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  \times 3) is equal to the combination of the unit digit with its 2-time as the tenth/hundredth digit.

Unit digit, x x \times 3 \times 7
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)