Congruence Conjectures?.....Please Help!!!!?
830270
a) not necessarily...\nb)if the bottom contains n cells, the number of stories is also n...\nand number of cards is 2n+3(n-1)+3(n-2)....+3(n-(n-1))....\nor 2n + 3+ 3(2) + 3(3)...3(n-1)  (writing from back...)\nthe second is a arithmetic series,with a =3 and d=3\nthe nth term is (3 +(n-1)*3)\nsince the formula for the sum upto n-1th term is \ns = [(n-1)*(2a+ (n-2)d)]/2 = [(n-1)*(6+(n-2)*3]/2 \n  = [3n^2 - 3n ]/2\nsummation of 2n and this is ...\n[4n + 3n^2 -3n]/2  =[ 3n^2 + n]/2 \n but you know total cards is 104.\nor    [3n^2 + n]/2  = 104\nor    3n^2 + n   = 208\n  3n^2 + n-208 = 0\nsolve this quadratic in n to get number of stories...\nwe get n =  8.16  or approx 8\nso we get number of cards used is\nfor bottom  = 2*n = 16\n1st storie  =  3* (n-1) = 21\n2nd storie  = 18\n3rd storie  = 15\n4th = 12\n5th = 9\n6th =6\n7th = 3\nadding  3+6+9+12+15+18+21+16, we get, 100...\nthus 4 cards go to waste for making the pyramid which has 8 stories (including bottom ) and uses 100 cards.\nyaaaaaaaaay  :D
294107
>> Prove that for each positive integer n\n>> there exist n consecutive positive integers\n>> none of which is an integral power of a\n>> prime number.\n\nFor example, I challenge you with n=5 and\nyou respond 16, 17, 18, 19, 20 and I win\nbecause 16 = 2^4, 17 = 17^1, 19 = 19^1\nare all integral powers of a prime.  But if\nI challenge you with n=3 and you reply with\n33, 34, 35 then you win because none of\nthose are prime powers.\n\nA proof (or winning strategy for the second\nplayer) is that given n, you can choose 2n\nprimes greater than n and use the CRT\n(Chinese Remainder Theorem) to find a\npositive integer x that simultaneously\nsatisfies\n\nx == -1 modulo prime_1\nx == -1 modulo prime_2\nx == -2 modulo prime_3\nx == -2 modulo prime_4\n[...]\nx == -n modulo prime_(2n-1)\nx == -n modulo prime_(2n)\n\nThen each of the n consecutive positive\nintegers x + 1, x + 2, ..., x + n is\ndivisible by at least two primes and\nso is not a power of a prime.
897429
This is the third time that this question has been asked!?!?!\n\nIt's a tough one, too.\n\nI'm taking it to work with me and asking my colleges to help me out on it.\n\nThis is as far as I got:\n(2B+1)^(D-E) - F^(3G+G2) = (10J)/A\n\nBut someone "did" crack this CHALLENGE Question.\n\nSo here's the answer. . .\n\nhttp://answers.yahoo.com/question/index;_ylt=AvnsMpyNhtwrHV1JVMvXrE4jzKIX?qid=20061130045858AAB57A7\n\nANSWER:  ABCDEFGHJ = 25,693,148\n\nA=2\nB=5 \nC=6\nD=9\nE=7\nF=3\nG=1\nH=4\nJ=8
500368
The links below have good info. The first briefly addresses your first question, and the second is a good brief history of the problem which provides some insight as to the difficulties encountered. I actually played around with this problem a little on an elementary level and it is interesting. Cheers.\n\nhttp://www.sciam.com/print_version.cfm?articleID=000DBA55-B698-1C71-9EB7809EC588F2D7\n\nhttp://www.geocities.com/fermatnow/flt/
892068
Well, you got me stumped!\n\n-------------------------------------\n\nI started working it out and got pretty far.\n\nThe above given information might be incorrect.  Maybe?  I don't know.\n\nI tried it several times.  I even double and triple checked my work!\n\n---------------------------------------------\n\nThis is as far as I got and then I got stumped:\n(D-E) (B+C) = (10J + AFGH) / A\n\n------------------------------------------------------\n\nGreat question!\n\nI look forward to the continuing discussion.\n\nI love the way you worded your question.  Right on!!!\n\nKeep those challenge questions coming!  ;o}
403642
just remeber, if you want to know if any number is divisible by 4 just check to see if the last two digets are divisible by 4... that should greatly help in solving this problem\n\nNow if you would please, stop asking all of your homework questions
55179
hold on let me look at that for a minute....(dya hate math like this as much as i do?)\n\n(1,4) is one for sure\nyea thats' the only one!!!
385685
Consider the numbers\nA1,\nA1+A2,\nA1+A2+A3\n..\nA1+A2+A3+...+An.\n\nDo the computations modulo n. If none of these is divisible by n, then we have n numbers and n-1 non-zero residue classes (mod n), so there are two of these numbers which are equivalent (mod n) by the pigeon hole principle. But then subtracting gives a sum of the required form that is 0 (mod n), in other words, so that n divides that sum.\n\nLooking at the proof, there is no need for the sequence to be monotonic.
1186148
There used to be a lot of them, but with computers these days they are being solved by brute force. Examples include Fermat's Last Theorem and the Four Color problem in topology. You can look those up to see what they are, and you will probably find sites listing other unsolved mathematical puzzles.
749939
Nobody can help you because it's false.\n\nLet a1 = 1, a2 = 4, b1 = 2, b2 = 3, and n = 7\nThen a1 + a2 = 5, b1 + b2 = 5\nBut a1*a2 = 4, while b1*b2 = 6.\n\nYou are talking about modular arithmetic, right?
593943
1) The sum of the positive 3-digit odd integers = 101 + 103 + 105 + ... + 995 + 997 + 999.  Adding the first of these to the last, we get 101 + 999 = 1100.  Adding the second of these to the next to last, we get 103 + 997 = 1100.  This pattern keeps repeating until 549 + 551 = 1100.  The question then becomes, how many of these 1100's do we have?  We have one for 101, 103, 105, ..., 549 (the later numbers share their 1100's with these).  How many numbers are in the set 101, 103, 105, ..., 549?  Well, there are 449 numbers in 101, 102, 103, ..., 549.  Since there is one more odd number than even number, we must have 225 odd numbers and 224 even numbers.  Thus, there are 225 odd numbers in the set 101, 103, 105, ..., 549.  Thus, we have 225 1100's to add together.\n\n225 * 1100 = 247500.\n\n\n\n2) This time we wish to add the numbers of the set 10, 11, 13, 14, ..., 94, 95, 97, 98.  Adding the first to the last we get 10 + 98 = 108.  Adding the second to the second to last we get 11 + 97 = 108.  The pattern keeps repeating until we get 53 + 55 = 108.  How many 108's do we have?  Well, we have one for 10, 11, 13, 14, 16, 17, ..., 52, 53 (again, the later numbers share their 108's with these).  How many numbers are in this set?  In the set 10, 11, 12, 13, ..., 52, 53, there are 44 numbers.  The multiples of three range from 12 to 51--from 3*4 to 3*17.  Thus, there are 14 multiples of three in this range.  This means that there are 30 numbers in the set 10, 11, 13, 14, 16, 17, ..., 52, 53.  Thus, we have 30 108's to add together.\n\n30 * 108 = 3240.\n\n\n\n3) Here, I am assuming that the raise of 10% also applies to the previous year's raise and not just to the "principal" of $18000.  Thus, I am assuming a compound raise instead of 5 simple raises.\n\nDuring the first year, you make $18000.  During the second year, you make 18000 * 1.1 = $19800.  During the third year, you make 19800 * 1.1 = $21780.  During the fourth year, you make 21780 * 1.1 = $23958.  Finally, during the fifth year, you make 23958 * 1.1 = $26353.80.\n\nGood luck.
476679
First do this for d=gcd(a,b)=1. Subtracting the equations for the solutions (x_0,y_0) and (x,y), get that a(x-x_0)+b(y-y_0)=0. If gcd(a,b)=1, this shows that a divides y-y_0, i.e. y=y_0+at. Conclude that x=x_0+bt.\n\nNow see what happens if gcd(a,b)=1 and d is not 1. Then do the general case.
1065017
I'm writing up a proof right now that this is not possible.\n\nCheck out the pdf at:\nhttp://www.geocities.com/euleratuo/
301463
You r wrong, Not for every positive number... Such as your referance is not true for 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53...its a long list... you cant sum them up using 2 prims...\nbut its true for all even number(because no such even number has found yet) and Goldbach's Conjecture says so..\n**Goldbach's Conjecture: For any even number n greater than or equal to 4, there exists at least one pair of prime numbers p1 and p2 such that n = p1 + p2.\n\noh now I see u also ment even Integer.. I am sorry..\n\nand 2nd part...\nif all even number greater than 3 is sum of 2 primes..\nthen all even number greater then 6(not 5)is sum of 3 primes..\nthat means n = 5+p;\nbut 5 = 2+3;\nso, n = 2+3+p
943730
This is Goldbach's conjecture and has NOT been proven\ntrue or false.  Good luck solving it!!\n\nI looked at the above answers, adn they dont seem to understand the question.  You shouldl have said:\n\nAny even number greater than 2 CAN be written as the sum of two primes. Prove true or false.
1146358
I’m not seeing how this is solvable.  I’m getting two possible answers.\n\nBut here’s how you should work the problem.  \n\nFirst of all, I think you pretty much have to assume that the x,y < 10.\n\nSo we know that Bill can’t tell what his two numbers are just by looking at the product.  As it turns out, there are very few products that allow this to happen: they are 12,16,18,24,36.  So you factor out all the number pairs that would make those products, and the sums are 7,8 (twice), 9, 10(twice), 11(twice), 12, and 13.\n\nSo we know that Albert is seeing one of those sums.  If he sees 7, then he knows that the only way for his numbers to add up to 7, and multiply up to either 12, 16, 18, 24, or 36 is for the numbers to be 3 and 4.  But he says nothing.  Therefore, he isn’t seeing 7.  Neither is he seeing 9, 12 or 13, for the same reason.  So, now we know the sum must be either 8,10, or 11.  And now Bill knows that.\n\nSo, if Bill were seeing 12 as the product, he would know that the numbers were 2 and 6, since they add up to 8.  But he says nothing, so he’s not seeing 12.  Likewise, if he were seeing 18, he would know that the numbers have to be 2 and 9.  But he doesn’t, so he’s not seeing 18, and he can’t be seeing 36 either.  So that means he must be seeing either 16 or 24.\n\nAnd now Albert knows this, so either Albert is seeing a sum of 8, and knows that the numbers are 4 and 4, or he’s seeing 11, and the numbers are 3 and 8.  I just don’t know how to eliminate either of those options.
1075715
Here are two sites that look into the problem:\n\nhttp://www.kw.igs.net/~jackord/bp/f6.html\nhttp://math.fullerton.edu/mathews/N310/projects/e6.htm