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 The Magic Cafe Forum Index » » Magical equations » » Why does this work (0 Likes) drkptrs1975 Elite user North Eastern PA 452 Posts  Posted: Jul 2, 2005 05:36 pm 0 Take any 3 digit number, reverse the digits, subtract smaller from the larger, then reverse the digits again, then add them up. You will always get 1089. Here is a few examples. 741 Reverse is 147 take smaller from the larger 741-147=594 Reverse them agian, this time add them 594 + 495 that will give you 1089 another one 758 857 857-758 = 99 099 990 099+990=1089 It works with all numbers, I cannot find a proof, nor a counter example. Why does this work. stanalger Special user St. Louis, MO 996 Posts  Posted: Jul 2, 2005 07:37 pm 0 Reverse the 3-digit number and subtract the smaller from the larger. (100*a + 10*b + c) - (100*c + 10*b + a) = 100*(a - c) + (c - a) Note that after the first step, the value of the central digit in the original number has already become irrelevant. Before we go on to step 2, let's rewrite: 100*(a - c) + (c - a) =100*(a - c) - 100 + 100 + (c - a) =100*(a - c - 1) + 100 + (c - a) =100*(a - c - 1) + 90 + 10 + c - a =100*(a - c - 1) + 10*9 + (10 + c - a) Step 2: Reverse and add. 100*(a - c - 1) + 10*9 + (10 + c - a) + 100*(10 + c - a) + 10*9 + (a - c -1) =100*(-1 + 10) + 10*(18) + (10 - 1) =100*(9) + 180 + 9 =900 + 180 + 9 =1089 Scott Cram Inner circle 2677 Posts  Posted: Jul 2, 2005 09:11 pm 0 Another little-used aspect of this trick concerns the number you get after the first subtraction. Whatever number you get, the digits will always add up to 18. About the only time I've seen this aspect used is by a book test created by Jim Steinmeyer. Grey Matters:Blog|Videos|Mental Gym|Presentation|Store graemesd Veteran user 369 Posts  Posted: Jul 21, 2005 06:29 pm 0 Also I guess we all know that after the first calculation the middle number is aways 9 and the 2 outer numbers always add upto 9 eg 495, 198 see Banachek 'psychoogical subleties' for presentation ideas idris New user St. Louis, MO 38 Posts  Posted: Jul 27, 2005 07:32 pm 0 Minor correction, the three digits must be different, at least the first and last digit. Usually the spectator is request to chose a number with three different digits. Obviously, if the first and last digits are the same the result of the first subtraction is 0. Jerry Jerry Parson Smith Inner circle 1939 Posts  Posted: Oct 5, 2005 12:40 am 0 This is an example of a root number of 9. A couple of interesting things about nine. Any # multiplied by 9 gives you a root of 9. Any root of 9 multiplied by ANY whole # results in a root of 9. Any string of #s added together and the sum subtracted from the original gives a root of nine. There are many, many things that can be done with this. Enjoy. Peace, Parson Here kitty, kitty,kitty. +++a posse ad esse+++ Jaz Inner circle NJ, U.S. 6112 Posts  Posted: Nov 6, 2005 02:06 pm 0 Just a thought Have someone do the calculation for the 1089 result. Have them add 6645 and turn the paper upside down. It looks like the word hELL. Maybe even force two 6s a 4 and a 5 card so the 6654 number seems random. Add some patter and who knows where it can take you. Hades likely. Mike Powers Inner circle Midwest 2897 Posts  Posted: Nov 21, 2005 04:33 pm 0 If you are familiar with MOD arithmetic, here's a way to look at it. Since 10 MOD 9 = 1, all powers of 10 MOD 9 also equal 1. Thus all place values i.e. 1, 10, 100, 1000 etc = 1 MOD 9. Thus any number is equivalent (MOD 9) to the sum of its digits. That's why all multilples of 9 have the property that the sum of their digits is a multiple of 9. (X MOD Y means the remainder when X is divided by Y. Thus 15 MOD 7 = 1 since 15 divided by 7 = 2 with a remainder of 1.) Mike Mike Powers http://www.mallofmagic.com The Magic Cafe Forum Index » » Magical equations » » Why does this work (0 Likes)
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