Talk:06-240/Homework Assignment 4

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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:

86415
105
  105/7=21
----
8631
21
  21/7=3
---
861
21
  21/7=3
--
84
84
  84/7=12
-
0
  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 Jim Loy's divisibility page[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