by mathematical induction, to prove  even number of persons  is odd number of shakehands?
826317
If you have two people, they shake hands once (odd).\nIf you add two more people, they will each have an even number of handshakes with the existing people, plus one more handshake for themselves.  odd + even + odd = even.  So they will change the parity from odd to even.  \n\nSo I've proven it false by contradiction.\n\nRepeat the process and you will add an even number of handshakes to the existing participants, plus 1 more handshake for the pair...again changing the parity from even to odd.\n\n2 people = 1 handshake (odd)\n4 people = 6 handshakes (even)\n6 people = 15 handshakes (odd)\n8 people = 28 handshakes (even)\n10 people = 45 handshakes (odd)\n12 people = 66 handshakes (even)\netc.\n\nThe general formula is n choose 2, or n(n-1) / 2.\n\nFor the case of:\nn = 2 --> 2 * 1 / 2 = 1\nn = 4 --> 4 * 3 / 2 = 6\nn = 6 --> 6 * 5 / 2 = 15\nn = 8 --> 8 * 7 / 2 = 28\nn = 10 --> 10 * 9 / 2 = 45\nn = 12 --> 12 * 11 / 2 = 66\netc.\n\nAre you sure you stated the problem correctly?  Perhaps you meant for groups of 2, 6, 10, etc. adding four people?\n\nEdit:  \nOkay, I've reread your question and the only interpretation that might work is if you are counting the handshakes for *one* person.  For example, if I am one of two people, then I shake hands with one person.  Now if you add two more people, I shake hands with three people.  If you add two more people, I shake hands with five people, etc.\n\nFirst you prove that with 2 people, I shake hands with one person (2-1).  This is pretty obvious.  Now assume you have n people and an odd number of handshakes.  This can be represented as 2k + 1 handshakes, where k is an integer.  You need to prove that with n+2 people, you still have an odd number of handshakes.  Since you are adding 2 new handshakes, this would be 2k + 3 handshakes which is also 2(k+1) + 1.  If k is an integer, so will k + 1.  As a result 2(k+1) + 1 will also be an odd number, so by mathematical induction you have proven that in a room with an even number of people, you will shake hands an odd number of times.
932028
Yes.\n\nThe argument is valid but not perfectly worded.  \n\nHere A, B, C can be variables as others have pointed out or they are statements that are fulfilled in direct progression.  In fact this is a form of the "Direct Proof"\n\nDirect Proof example:\n\nAs an example we can claim that if we have A:={the addition of two odd numbers x, y |  x+y} it will become an even number which can be statement C.\n\nSo by our knowledge that odd numbers can be written as \nx = 2m+1   and y = 2n+1  where m, n are any integers we can use that as statement B.  Hence B:={x = 2m+1   and y = 2n+1  where m, n are any integers}\n\nDoing a little algebra we have \nx+y = (2m+1) + (2n+1) = 2m + 2n + 2 = 2(m+n+1)\n\nNow we can find out that (m+n+1) is an integer as by definition and we can call it "p" s.t. p=(m+n+1),\nHence x+y = 2(m+n+1) = 2p, and by our knowledge of numbers 2p is an even number.  Hence C is fullfilled.\n\nThus like questioner's argument:\nIf A then B then C\nA \ntherefore C.\n\nAnd with my example and statements in quotations can be be replaced by letters A, B, and C: \nIf "you add two odd numbers" then "they are integers that can be written as the addition of two times another integer plus one which algebraically can be written as two times integer" then "you obtain a number which is even".\n"You add two odd numbers" \ntherefore "you obtain a number which is even".\n                                              Q.E.D.
506986
This is called the twin prime conjecture. The conjecture states that there are an infinite number of pairs of twin primes (p and p+2). It has not been proven there are an infinite number of such primes but it seems almost certain to be true.  While Hardy and Wright note that "the evidence, when examined in detail, appears to justify the conjecture," and Shanks states even more strongly, "the evidence is overwhelming," Hardy and Wright also note that the proof or disproof of conjectures of this type "is at present beyond the resources of mathematics."
579799
Independent Order of Odd Fellows
958370
We wish to show that:\n[k=1, n]∑kk! = (n+1)! - 1\n\nClearly, this is true for n=1, since 1*1! = 2! - 1 = 1. Suppose it is true for some n. Then for n+1:\n[k=1, n+1]∑kk!\n(n+1)*(n+1)! + [k=1, n]∑kk!\n(n+1)*(n+1)! + (n+1)! - 1\n(n+2)*(n+1)! - 1\n(n+2)! - 1\n\nWhich is what we would get if we substituted n+1 for n in our formula. So by induction, the formula holds for all n. Q.E.D.
446269
Just put the question from Mathmatters 2 whether it is even or odd on Yahoo! Answers and you will find the answer in different methods within seconds.
171134
As some of you noted, there are three cases to consider:\nGirls shake with Girls\nBoys shake with Girls\nBoys shake with Boys\n\nNotice the boys with boys shakes are 0, noted in the original question.  \n\nGirls shake with Girls:\nWell, each girl shakes with 49 other girls.  So 50*49 right?  Whats the problem?  if A shakes with B, then B must have shook with A also.  So each shake is counted twice.  Hence we need to divide the answer by two.  50*49/2 is your answer.\n\nBoys with Girls\nWe use the same technique.  Each girl shakes with 50 boys.  Thats 50 * 50.  But each shake is counted exactly once because we are only counting the girls side of things.\n\nThe final answer is:\n\n25*49 + 50*50 = 3725
92016
When dealing with the occurrence of more than one event, it is important to be able to quickly determine how many possible outcomes exist.\n\nFor example, if ice cream sundaes come in 5 flavors with 4 possible toppings, how many different sundaes can be made with one flavor of ice cream and one topping?  \n\nRather than list the entire sample space with all possible combinations of ice cream and toppings, we may simply multiply 5 • 4 = 20 possible sundaes.  This simple multiplication  process is known as the Counting Principle.
545283
there are two groups of 2 people , so take those 2 people as 1 unit. so there are total 8 units.\nthus number of ways to be arranged is 8! , that is 40320.
324840
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.
246807
I can't count, so don't blame me if I get this wrong.\nLet n be the number of participants.\nSince it's a close circuit, the people in front of him are in a sense also behind him so there ara (n-1) people both ahead and behind him.\nso (1/5+5/6)(n-1)=n\n31/30*n -31/30=n\n1/30*n=31/30\nthus n=31\n31 total people
922060
My suspicion upon seeing this question is that it can be solved using mathematical induction.  After all, couldn't we simply 'build' such a (maximally connected) graph starting with n=2 vertices and 'grow' it from there?\n\nRather than take on the proof myself -- mostly because I am unsure of the definition of 'components' -- I leave it as an exercise for you.  However, it bears noting that the formula shown is the commonly accepted "sum of the first n integers" formula used in almost every text on induction.\n\nGood luck!
831013
suppose c is even.  then this means that a is even and b is even, or a is odd and b is odd.  that is, a and b have the same parity.  we could not have a even, b even and c even because otherwise, a, b and c would not be coprime.  so a is odd and b is odd, if c is even.  I am unable to do this right now (I have forgotten my math, sorry), but I believe you could dispense with a proof which, loosely speaking, allows you to conclude that c must be odd on the basis of the fact that c^n = 0 (mod 2) is too restrictive (i.e. not all nth powers are even (except for those containing 2^n as a factor), or in other words, most nth powers are odd).  sorry, am not much help.\n\nincidentally though (if I may mention it here), a proof for FLT has already been published (and verified) in 1994.  the original proof comprised a whole edition of the Annals of Mathematics for that year and is composed of two papers, to wit: \n\nModular Elliptic Curves and Fermat's Last Theorem\n\nAND\n\n(again, it is easy to forget article titles, so please just search on the WWW web for it... :-))\n\nthe proof is subtle and surprisingly involved, and there is actually a page on the Internet dedicated to presenting the mathematics behind the proof to the general mathematical audience.  try googling "The Mathematics of Fermat's Last Theorem".\n\nhappy reading!
609465
Both sexes have an equal number. 24. \nChili's rocks!!!
1157222
You know there are 8 axioms in mathematics which we cannot prove\n\n1) if 'a' and 'b' are integers then 'a+b' is an integer and 'a.b' is an integer (Closure law)\n2) a + b = b + a and a..b = b.a (cumulative law)\n3) a + (b + c) = (a + b) + c (Associative law)\n4) a (b + c) = a.b + a.c (Distributive law)\n5) a + 0 = a = 0 + a (Additive Identity)\n6) a.1 = a = 1.a (Multiplicitive Identity)\n7) for all integer 'a' there is a unique integer (-a) such that a + (-a) = 0\n8) I can't remember...\n\nSo we can prove your requirement using above axioms, \n\na + (-a) = 0\n(-1)(a + (-a) = -1(0)  (Multiplying by same factor)\n-a + -(-a) = 0 (Distributive law over addition)\n\nadding +a both sides\n\na + (-a + -(-a)) = a + 0 \n(a+ (-a)) + -(-a) = a (Associative law)\n0 + -(-a) = a \n-(-a) = a (proved... huh...)
340631
First you must consider that the first digit can not be zero, otherwise it would not be a 6 digit number.\nThen you must consider how you can arrange odd or even within the digits. (O=ODD, E-EVEN)\n1.OOOEEE\n2.OOEOEE\n3.OOEEOE\n4.OOEEEO\n5.OEOOEE\n6.OEOEOE\n7.OEOEEO\n8.OEEOOE\n9.OEEOEO\n10.OEEEOO\n11.EOOOEE\n12.EOOEOE\n13.EOOEEO\n14.EOEOOE\n15.EOEOEO\n16.EOEEOO\n17.EEOOOE\n18.EEOOEO\n19.EEOEOO\n20.EEEOOO\nFor each of the #1-10 above we can consider 5 possible digits for the odds and 5 for the evens\nso for each there are 5*5*5*5*5*5=15625 different digit combinations.\nFor #11-20 we must consider that the first digit can not be zero, so there are 4*5*5*5*5*5=12500 different digit combinations for each.\nSo there is a total of 10*15625 + 10*12500=281250 different numbers with 3 odd and 3 even digits.
367524
using principle of mathematical induction u can prove this\n check for n=1  2^2 -3 -1=0 which is div by 9\nlet this be true for n=m\n2^2m -3 m -1 =9k  ,where k is any integer\n\nnow for n=m+1 \n2^2(m+1) -3(m+1) -1\n=4.2^2m-3m-4\n=(2^2m-3m-1)+3.2^2m -3\n=9k+3(2^2m-1), now (2^2m -1) is an odd integer divisible by 3 for all m >0\ntherefore the above expression becomes 9k +3.3g ,where g is any integer\n=9(k+g)=9i  where i is an integer \ntherefore the condition for divisibility by 9 is true for n=m+1\nhence by principle of mathematical induction it is true for all n>=1