Talk:06-240/Homework Assignment 4: Difference between revisions
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--[[User:Drorbn|Drorbn]] 12:09, 5 October 2006 (EDT) |
--[[User:Drorbn|Drorbn]] 12:09, 5 October 2006 (EDT) |
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I found the problem solved three different ways at [http://www.jimloy.com/number/divis.htm] |
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He mentions the problem as stated can be found in ''The Dictionary of Curious and Interesting Numbers''\ by David Wells. |
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I prefer his second method. Beginning from the right most digit, multiply corresponding digits by the following sequence of coefficients 1 3 2 6 4 5. For larger numbers sequence repeats. Alternatively the sequence 1 3 2 -1 -3 -2 could be used. |
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Using the given example <math>86415\Rightarrow 1(5)+3(1)+2(4)-1(6)-3(8)=-14\Rightarrow 7|86415</math> |
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Coefficients can be determined by taking multiplying the previous coefficient by 10 and take the modulus of this new number. |
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For n=17, sequence begins with 1 multiply by 10 then |
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<math>10\equiv 10 \mod(17),\quad 10\cdot 10\equiv 15\mod(17) \Rightarrow 15\equiv -2\mod(17)</math> |
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So for n=17, sequence is: 1, 10, 15, 14, 4, 6, 9, 5, 16, 7, 2, 3, 13, 11, 8, 12 |
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or alternatively: 1, -7, -2, -3, 4, 6, -8, 5, -1, 7, 2, 3, -4, -6, 8, -5 |
Revision as of 08:47, 10 October 2006
Divisibility by Prime Number
Pls correct me if I were wrong. The operation of cut away the unit digit is a distraction. If we consider the unit digit, the operation basically is a deduction of a number, and that number is divisible by 7. The whole operation is shown as follow:
8641 | |
10 |
105/7=21 |
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863
| |
2
|
21/7=3 |
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86
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2
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21/7=3 |
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8
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8
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84/7=12 |
-
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0
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0/7=0 |
Since it is an operation of series subtraction by multiples of 7, therefore the number we started from is divisible by 7 iff the resulting number is divisible by 7.
Moreover, there is a relationship between the unit digit, 2, and 7. The unit digit multiple by 21(7 3) is equal to the combination of the unit digit with its 2-time as the tenth/hundredth digit.
Unit digit, | |
---|---|
0 | 0 |
1 | 21 |
2 | 42 |
3 | 63 |
4 | 84 |
5 | 105 |
6 | 126 |
7 | 147 |
8 | 168 |
9 | 189 |
From the table above, I've induced the criterion for divisibility by 17 that is similar operation but the unit digit multiplies by 5 instead of 2. For divisibility by 13, the unit digit multiple by 9. Alright, I think it will be more fun if it's explained by other people. Wongpak 09:38, 5 October 2006 (EDT)
Excellent!
--Drorbn 12:09, 5 October 2006 (EDT)
I found the problem solved three different ways at [1] He mentions the problem as stated can be found in The Dictionary of Curious and Interesting Numbers\ by David Wells.
I prefer his second method. Beginning from the right most digit, multiply corresponding digits by the following sequence of coefficients 1 3 2 6 4 5. For larger numbers sequence repeats. Alternatively the sequence 1 3 2 -1 -3 -2 could be used.
Using the given example
Coefficients can be determined by taking multiplying the previous coefficient by 10 and take the modulus of this new number.
For n=17, sequence begins with 1 multiply by 10 then
So for n=17, sequence is: 1, 10, 15, 14, 4, 6, 9, 5, 16, 7, 2, 3, 13, 11, 8, 12
or alternatively: 1, -7, -2, -3, 4, 6, -8, 5, -1, 7, 2, 3, -4, -6, 8, -5