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06122010, 12:05 PM  #1 (permalink) 
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 How to multiply 177 x 23 After writing down the numbers halve the first one and double the second, writing the new numbers below the preceding ones. If the number being halved is odd, just ignore the remainder. Repeat this operation as long as you can: 177 x 23 88 . 46 44 . 92 22 . 184 11 . 368 5 . 736 2 . 1472 1 . 2944 Now remove from the second column all numbers where the corresponding number in the first column is even: 177 x 23 88 . 46 < remove 44 . 92 < remove 22 . 184 < remove 11 . 368 5 . 736 2 . 1472 < remove 1 . 2944 Then add up the remaining numbers: 23 + 368 + 736 + 2944 = 4071 This works for any two numbers Tess 
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06122010, 03:37 PM  #2 (permalink) 
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06132010, 07:44 PM  #3 (permalink) 
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Business: DG Productions Client: Firestorm Blog Entries: 5  So is there an analysison the web or elsewhereof why this works?
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06132010, 10:19 PM  #4 (permalink) 
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 This is called "Russian Peasant" arithmetic. If you think about it a bit it's pretty obvious what's happening. What you're basically doing is converting the first number to binary, lowest bit first, and multiplying the second number by the corresponding powers of two. Where the lowest bit is 0 (it's even) you leave that row out, where the lowest bit is 1 (it's odd) you keep it.
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06162010, 07:22 AM  #7 (permalink)  
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06172010, 12:12 PM  #9 (permalink) 
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06172010, 12:18 PM  #10 (permalink)  
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06172010, 01:15 PM  #11 (permalink) 
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 OK, start with the number 177. In binary that's 10110001. Dividing by two is equivalent to shifting one bit to the right. 177/2 is 88, that's 1011000. So, here's what you're doing in binary: Code: LHS RHS LO BIT OF LHS RHS = 23 * 2^N 10110001 10111 1 23 * 2^0 = 23 * 1 = 23 1011000 101110 0 101100 1011100 0 10110 10111000 0 1011 101110000 1 23 * 2^4 = 23 * 16 = 368 101 1011100000 1 23 * 2^5 = 23 * 32 = 736 10 10111000000 0 1 101110000000 1 23 * 2^7 = 23 * 128 = 2944 At the same time you're calculating 2^n * LHS by multiplying it by two each time. So what we've done is this: 177 * 23 is (2^0 + 2^4 + 2^5 + 2^7) * 23 By pulling out the values where it's odd, you've extracted: 2^0 * 23 + 2^4 * 23 + 2^4 * 23 + 2^7 * 23 And because multiplication is distributive... that's the value you want. Last edited by Argent Stonecutter; 06172010 at 01:21 PM. Reason: man it's hard to line these tables up... 
06172010, 01:26 PM  #12 (permalink) 
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Business: DG Productions Client: Firestorm Blog Entries: 5  So much for being lazy. The pointer to the Russian Peasant Method was a good one. (Thx Argent) . . . and the web is rife with articles. Here's one of the more succinct, clearer explanations: ================================== Now, let's see why the algorithm works. Since halving and doubling play such an important role in the algorithm, it should not come as a great surprise that its real foundation lies in the binary system. To obtain the binary representation of a number, the number is to be repeatedly divided by two, the remainders recorded and then written in the reverse order. Check now the "Show binary" box: We see two additional columns: the first indicates the step (and a power of 2), the second contains the binary digits of 85 written from the bottom upwards: 85 = 10101012, which essentially means that 85 = 10101012 = 1·26 + 0·25 + 1·24 + 0·23 + 1·22 + 0·21 + 1·20 = 64 + 16 + 4 + 1. Note that a binary digit is the remainder of division by 2: it is 1 for odd numbers and 0 for even numbers. Which bring forth the connection between the binary digits of the first multiplicand, 85 in our case, and the parity of the number in the column beneath it. The numbers are odd exactly when the remainder that appears to their left is 1. This makes the algorithm tick: 85×18 = (64 + 16 + 4 + 1)×18 = 1152 + 288 + 72 + 18 = 1530. For the product 18×85 we get: 18 = 100102 = 1·24 + 0·23 + 0·22 + 1·21 + 0·20 = 16 + 2. and subsequently 18×85 = (16 + 2)×85 = 1360 + 170 = 1530. =================================== Last edited by Rema Quandry; 06172010 at 01:38 PM. 
06172010, 01:29 PM  #13 (permalink)  
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I like your explanation better.  
06172010, 02:15 PM  #14 (permalink)  
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