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 Aptitude  |  Question 5 of 41    Print  

(998-1)(998-2)(998-3)…………..(998-n)=------- when n>1000

zero


  
Total Answers and Comments: 13 Last Update: March 08, 2008   
  
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June 17, 2005 14:40:29   #1  
Sandep Kumar        

RE: (998-1)(998-2)(998-3)…………..(998-n)=------- when n>1000
It is surely 0 as when n=998, 
 
then 998-998 =0 so all is 0

 
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June 26, 2006 09:51:31   #2  
Akash        

RE: (998-1)(998-2)(998-3)…………..(998-n)=------- ...
Zero Because at some point n = 998 and 998-998 =0
 
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August 13, 2006 11:43:18   #3  
thiyagarajanganesan Member Since: August 2006   Contribution: 4    

RE: (998-1)(998-2)(998-3)…………..(998-n)=------- ...
the questionr answer is wrong.how 998 comes when n is greater than 1000.explain me please.
 
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November 30, 2006 22:29:44   #4  
ss        

RE: (998-1)(998-2)(998-3)…………..(998-n)=------- ...

if u take 998 common then eq. will be

(998)^n(1-1)(1-2)....................(1-n)=(998)^n(0)(-1)......

=0


 
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March 09, 2007 02:25:51   #5  
Hemanshu        

Answer is 55
Ans == 55

Here are all the combinations

009 019 029 039 049 059 069 079 089 099
109 119 129 139 149 159 169 179 189
209 219 229 239 249 259 269 279
309 319 329 339 349 359 369
409 419 429 439 449 459
509 519 529 539 549
609 619 629 639
709 719 729
809 819
909

 
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April 17, 2007 04:49:06   #6  
ram        

RE: (998-1)(998-2)(998-3)…………..(998-n)=------- ...
as n>1000.in the series as u go there will be (998-998)=0 and the whole product is zero.1, 2,3 .......................................998,999,1000.
 
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May 07, 2007 07:32:01   #7  
Boopalan        

RE: (998-1)(998-2)(998-3)…………..(998-n)=------- ...
The condition given in the question is n>1000, therefore the consideration n=998 is not valid.
 
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May 12, 2007 02:26:37   #8  
kiran        

RE: (998-1)(998-2)(998-3)…………..(998-n)=------- ...


   The above problem result is zero bcz n=998 in one position.
             our condition is n>1000
  but we give up to so on (998-1)(998-2)(998-3)(998-4)………..(998-998)(998-999)(998-1000) (998-n)
                      n>1000
1 is not greater then 1000
  how it is possible (998-1)


 
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May 12, 2007 03:05:11   #9  
MANISHA        

RE: (998-1)(998-2)(998-3)…………..(998-n)=------- ...
As n is greater than 1000. but you have not observed that n is getting reduced by one in each step. so at certain point n will become 998. And 998-998 will be '0'. Therefore the answer is '0'.
 
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July 20, 2007 13:20:05   #10  
sameer2386 Member Since: April 2007   Contribution: 3    

RE: (998-1)(998-2)(998-3)…………..(998-n)=------- ...
The Q is (998-1)(998-2)(998-3)…………..(998-n)=------- (n>1000)

as u solve it u will be in a position that
(998-1)(998-2)(998-3)…………..(998-998)(998-999)(998-1000)------(998-n)

so means at some position it is (998-998)
means answer will be zero

 
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