Talk:06-240/Homework Assignment 4 - Revision history

Drorbn: Reverted edit of 62.231.243.136, changed back to last version by Drorbn

2007-06-05T13:24:04Z

Reverted edit of 62.231.243.136, changed back to last version by Drorbn

@@ Line 9: / Line 9: @@
 |-
 |<div align="right"> 10<s>5</s>
+|-

62.231.243.136 at 11:03, 5 June 2007

2007-06-05T11:03:34Z

← Older revision		Revision as of 07:03, 5 June 2007
Line 9:		Line 9:
	\|-		\|-
	\|<div align="right"> 10<s>5</s>		\|<div align="right"> 10<s>5</s>
		⚫	\|
	\|  105/7=21
⚫	\|-
	\|<div align="right"> ----
	\|-
	\|<div align="right">863<s>1</s></div>
	\|-
	\|<div align="right">2<s>1</s></div>
	\|  21/7=3
	\|-
	\|<div align="right"> ---</div>
	\|-
	\|<div align="right">86<s>1</s></div>
	\|-
	\|<div align="right">2<s>1</s></div>
	\|  21/7=3
	\|-
	\|<div align="right"> --</div>
	\|-
	\|<div align="right">8<s>4</s></div>
	\|-
	\|<div align="right">8<s>4</s></div>
	\|  84/7=12
	\|-
	\|<div align="right"> -</div>
	\|-
	\|<div align="right">0</div>
	\|  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 <math> \times </math> 3) is equal to the combination of the unit digit with its 2-time as the tenth/hundredth digit.
	{\|border="1"
	!Unit digit, <math>x</math> !! <math>x \times 3 \times 7</math>
	\|-
	\|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. [[User:Wongpak\|Wongpak]] 09:38, 5 October 2006 (EDT)

	== Excellent! ==

	--[[User:Drorbn\|Drorbn]] 12:09, 5 October 2006 (EDT)

	I found the problem solved three different ways at Jim Loy's divisibility page[http://www.jimloy.com/number/divis.htm]
	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 <math>86415\Rightarrow 1(5)+3(1)+2(4)-1(6)-3(8)=-14\Rightarrow 7\|86415</math>

	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
	<math>10\equiv 10 \mod(17),\quad 10\cdot 10\equiv 15\mod(17) \Rightarrow 15\equiv -2\mod(17)</math>

	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

	== Nice, but... ==

	but for each divisibility you have to memorize (or calculate again and again) a whole sequence of numbers, rather than just one. --[[User:Drorbn\|Drorbn]] 10:30, 10 October 2006 (EDT)

Drorbn: Reverted edit of 61.132.90.174, changed back to last version by Drorbn

2007-05-08T12:20:02Z

Reverted edit of 61.132.90.174, changed back to last version by Drorbn

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61.132.90.174 at 11:58, 8 May 2007

2007-05-08T11:58:22Z

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Drorbn: Reverted edit of 218.233.31.250, changed back to last version by Drorbn

2007-05-04T14:34:04Z

Reverted edit of 218.233.31.250, changed back to last version by Drorbn

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218.233.31.250 at 12:34, 4 May 2007

2007-05-04T12:34:41Z

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Drorbn: Reverted edit of 58.231.98.91, changed back to last version by Drorbn

2007-04-28T13:46:54Z

Reverted edit of 58.231.98.91, changed back to last version by Drorbn

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58.231.98.91 at 13:40, 28 April 2007

2007-04-28T13:40:32Z

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Drorbn: Reverted edit of 200.238.102.170, changed back to last version by Drorbn

2007-04-10T17:00:21Z

Reverted edit of 200.238.102.170, changed back to last version by Drorbn

@@ Line 9: / Line 9: @@
 |-
 |<div align="right"> 10<s>5</s>
+|-

200.238.102.170 at 16:48, 10 April 2007

2007-04-10T16:48:25Z

← Older revision		Revision as of 12:48, 10 April 2007
Line 9:		Line 9:
	\|-		\|-
	\|<div align="right"> 10<s>5</s>		\|<div align="right"> 10<s>5</s>
		⚫	\|
	\|  105/7=21
⚫	\|-
	\|<div align="right"> ----
	\|-
	\|<div align="right">863<s>1</s></div>
	\|-
	\|<div align="right">2<s>1</s></div>
	\|  21/7=3
	\|-
	\|<div align="right"> ---</div>
	\|-
	\|<div align="right">86<s>1</s></div>
	\|-
	\|<div align="right">2<s>1</s></div>
	\|  21/7=3
	\|-
	\|<div align="right"> --</div>
	\|-
	\|<div align="right">8<s>4</s></div>
	\|-
	\|<div align="right">8<s>4</s></div>
	\|  84/7=12
	\|-
	\|<div align="right"> -</div>
	\|-
	\|<div align="right">0</div>
	\|  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 <math> \times </math> 3) is equal to the combination of the unit digit with its 2-time as the tenth/hundredth digit.
	{\|border="1"
	!Unit digit, <math>x</math> !! <math>x \times 3 \times 7</math>
	\|-
	\|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. [[User:Wongpak\|Wongpak]] 09:38, 5 October 2006 (EDT)

	== Excellent! ==

	--[[User:Drorbn\|Drorbn]] 12:09, 5 October 2006 (EDT)

	I found the problem solved three different ways at Jim Loy's divisibility page[http://www.jimloy.com/number/divis.htm]
	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 <math>86415\Rightarrow 1(5)+3(1)+2(4)-1(6)-3(8)=-14\Rightarrow 7\|86415</math>

	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
	<math>10\equiv 10 \mod(17),\quad 10\cdot 10\equiv 15\mod(17) \Rightarrow 15\equiv -2\mod(17)</math>

	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

	== Nice, but... ==

	but for each divisibility you have to memorize (or calculate again and again) a whole sequence of numbers, rather than just one. --[[User:Drorbn\|Drorbn]] 10:30, 10 October 2006 (EDT)