This website that has mind jokes on it has a esp joke.It askes you to think of any number and concentrate on it.Then as you click the next steps it has you add numbers to it and subtract.The end result is it shows your original number or it says what the last number you came up with in your math.Now in the beginning it says you may use a pen and paper.What is it that makes this work?Im blown away, creative to say the least.
217227
Properties of mathematics makes it work. I'm not sure exactly what it has you do but essentially it does this.\n\nIf your number is X. You add 5 to it then subtract 5 from it and you get your original number. Of course the nice ones hide the fact that your doing exactly this. \n\nDavid Copperfield did a TV bit once where he had a 3X3 square. You start out at any square you want and move according to his directions. After each move he took away a square. In the end you always end up on one square. That was cool. Anyone heard of it?
361493
I think the easiest way to know whether or not you add is by picturing a number line that is negative on the left and positive on the right with zero in the middle.\nif you subtract, then you move left\nif you add, then you move right.\n\n\n\nMemorize this!\n+ + makes +\n+ - makes -\n- + makes -\n- - makes +\n\nexample:\n1+(+2)=1+2\n1+(-2)=1-2\n1-(+2)=1-2\n1-(-2)=1+2\n\nanother example:\n(-1) + (+2) = (-1) + 2 = 2 - 1\n(-1) + (-2) = (-1) - 2 = -(1+2)\n(-1) - (+2) = (-1) - 2 = -(1+2)\n(-1) - (-2) = (-1) + 2 = 2 - 1\n\ntry these, and we'll tell you if you got it\n4 - (-5) =____\n6 - 7 = _____\n-9 - 8 = ______\n-3 - (-5) = _______
417130
luv phy is doing the calculations as if each letter could be placed in any one of the 8 envelopes, even after some of the envelopes have been used for other letters. In other words, he is treating the placing of the letters as 8 independent events. But they are not independent, as each event is affected by what has happened before.\n\nThe probability that no one will get the write letter is equal to the number of ways that can happen, divided by the total number of ways that the 8 letters could possibly be arranged. The second number (the denominator) is easy: it's 8! = 8x7x6x5x4x3x2x1 = 40,320.\n\nTo get the numerator, you have to do a calculation. There may be a standard formula, but one rule that works is this:\nThe number of ways that n envelopes can all contain the wrong letter (call this number W(n) ... "W" for "wrong") is\n(n-1) x (W(n-1) + W(n-2)).\nThis applies when n is 3 or more.\nFor n = 1, W(1) = 0 (i.e., you can't screw up if there's just one letter and one envelope).\nFor n = 2, W(2) = 1 (i.e., there's only one way to screw up 2 letters: you get them reversed).\n\nSo, applying the formula, \nW(3) = (3-1) x (1 + 0) = 2\nW(4) = (4-1) x (2 + 1) = 9\nW(5) = (5-1) x (9 + 2) = 44\nW(6) = (6-1) x (44+ 9) = 265\n\nI'll let you work out the values for W(7) and W(8).\nThen you need to divide W(8) by 40,320 to get the probability requested in (a) of the problem.\n\nOnce you solve (a), then (b) is simple. It's just 1 - (a). Either no one gets the right letter, or someone (at least one person) gets the right letter. So the answers to (a) and (b) have to add up to 1.\n\nGood luck!
275020
I have written a math book that explains math in plain English.  It is called Mauk Math.  I teach math and I could not find any math book that I liked so I decided to write my own.  Math does NOT have to be difficult.  It CAM be explained in simple terms and it CAN be easy and even FUN.  I use music to learn certain concepts and help retain it.  I use songs to help break it down and make it easier.  For instance, when I teach how to make negative exponents positive I use Will Smith's song, SWITCH. We play the song in class and sing it.  Because that's how you make a negative exponent positive.... you switch it !  If it is on the top of a fraction... you switch it to the bottom.  If it is on the bottom...you switch it to the top...EASY!  In typical math books, they make it difficult by explaining it like this: a number taken to the -nth power in the denominator can be make positive by rewriting the fraction with the denominator as the numerator taken to the nth power.  Ridiculous...especially for the struggling student.  MAUK MATH uses M.C. Hammer's Hammertime to teach the order of operations.  You always do everything in parentheses first.  If something is outside the parentheses....CAN'T TOUCH THAT na na na na. break it down.  Math can be easy and fun with the right book.... my book.  LOL
561342
It uses the logical part of your brain, as opposed to the creative part. Of course, in the developed world it is called Maths.
417645
If the number begins with 5, there are still three of the odd digits available to fill the last place.  If a 3 is used for the last place, the others are filled using 1, 2, 3, 4, and so there are 4! arrangements.\nIf 1 is used for the last place, that leaves 2, 3, 3, 4 for the others, and so the number of arrangements is  4!/2! because of the repeated 3.  So that gives \n4! + 4!/2! = 36\n\nIf it begins with a 3, the last place can be filled with 1, 3, or 5, and  the numbers left to fill the remaining four places are all different and so the number of possibilities is\n3*4! =72\n\nIf it begins with a 2 or 4 ( two choices), and the last with 1 or 5 (two choices), there are 4!/2! ways of filling the other places because 3 is repeated.\n2*2*4!/2! = 48\n\nIf it begins with 2 or 4 and ends with 3, then the other four digits are all different, and the number of possibilities is\n2*4! = 48\n\nAdding all these together gives 204.  Yes.
123220
such a no. cannot exist but hey dude a great question lets look at it this way\nthe last digit of the given no. will be 6\nand the resulting no will be 4 times that no\nlast digit of 4*6 is 4 so the resulting nos last digit digit has to 4 \nthat would mean the first digit of the original no will  be 4\nso just as an eg let us consider the no. 46\n46*4 will be 184\nthat is the first digit will be 1(or 2 or 3 or 4 but never 6)\nthis will be followed by every no \nso u c such a no just canno exist
553608
It's when you make mistake in addition. But I won't say the sum can't be put correctly but It can be made correct by changing the sign You could change the sign from + to - . Now can I have my ten points ? Please ???!
82972
It is true that the ones digit for 5! and up is 0.  The reason for this is that when the 5 is included, the number would have to end in either a 0 or 5.  Then when you multiply by the 2, it has to be even, so it's the 0.  So you would only need to add up 4! + 3! + 2! + 1! + 0!\n\nDon't forget that 0! is 1.\n\n24+6+2+1+1 = 34, so the one's digit will be 4.
20508
in many cases, you need innovative thinking.. Consider this:\n\n250*844 = 250*(4*411) = (250*4)*411 = 1000*411 = 411000\n251*844 = (250+1)*844 = 250*844 + 1*844 = 411000 + 844 = 411844\n\n37*123 = 37*3*41 = 111*41 = (100+10+1)*41 = 4100+410+41 = 4551\n\nsometimes, do not oversimplify...\n105*598 = (100+5)(600-2) = 100*600 - 100*2 + 5*600 - 5*2 = 60000 - 200 + 3000 -10 = 62890 (not good)\n\n105*598 = 105*(600-2) = 63000 - 210 = 62890 (better)\n\nThe key is, you have gotta practice a lot.\nFor example, when I see 111*41, I wrrite down in a peice of paper like this:\n\n(* = space)\n41**\n*41*\n**41\n----\n4554
225599
this really only works when its done verbally...when people are looking at the numbers, its much easier to realize that the answer is 4100 instead of 5000, pretty neat one isnt it
451425
this is also true with any odd number (3...7,9,11,etc) not just 5.\n\nThe reason is because you are taking the average of the 5 numbers.  the average of a number is the exact middle of the set of numbers.  Look at the numbers in a line..\n 1+2+3+4+5\n\nnow if you look at the middle number in that line, it is 3.  if you add all the numbers together and divide by 5, you will get 3.  \n\nnow try it with 7 digits\n 1+2+3+4+5+6+7\n\nthe middle number is 4.  now add them up, and you get 28. divide by 7 and you get 4.  MAGIC!!\n\nnow try it with 4 digits.  NOTE: this is an even number, you will get a decimal.\n1+2+3+4\nthe middle number is...well..there really isn't one. its is between 2 and 3, isn't it? ok, now add it up, and you get 10. divide by 4 and you get 2.5.  isn't 2.5 right in the middle of 2 and 3??\n\nthere you go, i hope that works for ya\n\n(btw,this trick will only work if the numbers are consecutive)
405406
Hey, I seen what you need on TV last night, real cool. It these pens that think for you. They do Math, English, Spanish, and more. I not catch it all, but, I guess you need a pen for each subject. It really cool. Showed the guy do a math problem and the pen  told him how to work it! Tell me that not something. Heck, in another 5-10 years we not need school anymore, just the correct pen for the job we do! :-D
70337
There is no flaw as long as you recognize that .99999 repeats forever.  If you were to end it at some point and subtract, then as a previous poster said, you'd actually get 8.99....1.  But since it repeats, there is always another 9 and so it works.\n\nAnd consider this - if .9999999repeating did NOT equal 1, then there would be another number between them.  What would that number be?\n\n~~~~~~~~~~~~~~~~~~~~\nNo, the number in question is not infinitessimal, it doesn't exist.  You can never get to the end of the string of nines to try and squeeze another number in there :)
454798
study + no study = no failed + failed\nstudy [ 1 + no ] = failed [ no + 1 ]\n\ndivide [ 1 + no ] from each side :\n\nstudy = failed\n\neyeonthescreen you have no sense of humor. However, if "study", "failed", and "no" are variables, then my equations do in fact work (assuming no≠-1)!  Also, it is "no failed" not "not failed"; therefore not Boolean either!
67894
How about "Math -  Does it really add up?"
491060
Discounting rotations and mirror images, there is only one way to make a magic square using the digits 1 through 9.  \n\nFirst, we can prove the 5 has to go in the middle:\n\nLet the numbers in the top row from left to right be a, b, and c.  Let the numbers in the middle row from left to right be d, e, and f.  (Note e is the middle box.)  Let the numbers in the bottom row from left to right be g, h, and i.     \n\nWe know a + b + c = 15 and g + h + i = 15\n\nWe also know a + e + i = 15, b + e + h = 15, and c + e + g =15.  Therefore, (a + e + i) + (b + e + h) + (c + e + g) = 15 * 3 = 45.  We can rearrange this to (a + b + c) + (g + h + i) + 3e = 45.  Since a + b + c = 15 and g + h + i = 15, that reduces to 15 + 15 + 3e = 45, or 3e = 15.  Therefore e (the middle box) must equal 5.   \n\nKnowing the middle box is 5, try 9 in a corner and you'll notice a problem - the 6, 7 and 8 can't all be placed in the square without the row or column with the 9 exceeding 15.  So, the 9 has to occupy a side box.  Since we are discounting rotations and mirror images, we can arbitrarily select box b for the 9.  This forces the placement of the 1 in box h.   \n\nWith the 9 and 1 now placed, the 8 and 6 have to go in the row with the 1.  So, we'll put the 8 in box g and the 6 in box i.  (We still haven't done anything to create a second square if rotations and mirror images are discounted.)  That forces the placement of the final four numbers.