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Here's my new signature: Micasta (talk) 03:19, 5 March 2009 (UTC)
Red is not my color. Micasta (talk) 07:11, 11 March 2009 (UTC)
BB Daily Gubbins: April 14 2009
From BrainBashers Daily Gubbins:
Arrange the digits 1-9 inclusive so as to form two numbers, one of which is the square of the other.
Made by: Micasta (talk) 11:14, 19 April 2009 (UTC)
- Let X = x2.
- Squares can only end in 1, 4, 9, 6, 25, 00. Both X and x don't have the digit 0 in them. X cannot end in 25, because x cannot end in 5 at the same time. Therefore, X can only end in 1, 4, 6 or 9. x cannot end in either 1 or 6 because this will result in X also having 1 or 6 as its last digit. Therefore, x can only end in 2, 3, 4, 7, 8, 9.
- The square of a 1-digit number can have either 1 or 2 digits. The square of a 2-digit number can have either 3 or 4 digits. The square of an n-digit number can have either 2n-1 or 2n digits. Since there are 9 digits to be distributed between X and x, x has 3 digits and X has 6 digits.
- The largest 3-digit number whose square has exactly 5 digits is 316, whose square is 99,856. Since X has 6 digits, the first digit of x cannot be smaller than 3.
- The digital roots of a square can only be 1, 4, 7 or 9.
|3, 6, 9||9|
- Let's play Kakuro. Three non-repeating digits can add up to anywhere from 1+2+3=6 to 9+8+7=24. Six non-repeating digits can add up to anywhere from 1+2+3+4+5+6=21 to 9+8+7+6+5+4=39. Digital roots limit the sum of the digits of X to 22, 25, 27, 28, 31, 34, 36. Since the sum of digits 1 to 9 is 45, the sum of the digits of x can be: 23, 20, 18, 17, 14, 11, 9, respectively.
- When the relation between the digital roots of X and x is considered, the list is narrowed down to the following:
|Sum of digits
|Sum of digits
|Possible digits of x|
|18||27||1+8+9, 2+7+9, 3+6+9,
3+7+8, 4+5+9, 4+6+8,
|17||28||1+7+9, 2+6+9, 2+7+8,
3+5+9, 3+6+8, 4+5+8,
|9||36||1+2+6, 1+3+5, 2+3+4|
- Imposing the limits on the first and last digits of x and on the last digit of X limits the list even further. For example, in the case of 1+8+9, the first and last digits cannot be 1. Since there is a digit 1 in x, the last digit cannot be 9; else, X will have a last digit of 1. Therefore, the only possible value for x using this combination is 918 (which has a square of 842,724 and cannot be the answer).
Anyway, the answers are 567 with 321,489, and 854 with 729,316.
If you think there are better ways to solve this puzzle, kindly leave a message in this talk page. I would love to learn from you! Puzzle is from BrainBashers.
WikiProject change for Talk:Psychrometrics
After I changed the Wikiproject from Chemical and Bio engineering to engineering, I noticed that this was a relatively recent edit by you. I started a thread on the talk page if you would like to discuss. I am a mechanical EIT so it is where my mind goes, but if the field is used in chemical engineering as well I don't want to bury that (AFAIK there is no Wikiproject Mechanical Engineering). Thanks! Jminthorne (talk) 01:09, 24 August 2009 (UTC)
Wikipedia Philippine Chapter in Philippine Youth Congress
I am glad to announce to you that we will be debuting as an organization at the Philippine Youth Congress in Information Technology on September 14 to 17, 2010 at the University of the Philippines, Diliman. Jojit will be Wikimedia Philippines resource speaker at the second day of the conference at the UP Film Center. He will be speaking about Wikipedia and how it revolutionizes the World Wide Web. That will be at 9:00 to 10:00 am. We will also set up a booth at the UP Bahay ng Alumni and we will showcase our existing and future projects. We encourage you to participate in our first major project. We have prepared food and refreshments for you. Please let us know so that we can enlist you to our delegation.--Exec8 (talk) 18:59, 2 September 2010 (UTC)
You are invited to the 3rd Philippine Wiki Conference (WikiCon) on May 26, 2012 9am-1pm at the co.lab.exchange in Pasig City. Please fill this form should you signify interest. --Exec8 (talk) 17:45, 6 May 2012 (UTC)