Imaginary Thirteen - Revision history

en>TheChampionMan1234: removed Category:Solitaire card games using HotCat

2014-12-23T03:27:29Z

removed Category:Solitaire card games using HotCat

← Older revision		Revision as of 05:27, 23 December 2014
Line 1:		Line 1:
	The ~~'''magic gopher'''[http://learnenglish.britishcouncil.org/en/word-games/magic-gopher] is an interactive [[Adobe Flash]] game published online by the [[British Council]].~~		The title of the writer is Nestor. Delaware is our beginning location. What she enjoys doing is taking part in croquet and she is attempting to make it a profession. Interviewing is how he supports his family but his marketing never comes.<br><br>Also visit my site; extended car warranty - [http://Jamsuham.net/xe/joinqna/670399 please click the next website page] -

	~~==Overview==~~

	The game involves the 'magic gopher' asking the player to pick a random two-digit number. The player is then asked to sum the digits of the number and subtract them from the original number. The gopher then presents a list of symbols, of which the player finds the one corresponding to this new number. The gopher then proceeds to correctly guess the player's symbol, supposedly after reading his or her mind.

	~~==The trick==~~

	The trick to the game is that the gopher uses the same symbol for every multiple of 9, from 0 through 81 (it is impossible to get a higher number using only two digits). The gopher assigns the same symbol to the multiples of 9 as to other non-multiples of ~~9 in order to cover up the trick;~~ the ~~symbol picked for each game~~ is ~~randomized~~. ~~No matter which two-digit integer the player chooses, when the subtraction~~ is done, the resulting number will always be a multiple of 9. In fact, no matter which (nonnegative) integer the player chooses, the result will always be a multiple of 9. The former can be proven using elementary algebra. Be aware that the proofs for ''n'' digits rely on that the digits of the number <math>n\,</math> are indexed with the rightmost digit being assigned position 0, with the adjacent digits (to the left) having an index increasing by one each time. ~~For example, for the number '261', the digit '1'~~ is in ~~position 0, '6' is in position 1,~~ and ~~'2'~~ is ~~in position 2.~~

	~~===Proof for 2 digits===~~

	Let ''n'' be a 2-digit integer. Additionally, let ''a'' be the first digit of ''n'' and ''b'' be the second digit of ''n''. Finally, let ''c'' equal the sum of the digits of ''n'', so ''c'' = ''a'' + ''b''..

	~~An equivalent form for ''n'', by virtue of using a decimal [[numeral system]], is ''n'' = 10''a'' + ''b''.~~

	~~The resulting number, ''z'', is given by ''z'' = ''n'' − ''c'' = (10''a'' + b) − (''a'' + ''b'') = 9''~~a''. ~~Hence, ''z''~~ is ~~always a multiple of 9. [[Q.E.D.]]~~

	~~===Proof for ''n'' digits===~~

	~~Proving that no matter~~ how ~~large <math>n\,</math> is (and how many digits <math>n\,</math> has), it is always a multiple of 9 is slightly trickier~~. ~~The following proof makes use of [[modular arithmetic]]:~~<br>

	~~Let <math>n\,</math> be an integer with <math>m\,</math> digits and let <math>n_m\,</math> represent the <math>m^{th}\,</math> digit of <math>n\,</math>.~~<br>
	~~Thus, <math>n\ = 10^m \cdot n_m + 10^{m~~-~~1} \cdot n_{m-1} + \cdots + 10^0 \cdot n_0</math>.<br>~~
	~~Let <math>c\,</math> be the sum of the digits of <math>n\,</math>. So, <math>c\ = n_m + n_{m-1} + \cdots + n_0</math>.<br>~~
	~~Since <math>10^m \equiv 1 \mod 9, n = n_m + n_{m-1} + \cdots + n_0 = c \mod 9</math>.<br>~~
	~~Hence <math>n - c \equiv 0 \mod 9</math> so the resulting number <math>z = n - c\,</math> is a multiple of 9.<br>~~
	~~Q.E.D.~~

	~~====Alternative proof for ''n'' digits====~~

	~~This alternative proof is less mathematically rigorous, relying on some common sense and intuition, but it is still sufficient to demonstrate the same as the above.~~

	~~Again, let ''n'' be an integer with ''m'' digits and let ''n''<sub>''m''</sub> represent the ''m''<sup>th</sup> digit of ''n''.~~

	~~Thus,~~

	~~: <math>n = 10^m \cdot n_m + 10^{m-1} \cdot n_{m-1} + \cdots + 10^0 \cdot n_0.\,</math>~~

	~~Let ''c'' be the sum of the digits of ''n''. So~~

	~~: <math>c\ = n_m + n_{m-1} + \cdots + n_0.</math>~~

	~~Now, let~~

	~~: <math>z = n - c = (10^m \cdot n_m + 10^{m-1} \cdot n_{m-1} + .. + 10^0 \cdot n_0) - (n_m + n_{m-1} + \cdots + n_0).</math>~~

	~~This can be written as~~

	~~: <math>z = (10^m \cdot n_m - n_m) + (10^{m-1} \cdot n_{m-1} - n_{m-1}) + \cdots + (10^0 \cdot n_0 - n_0).</math>~~

	~~By factoring, we obtain~~

	~~: <math>z = n_m(10^m - 1) + n_{m-1}(10^{m-1} - 1) + \cdots + n_0(10^0 - 1).\,</math>~~

	Now, <math>10^m - 1\,</math> will give a number with <math>m - 1\,</math> nines, hence, each individual digit of ''n'' is being multiplied by a multiple of 9 as the numbers given by <math>10^m - 1\,</math> are implicitly multiples of 9. Since the sum of any number of multiples of 9 is always divisible by 9, we conclude that whichever number is picked for ''n'', it will always be a multiple of 9.

	~~If we use the resulting formula for ''z'' assuming that ''n'' will only be 2 digits in length, we obtain the same formula as with the proof for a 2 digit ''n'':~~

	~~: <math>z = n_1(10^1 - 1) + n_0(10^0 - 1) = n_1 \cdot 9 + n_0 \cdot 0 = 9n_1</math> where <math>n_1 = a.\,</math>~~

	~~==External links==~~
	* [http://~~learnenglish.britishcouncil~~.~~org~~/en/~~word-games~~/~~magic-gopher Original Magic Gopher site~~]
	* [http://jeffki.blogspot.com/2007/05/magic-~~gopher-explained.html Very Simple Explanation]~~

	~~[[Category:Adobe Flash]]~~
	~~[[Category:Mathematical games]]~~
	~~[[Category:Articles containing proofs]]~~

en>Frietjes at 23:11, 11 October 2013

2013-10-11T23:11:11Z

← Older revision		Revision as of 01:11, 12 October 2013
Line 1:		Line 1:
	Let me first ~~start~~ by ~~introducing myself~~. ~~My name~~ is ~~Boyd Butts even though~~ it is ~~not~~ [~~http~~://~~www~~.~~dailyroads~~.~~com~~/~~profile_info~~.~~php?ID~~=~~2363798 over~~ the ~~counter std test] name~~ on ~~my beginning certificate~~. ~~My working day occupation~~ is a ~~meter reader~~. ~~Playing baseball~~ is the ~~hobby he~~ will ~~never quit performing. Years~~ in the ~~past we moved to North Dakota~~.		The '''magic gopher'''[http://learnenglish.britishcouncil.org/en/word-games/magic-gopher] is an interactive [[Adobe Flash]] game published online by the [[British Council]].

			==Overview==

			The game involves the 'magic gopher' asking the player to pick a random two-digit number. The player is then asked to sum the digits of the number and subtract them from the original number. The gopher then presents a list of symbols, of which the player finds the one corresponding to this new number. The gopher then proceeds to correctly guess the player's symbol, supposedly after reading his or her mind.

			==The trick==

			The trick to the game is that the gopher uses the same symbol for every multiple of 9, from 0 through 81 (it is impossible to get a higher number using only two digits). The gopher assigns the same symbol to the multiples of 9 as to other non-multiples of 9 in order to cover up the trick; the symbol picked for each game is randomized. No matter which two-digit integer the player chooses, when the subtraction is done, the resulting number will always be a multiple of 9. In fact, no matter which (nonnegative) integer the player chooses, the result will always be a multiple of 9. The former can be proven using elementary algebra. Be aware that the proofs for ''n'' digits rely on that the digits of the number <math>n\,</math> are indexed with the rightmost digit being assigned position 0, with the adjacent digits (to the left) having an index increasing by one each time. For example, for the number '261', the digit '1' is in position 0, '6' is in position 1, and '2' is in position 2.

			===Proof for 2 digits===

			Let ''n'' be a 2-digit integer. Additionally, let ''a'' be the first digit of ''n'' and ''b'' be the second digit of ''n''. Finally, let ''c'' equal the sum of the digits of ''n'', so ''c'' = ''a'' + ''b''..

			An equivalent form for ''n'', by virtue of using a decimal [[numeral system]], is ''n'' = 10''a'' + ''b''.

			The resulting number, ''z'', is given by ''z'' = ''n'' − ''c'' = (10''a'' + b) − (''a'' + ''b'') = 9''a''. Hence, ''z'' is always a multiple of 9. [[Q.E.D.]]

			===Proof for ''n'' digits===

			Proving that no matter how large <math>n\,</math> is (and how many digits <math>n\,</math> has), it is always a multiple of 9 is slightly trickier. The following proof makes use of [[modular arithmetic]]:<br>

			Let <math>n\,</math> be an integer with <math>m\,</math> digits and let <math>n_m\,</math> represent the <math>m^{th}\,</math> digit of <math>n\,</math>.<br>
			Thus, <math>n\ = 10^m \cdot n_m + 10^{m-1} \cdot n_{m-1} + \cdots + 10^0 \cdot n_0</math>.<br>
			Let <math>c\,</math> be the sum of the digits of <math>n\,</math>. So, <math>c\ = n_m + n_{m-1} + \cdots + n_0</math>.<br>
			Since <math>10^m \equiv 1 \mod 9, n = n_m + n_{m-1} + \cdots + n_0 = c \mod 9</math>.<br>
			Hence <math>n - c \equiv 0 \mod 9</math> so the resulting number <math>z = n - c\,</math> is a multiple of 9.<br>
			Q.E.D.

			====Alternative proof for ''n'' digits====

			This alternative proof is less mathematically rigorous, relying on some common sense and intuition, but it is still sufficient to demonstrate the same as the above.

			Again, let ''n'' be an integer with ''m'' digits and let ''n''<sub>''m''</sub> represent the ''m''<sup>th</sup> digit of ''n''.

			Thus,

			: <math>n = 10^m \cdot n_m + 10^{m-1} \cdot n_{m-1} + \cdots + 10^0 \cdot n_0.\,</math>

			Let ''c'' be the sum of the digits of ''n''. So

			: <math>c\ = n_m + n_{m-1} + \cdots + n_0.</math>

			Now, let

			: <math>z = n - c = (10^m \cdot n_m + 10^{m-1} \cdot n_{m-1} + .. + 10^0 \cdot n_0) - (n_m + n_{m-1} + \cdots + n_0).</math>

			This can be written as

			: <math>z = (10^m \cdot n_m - n_m) + (10^{m-1} \cdot n_{m-1} - n_{m-1}) + \cdots + (10^0 \cdot n_0 - n_0).</math>

			By factoring, we obtain

			: <math>z = n_m(10^m - 1) + n_{m-1}(10^{m-1} - 1) + \cdots + n_0(10^0 - 1).\,</math>

			Now, <math>10^m - 1\,</math> will give a number with <math>m - 1\,</math> nines, hence, each individual digit of ''n'' is being multiplied by a multiple of 9 as the numbers given by <math>10^m - 1\,</math> are implicitly multiples of 9. Since the sum of any number of multiples of 9 is always divisible by 9, we conclude that whichever number is picked for ''n'', it will always be a multiple of 9.

			If we use the resulting formula for ''z'' assuming that ''n'' will only be 2 digits in length, we obtain the same formula as with the proof for a 2 digit ''n'':

			: <math>z = n_1(10^1 - 1) + n_0(10^0 - 1) = n_1 \cdot 9 + n_0 \cdot 0 = 9n_1</math> where <math>n_1 = a.\,</math>

			==External links==
			* [http://learnenglish.britishcouncil.org/en/word-games/magic-gopher Original Magic Gopher site]
			* [http://jeffki.blogspot.com/2007/05/magic-gopher-explained.html Very Simple Explanation]

			[[Category:Adobe Flash]]
			[[Category:Mathematical games]]
			[[Category:Articles containing proofs]]

en>SmackBot: remove Erik9bot category,outdated, tag and general fixes, added orphan tag

2009-12-16T10:53:26Z

remove Erik9bot category,outdated, tag and general fixes, added orphan tag

New page

Let me first start by introducing myself. My name is Boyd Butts even though it is not [http://www.dailyroads.com/profile_info.php?ID=2363798 over the counter std test] name on my beginning certificate. My working day occupation is a meter reader. Playing baseball is the hobby he will never quit performing. Years in the past we moved to North Dakota.