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Botany

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20 Feb 2010 23:45:48 IST

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453

Find no. of 5 digit numbers divisible by 3

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Find no. of 5 digit numbers divisible by 3

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Abhijeet Srivastava

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21 Feb 2010 00:11:30 IST

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Do u know answer...or just I tell u the technique ??????

Guruprasad Raghavan

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21 Feb 2010 00:15:46 IST

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tell the technique also

Abhijeet Srivastava

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21 Feb 2010 00:23:19 IST

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V know that the divisibility for 3 is that the sum of numbers should be divisible by 3....

So,in this question we have to find the number of integral solution as shown below:

(x¹+x²+......x⁹)for first position and (x⁰+x¹+x²+......x⁹)⁴for rest of 4 position.....this means:

(x¹+x²+......x⁹)(x⁰+x¹+x²+......x⁹)⁴=x³,x⁶,x⁹.......upto x⁴⁵.....becoz last term will be 9in all 5 places.&that sums up to 45.

If u have don't agree then u can tell me......

Salonii Sharma

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21 Feb 2010 00:28:57 IST

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Is it 30000??

Abhijeet Srivastava

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21 Feb 2010 00:34:43 IST

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Tell me whether u agree wid my technique ????

Salonii Sharma

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21 Feb 2010 00:38:57 IST

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I am sorry. I didn't understand your technique.

Abhijeet Srivastava

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21 Feb 2010 00:43:06 IST

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first of all....tell me if u know finding number of integral solution which we studied in Binomial Chapter ????

Salonii Sharma

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21 Feb 2010 00:45:33 IST

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I am quite well acquainted with binomial theorem of class eleventh.

Abhijeet Srivastava

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21 Feb 2010 00:49:03 IST

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by the way I was asking Guruprasad. that if he approve my method ????

Salonii Sharma

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21 Feb 2010 00:50:19 IST

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Even I am interested in knowing the answer and how you solved with your unique method!! Dude this is CLEARLY an A.P. question.

Abhijeet Srivastava

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21 Feb 2010 00:53:55 IST

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very Nice....so there will no be such topic of PnC then....we'll do wid AP & GP!!!!

Salonii Sharma

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21 Feb 2010 00:56:12 IST

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Arre yaar this is CLEARLY an A.P. question which is simple to death. I don't know why you're using your weird confusing P n C and Binomial theorem ke tactics. :-P

Abhijeet Srivastava

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21 Feb 2010 00:57:44 IST

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first u tell ur method.....

Salonii Sharma

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21 Feb 2010 01:02:16 IST

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My method?? I have told you it's an A.P. problem. Isn't that sufficient??? For your satisfaction, I will.

We know that five digit numbers are 100000-99999.

Right?? Jismei

"a" should be 10002 which is divisible by three and "an" i.e. the last digit is 99999. now solve for "n" with the formula an=a=(n-1)d.

Yasin

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21 Feb 2010 01:40:53 IST

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why shud we use binomial theorem here? can't we use th simpe thing: highest 5 digit no. - lowest 5 digit no. divided by three will give us th approximate, then we can think a bit n do th rest

Salonii Sharma

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21 Feb 2010 03:42:39 IST

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Arre yaaaaaaaaarrr why will you do that??????????????? It's a CLEAR A.P. question!!!!!!!!!!!!!!!!!!!!!!!!!!!

Aadithya Narayanan

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21 Feb 2010 13:36:18 IST

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we know every third number is a multiple of three starting from 10002 the greatest 5 didit number is 99999..we just need to find the numbers so consider an A.P. with first digit 10002 And nth term

99999...

therefore the answer is simply 30,000

Salonii Sharma

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21 Feb 2010 13:38:44 IST

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Toh mei kya SPANISH bol rahi hun???!!!!!!!!!!!!!! God people are sooo strong headed here!!!!!

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