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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Nov 2007 20:11:55 IST
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prove that 18! is divisible by 23
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24 Nov 2007 20:31:31 IST
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it is not divisible by 23 23 is prime and 17! doesn't have 23 as a prime factor
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24 Nov 2007 22:16:01 IST
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yup it is not divisible man.. question is wrong.
unless it is a silly riddle.
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* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
  
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24 Nov 2007 22:31:09 IST
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Yes 23 is a prime number and 18! doesn't have 23 as its factor so its not divisible.
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Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur
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25 Nov 2007 01:12:07 IST
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0.782608696 lol tis is d answer..soo its divisible hehe
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2 Dec 2007 16:37:22 IST
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ha ha
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S.Raghudevan
Everything that's happening as it should be happening, because of the simple fact that it's supposed to be happening just as it is happening. |
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5 Dec 2007 23:26:53 IST
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On dividing 18! from 23 we get, 278364074162086.95652173913043478.
So its divisible.
Hahahaha
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10 Dec 2007 08:22:29 IST
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Yes 23 is a prime number
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" You get What u deserve and not what you desire"
R.darshan kumar |
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16 Dec 2007 22:16:33 IST
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i think im usng calcy!!! i'm sure dat u can't tell it widout it. and to make chuckles like others- hohohoho hahahaha heheheheheeeee
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19 Dec 2007 23:22:27 IST
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impossible!!!!!!!!!!!!!!!!!!!!!!!!
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20 Dec 2007 12:12:12 IST
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okay, that was boring. here is a real good one. i mean, there are vairious ways to do it, but the most brilliant answer gets a salute from me. prove that any number of the form ABCABC is divisible by 7,11 and 13.
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Ratings do not necessarily signify that someone is good, or bad. I'm here to learn and help others learn, and a person unlocking the mysteries with the help of my solution, to a nagging problem, means more pleasure to me than ratings can ever make me feel. |
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22 Dec 2007 23:43:29 IST
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hahahahahaaaaaaaaa
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24 Dec 2007 12:40:59 IST
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oye, its not a joke prove that any number of the form ABCABC, where A,B and C are single digit numbers, is divisible by 7,11 and 13.
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Ratings do not necessarily signify that someone is good, or bad. I'm here to learn and help others learn, and a person unlocking the mysteries with the help of my solution, to a nagging problem, means more pleasure to me than ratings can ever make me feel. |
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24 Dec 2007 13:00:15 IST
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@TKA I CAN PROVE THAT NO. IS DIVISIBLE BY 11. ADD ALL DIGITS AT EVEN PLACES. THEN ADD ALL DIGITS AT ODD PLACES THEN SUBTRACT THESE TWO ADDN .IF THIS COMES OUT TO BE DIVISIBLE BY 11,THE NO. IS DIVISIBLE BE 11.IN THIS CASE IT COMES OUT TO BE 0(A+C+B-B-A-C).HENCE NO. IS DIVISIBEL BY 11. FOR 7 &13 I CAN GIVE U TESTS. Test for divisibility by 7. Double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7, then so was the original number. Apply this rule over and over again as necessary. Example: 826. Twice 6 is 12. So take 12 from the truncated 82. Now 82-12=70. This is divisible by 7, so 826 is divisible by 7 also. Test for divisibility by 13. Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. Apply this rule over and over again as necessary.
Example: 50661-->5066+4=5070-->507+0=507-->50+28=78 and 78 is 6*13, so 50661 is divisible by 13.
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24 Dec 2007 13:05:54 IST
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