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[Post New]17 May 2007 18:23:57 IST
    Subject: try this..!
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dinesh_ddt (163)

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the last three digits in 17256 is????? 
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[Post New]17 May 2007 18:43:22 IST
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lazycol (711)

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last digit is 1,but other 2??!!
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[Post New]17 May 2007 18:44:48 IST
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sherin (39)

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is it ..... .......373
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[Post New]17 May 2007 18:44:49 IST
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dinesh_ddt (163)

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wat r u talking abt im asking all last three digits...
like 7 8 9 etc.......!
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[Post New]17 May 2007 18:46:02 IST
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dinesh_ddt (163)

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no sherin its incorrect
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[Post New]17 May 2007 18:53:48 IST
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dinesh_ddt (163)

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plz guys..! try this
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[Post New]17 May 2007 19:00:01 IST
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lazycol (711)

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last 2 digits are 81 ,isn't it?
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[Post New]17 May 2007 19:36:53 IST
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catch_arnnie (521)

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the ans. is 081
 
solution:-->
 
17256 can be re-written as (7+10)256
 
using binomial expansion & writting just the 1st three terms becoz' just the 1st three terms will give the answer --->
 
256C0 7256 100 + 256C1 7255 101 + 256C2 7254 102 +.....
 
now writing the last digit of powers of 7
74K     => 01,
74K+1 => 07,
74K+2 => 49,
74K+3 => 43,  where 'K' is any whole number.
 
now from expansion,
the 1st term gives us 601 as last 3 digits, so last digits of 17256 is 601
 
the 2nd term gives us 080 as the last digits , so last digits of 17256 are 580 + 601 = 681
 
similarly, from the 3rd term, we get zero as last digit when we solve 256C2 7254 & therefore, the last 3 digits will be 000
=> last 3 digits of 17256 will be 681
 
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[Post New]17 May 2007 20:46:12 IST
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shine (262)

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hey i m geeting the last digits as
089
is tht correct
there is no right way 2 do something wrong !!!!!!!!
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[Post New]17 May 2007 21:29:35 IST
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nano0101 (44)

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No it must end with 451 ......... if this is right i will post the proof.
In the process of learnin..............blunders do happen !!!
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[Post New]17 May 2007 21:37:09 IST
    Subject: Re:try this..!
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prathima (49)

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081 is right
plz post ur solution
we can find out the mistake

GBXDHNXDFNHBHRSDRGWEASGSEDH
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[Post New]17 May 2007 21:52:04 IST
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lazycol (711)

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u can't find 3rd last digit from 256 C2 x100x7 254 becoz 3rd last digit of 
256C0x7256 also have to be considered. 
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[Post New]17 May 2007 22:00:51 IST
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nano0101 (44)

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here is the proof
we can expand (7+10)256 binomially

since a= 10 every other term other than the first and the second will have more than 2 zero.

so lets see the first term
nC07256  = 7256
 the pattern goes like 7 to the power 1k,5k,9k .......(4n-3)  end with_07
                               7 to the power 2k,6k,10k ......(4n-2) end with _49
                               7 to the power 3k,7k,11k ......(4n-1) end with_43
                               7 to the power 4k,8k,12k ...... (4n-0) end with_01
Thus 7^256 must end with _01............. again by recognising the patttern i came to a conclusion the the last three digits are  601

now the second term:
nC1725510 = 2550 x 7255
the last three digits of this will be 850

ADDind the two 850+601 we get the last digits to be 451
In the process of learnin..............blunders do happen !!!
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[Post New]17 May 2007 22:05:56 IST
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nano0101 (44)

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ya the third term
nC2 100 x 7254  = 256 x 255 x 100 x 7254
                       =6528000 x 7254
so no question of considering it as there are 3 xeros

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[Post New]18 May 2007 02:12:59 IST
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arvind.b (146)

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