  So we decided that the conjecture is true. Proof: Let a,band c be integers. We will prove the conjecture by contrapositive. So we assume that c doesn't divide a and that c doesn't divide b. We will show that c doesn't divide a*b. Since c doesn't dived a, then we know that a doesn't = c*m. Also, c doesn't divide b, so we know that b doesn't = s*n. So ab doesn't = (cm)(cn)=c(cmn). Now, cmn is an integer q(since c,m and n are all integers).Thus ab doesn't = cq for any int q. So c doesn't divide ab.
137705
Well, the basic problem you've run into is that the conjecture isn't true.  If a = 6, b = 3, and c = 9, then a*b = 18, and 18/9 = 2, but neither 6 nor 3 is divisible by 9.\n\nTo prove that something is true you must prove it is true for all possible values of the variables, in this case a, b, and c.  However, to prove something is false, you need only supply a single counter-example.  Thus the above example disproves the conjecture that for all integers a, b, and c if a*b is divisible by c then a or b is also divisible by c.\n\nNote that a ≠ cm and b ≠ cn does not imply that ab ≠ c(cmn).  For example, if a = 6, b = 8, c = 2, m = 4, and n = 3, then cm = 2*3 = 8 ≠6 so cm ≠ a, and cn = 2 * 3 = 6 ≠ 8 so cn ≠ b.  However, (cm)(cn) = 2 * 4 * 2 * 3 = 48, and a * b = 6 * 8 = 48, so ab = (cm)(cn).  This proves that you cannot multiply not-equalities (i.e. statements with "doesn't equal).\n\nAlso, concluding that if c does not divide a or b implies that c does not divide ab is not the same as concluding that if c does divide ab it must also divide a or b.  This logical fallacy is equivalent to the statement that for all plants, not a tree implies not a rock, so being a rock implies being a tree.  The first statement is true: all plants that are not trees are also not rocks.  However, the second statement is clearly false, as no rocks are also trees.
188714
0 is the additive identity, which means that:\n\na + 0 = a (which implies that a - a = 0)\n\nMultiplying both sides by b gives:\n\nab + 0b = ab\n\nSubtracting ab from both sides gives.\n\n0b = ab - ab\n\nBut because a - a = 0, we also know that\n\n0b = ab - ab = 0\n\nSo,\n\n0b = 0\n\nMultiplying by 0 is straightforward.\n\nHowever,\n\nIf 0/0 = a\n\nThen 0 = 0a.\n\nThis is true for all a, so 0/0 is not well-defined.\n\nIn your example, suppose you had:\n\n1 / 1 = 1\n.1 / .1 = 1\n.01 / .01 = 1\n.001 / .001 = 1\n\nAs you approach dividing 0/0 this way, the number always stays the same.  There are really many ways to approach dividing 0/0.
506147
It looks like you are trying to solve a literal equation, which is an equation with more than one variable, and you have to solve for a specified variable.  You can use the SAME rules to solve this kind of equation as you do with regular equations (those with numbers).  For example, if you had 8=2(b)(9), you would divide by 2 and then again by 9 (which is like dividing by 18).  \n\nIf your problem is a=(1/2)bc^2, then you would divide by 1/2 and then again by c^2.  Remember that to divide by a fraction, you multiply by the reciprocal.
483282
Lim (as x --> 1) of (x^4 - 3x^3 + 2) / (x^3 - 5x^2 + 3x + 1)\n\nNotice that when x = 1, both the numerator and the denominator equal zero. Therefore, (x - 1) must be a factor of both terms.\n\nDividing out this factor in both terms gives :\n\n(x^3 - 2x^2 - 2x - 2) / (x^2 - 4x -1)\n\nNotice now, that when x = 1, both the numerator and the denominator do not equal zero. Therefore, there are no more factors in either term that equals (x - 1).\n\nSo now we can let x = 1 and substitute. This gives :\n\n[1^3 - 2(1^2) - 2(1) - 2] / [1^2 - 4(1) - 1]\n= -5 / -4\n= 5 / 4, which is the limit as x --> 1.
167949
As the terms you give are clearly an arithmetic progression, with the difference between successive terms incrementing by 2 each time\n\n7 - 4 = 3\n12 - 7 = 5\n19 - 12 = 7\n28 - 19 = 9\n\nIt is clear that the next few terms are 39, 52, 67, 84, 103, 124 etc.\n\nI recognise this as rather similar to 36, 64, 81, 100, 121 because I am familiar with the squares of the positive integers having the property of being in Arithmetic Progression, with the difference incrementing by 2 (adding a constant, c, won't alter that property), so I can see where this is heading, but let's work the simultaneous equations through to prove what my observation and intuition is telling me ...\n\nThere are values of a b c to be found for which, with n increasing by 1 each time\n\na + b + c = 4 (1)\n4a + 2b + c = 7 (2)\n9a + 3b + c =12 (3)\n\nsubtracting (1) from (2)\n\n3a + b = 3 (4)\n\nsubtracting (2) from (3)\n\n5a + b = 5 (5)\n\nSubtracting (4) from (5)\n\n2a = 2\na = 1\n\nThen, substituting in (4),  b = 0 and substituting in (1) c = 3\n\nAnd it can be seen that this is indeed a sequence of the squares of the positive integers, with 3 added to them, each time.\n\nT(n) = n^2 + 3
498016
The only case that I could think of that A is not equal to B where AC=BC is when C=0.  Other than that A must equal B for the expression to be valid
33423
i got it!\n\nit's 3(2c+7)(c-4d)(c+d)!!!!\n\n6c^3 + 21c^2 - 18dc^2 - 63cd - 24cd^2 - 84d^2\nfirst we take out factor 3:\n3(2c^3 + 7c^2 - 6dc^2 - 21cd - 8cd^2 - 28d^2)\nthen we take out (2c+7) from each pain inside the parenthesses:\n3[(2c+7)c^2 - (2c+7)3cd - (2c+7)4d^2)]\nwe take (2c+7) out:\n3(2c+7)[c^2 - 3cd - 4d^2]\nwe factor [c^2 - 3cd - 4d^2] trinomially:\nc^2 - 3cd - 4d^2 = (c-4d)(c+d)\nthen we get:\n3(2c+7)(c-4d)(c+d)\n\nam i right??\n\n:) yeah i didn't believe myself...\nbut if u really write down all your factors, it's much more easier to take them out. and it WAS fun :))
483354
I enjoy algebra, and this is easily my favorite type of problem.  So I hope you, too, one day enjoy them for their inner beauty.\n\nFor now, let's tackle this "factoring problem."  We may wish to use the method of "synthetic division" which is arguably easier than "long division of polynomials."\n\nThink back to grade school where you learned that 6 = 2 * 3.  This simple statement enabled you to 'deconstruct' the number 6 as either "two times three" or "three times two".  Even if you forgot one of the 'factors' you could conjure it up again quickly by asking yourself, "What is six divided by two?" for example.\n\nIn problem (a), you are essentially asking, "What are the coefficients of P(x) if the only factor we know is (x-1)?"\n\nLet's set up synthetic division (see link below), by placing all the coefficients of P(x), and using 1 (the zero implied by (x-1)) as the divisor:\n\n1 ||  a  b  c  d\n___________\n.....a\n\n1 ||  a  b  c  d\n_____a_____\n.....a (a+b)\n\n1 ||  a  b        c       d\n_____a___(a+b)____\n.....a (a+b) (a+b+c)\n\n1 ||  a  b        c                d\n_____a___(a+b)___(a+b+c)_\n.....a (a+b) (a+b+c) (a+b+c+d)\n\nNote that this last term must equal 0 since it is assumed that (x-1) is a factor ... thus, the "remainder" term must equal zero!\n\nThere's the result you were after:  a+b+c+d = 0\n\nHope this helps you tackle the others!
657017
Wow, what an interesting question. I'm surprised no one has solved it yet, but that may be a consequence of the fact that no one has understood it yet. Clearly, schools need to focus more on reading comprehension ^_^.\n\nTo answer your question - not only can we prove that there exist integers a and b satisfying your assumptions, but we can even show a method of finding them. The proof is long, so sit back, relax, grab your favorite drink, and prepare to read.\n\nFirst, consider that any solutions of these equations must obey the quadratic formula: x=(-b±√(b²-4ac))/2a. Substituting the coefficients of the  actual equations in, we find that the solutions of the first equation are (-p±√(p²-4q))/2 and the solutions of the second equation are (-p±√(p²+4q))/2. Let us consider only the greater of the solutions in each pair - that is, replace each ± with a +. The difference between these solutions is then (-p+√(p²-4q))/2 - (-p+√(p²+4q))/2. Call this number K. We can then perform some simple algebra as follows:\n\n(-p+√(p²-4q))/2 - (-p+√(p²+4q))/2=K\nmultiply both sides by 2 and simplify\n√(p²-4q)-√(p²+4q)=2K\nsquare both sides\n(p²-4q) - 2√((p²-4q)(p²+4q)) + (p²+4q)=4K²\nsimplify\n2p²-2√(p^4-16q²)=4K²\ndivide both sides by 2\np²-√(p^4-16q²)=2K²\nsubtract 2K² and add √(p^4-16q²) to both sides\np²-2K²=√(p^4-16q²)\nsquare both sides\np^4-4K²p²+4K^4=p^4-16q²\nmultiply both sides by -1 and simplify\n16q²=4K²p²-4K^4\ndivide both sides by 4\n4q²=K²p²-K^4\nadd K^4 to both sides\nK^4+4q²=K²p²\ndivide both sides by K² and rearrange\np²=K²+4q²/K²\n\nThus, if we let a=K and b=2q/K, then p²=a²+b² as required by the problem. All that remains is to show that a and b are nonzero integers.\n\nLemma 1: K is an integer\nProof: All solutions of both quadratic equations are integers. K is the difference of two of these solutions - that is, K is the difference of two integers. Therefore, K is an integer.\n\nLemma 2: K is nonzero\nProof: K is the difference between a solution of the first equation and a solution of the second equation. Suppose K=0. Then these equations have a common solution - that is, for some x, x²+px+q=0=x²+px-q. This would imply that q=-q. This can only be true if q=0. But q≠0. Therefore, the equations do not have a common solution and K≠0.\n\nLemma 3: 2q/K is an integer\nProof: Earlier we showed that √(p²-4q)-√(p²+4q)=2K. We can perform some simple algebra as follows:\n\n√(p²-4q)-√(p²+4q)=2K\nmultiply both sides by (√(p²-4q)+√(p²+4q)) and simplify\n2K(√(p²-4q)+√(p²+4q)) = (p²-4q)-(p²+4q)\nsimplify further and rearrange:\n-8q=2K(√(p²-4q)+√(p²+4q))\ndivide both sides by -4:\n2q=-K(√(p²-4q)+√(p²+4q))/2\ndivide both sides by K:\n2q/K=-(√(p²-4q)+√(p²+4q))/2\n\nNote that 2K=√(p²-4q)-√(p²+4q) Adding 2√(p²+4q) to both sides of that identity gives 2K+2√(p²+4q) = √(p²-4q)+√(p²+4q). We can then substitute to get:\n\n2q/K=-(2K+2√(p²+4q))/2\nSimplifying at last yields:\n2q/K=-K-√(p²+4q)\n\nWe already know K is an integer, so if √(p²+4q) is also an integer, then 2q/K is the difference of two integers and therefore an integer. Now we know that the solutions of the second equation are (-p+√(p²+4q))/2 and (-p-√(p²+4q))/2. The difference between them is √(p²+4q). However, since both of the solutions to the second equation are integers, their difference must be an integer. Therefore √(p²+4q) is an integer, and by extension, 2q/K is an integer. Q.E.D\n\nLemma 4: 2q/K is nonzero\nWe know 2q/k=-K-√(p²+4q). Suppose 2q/K=0. Then -K-√(p²+4q)=0 From this we would get:\n\n-K-√(p²+4q)=0\nadd K to both sides and rearrange:\nK=-√(p²+4q)\nmultiply both sides by 2\n2K=-2√(p²+4q)\nremembering 2K=√(p²-4q)-√(p²+4q), we substitute:\n√(p²-4q)-√(p²+4q)=-2√(p²+4q)\nadd √(p²+4q) to both sides and rearrange:\n-√(p²+4q)=√(p²-4q)\n\nHowever we specified in the beginning t
532421
a+b-c=0\nDivision by a+b-c is division by zero, an "illegal" operation.\n\nI think that the "proof" is saying that the only way to make the process true is if a+b=0.  The "proof" stands apart from the assignment of values (a=4000, b=6000).
621622
your equation (1) a^(n) = (c+b)(e-d) is fine but e-d cannot be an integer since a^(n)/(c+b) is not an integer whereas a^(n)/(c-b) is an integer. In (1) if you select e and d to be c and b then you have a^(n) = c^2-b^2 but this is not adressing the original fermat equation. What you need to do is rewrite your argument in a more complete step-by-step form so it is perfectly clear what you are saying with respect to using the difference of squares. I realize you can't do that here with the limited sapce allowed. As it is, it is not clear or rigorous. \n\nYou have a creative flair but  if a^(n)= c^(n)-b^(n) where n is odd (in fact n is an odd prime) then a^(n) = (c^(n/2)-b^(n/2))(c^(n/2)+b^(n/2)). Although a^n or any odd number can be expressed as the difference between 2 squares: (2^2)-(1^2) = 3; (3^2)-(2^2) =5; (4^2)-(3^2)= 7 etc. the values of any intergers q,r such that a^(n) = q^2-r^2 cannot be substituted for c and b. In fact if a^(n)= c^(n)-b^(n) then c-b divides a^n. In fact a^n = (c-b)(c^(n-1)-bc^(n-2)+b^2c^(n-3)-........). From your c=x+e and b=x+d it follows (c-b) =e-d = what I've written in brackets and we don't have a difference of squares. \nA better elementary approach may be to consider the symmetric equation a^(n)+b^(n) = c^(n)+d^(n). For example 12^3+1^3 = 9^3+10^3 = 1729 and use the mutual divisibility properties of both sides. For larger n perhaps it may be shown that differences between the two sides may have a lower bound as a function of n, which would imply that 3 terms can only get so close. Just a thought.\nBut I encourage you to keep working on things you are interested in because you learn, and who knows you may be able to prove some things that have not been proven.
125236
let a = b  \nMultiply both sides by a \na² = ab  \n Add (a² - 2ab) to both sides \na² + a² - 2ab = ab + a² - 2ab  \n Factor the left, and collect like terms on the right \n2(a² - ab) = a² - ab  \n Divide both sides by (a² - ab)\n2 = 1\n(The place where the proof goes wrong is the very last step.  If a² - ab = 0, then you would be dividing by zero, which is undefined.)
370958
To see what conjecture they are interested in, it really is necessary to consider a number of examples, exactly as they said. So lets set up a table. We will call the product in question N.\nRemember that we are only interested in prime numbers for p.\n\np               N                N/p\n---------------------------------\n2                    2               1\n3                    3               1\n5                  25               5\n7                721           103\n11      3628801     329891\n13  479001601 36846277\n\nNotice that the numbers N so far have all been divisible by p. So, if we represented  them on a p-clock, they would all be p o'clock (or 0 o'clock if you prefer). That is the conjecture they are interested in. You are supposed to conjecture that the value of 1*2*3*...*(p-1)+1 in p-hour arithmetic is always p o'clock.
35254
listen the origine of this is difficult but if want to know contact me\n\nI can tell how you can get them:\nAny year that can satisfy 2 of this 3 conditions (1&2) or\n(1&3) and fail in the third\n1) the year must be divisible by 4\n2) the year must NOT be divisible by 100\n3) the year must be divisible by 400\n\nExample: 1996 this satisfy condtion 1 and 2 >> 1996 is leap\nExample: 2000 this satisfy condtion 1 and 3 >> 2000 is leap\nExample: 1900 this satisfy condtion 1 only >> 1900 is not leap [[[1900 is divisible by 4 and  it is not leap because it is divisible by 100 and not by ]]]\n\nWhy? you can contact me rarebonheur@yahoo.com
406747
ABCDEFGHJ = 256973148\n\n\nFirstly...\nC - B = 1.......C = B + 1\nH - G = 3.......H = G + 3\n\nThe equation can be simplified as:\n\nA * [ (2B + 1)^(D - E) - F^(G(G+3)) ] / J = 10\n\n1. The difference between the numbers within the square brackets cannot be more than 100 since dividing by J, which is a single digit will not give 10 (assuming that A was 1). Therefore both numbers are supposed to be weighted similarly.\n\n2. Consider the value of 'A'. If the difference were even, then it must be multiplied by 5 before dividing in order to get 10. If it were 5, then 'A/J' has to scale it down accordingly. If it were odd, there are no possible values 'A' can take. Thus the difference can be either be even or a multiple of 5.\n\n3. I assume F is an odd number! The reasoning behind this is that the difference between the 2 numbers can be even [Exception is if the difference were an odd multiple of 5. If it doesn't work then consider the other alternative]. Since (2B + 1) raised to any power is odd, then F must be odd to obtain an even difference.\n\n4. Look at all powers that have values in the range of 1 - 256, since I assume that the values are small and close in value. I consider 1^2 to 15^2, 1^3 to 6^3, 1^4 to 4^4 and 1^5 to 3^5. Look at those with odd bases [also easier to start off with since trial and error must take place now].\n\n5. I obtain and test 11^2 - 3^4 = 121 - 81 = 40. From this, (2B+1) = 11-->B = 5, D-E=2, F = 3 and G^2 + 3G = 4 --> G=1\n\nusing 40...A = 2 and J = 8 in order to get a solution of 10\n\nIf G = 1 then H = 4 from original equation\nIf B = 5 then C = 6 from original equation\n\nThe remaining numbers are 7 and 9, which allows D = 9 and E = 7 to satisfy the final equation of unknowns [D-E=2]\n\nI hope you understand and it helps!
651307
The set {1, 2, 3, ...,} has been closed under multiplication, because any product belongs to that set.\n24 and 25 belong to {1, 2, 3, ...,}. 24 x 25 = 600 as well.\n\nThe identity of the multiplication is 1, because n x 1 = 1 x n = n.\n\nAll numbers n have the inverse 1/n except n=0.\n\nAll the precious answers have no sense.
510108
slider is correct.  \n\nRemember the distributive property?\na(b+c) = ab + ac\n\nThis is using it in reverse.  \n\nIn part A, notice that there is an a in each term, so you can factor out the a's.  \n3a² + 2a = (3a)*a + 2 * a = a(3a + 2).\n\nIn part B, there aren't any letters in common but you can factor out numbers too.  \n12d²+6 = 6*2d² + 6*1 = 6(2d²+1)\n\nThe other parts use similar reasoning.