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hdvforever
Joined: 13 Mar 2006
Posts: 2
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 Posted: Mon Mar 13, 2006 1:02 pm Post subject: number systems |
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The largest number amongst the following that will perfectly divide 101^100 - 1 is
(1) 100
(2) 10,000
(3) 100100
(4) 100,000
Correct Answer - (2)
I want a foolproof and shortcut mthd for these type of problems |
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shaji
Serious about CAT
Joined: 22 May 2005
Posts: 71
Location: UK
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| Posted: Wed Mar 15, 2006 2:49 pm Post subject: Re: number systems |
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| hdvforever wrote: | The largest number amongst the following that will perfectly divide 101^100 - 1 is
(1) 100
(2) 10,000
(3) 100100
(4) 100,000
Correct Answer - (2)
I want a foolproof and shortcut mthd for these type of problems |
Hi;
The quickest fix is to look for the number of zeros trailing 101^100 -1.
101^1-1=100
101^2-1=10200
101^3-1=1030300
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101^10-1= .....000(Three zeroos)
101^100-1= .......(4 Zeros)
So the largest has to have 4 zeros, hence 10000.
Regards;
Shaji. |
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CZ@@@H@@@R
Serious about CAT
Joined: 14 Jan 2006
Posts: 64
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| Posted: Fri Mar 17, 2006 4:29 pm Post subject: Solution |
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(101^100)-1 = (1+100)^100 - 1
Using Binomial Theorem
(1+100)^100 = 1 + 100 x 100 + (Terms containing > 4 zeros)
Hence (1+100)^100 - 1 = 10000 + (Terms containing > 4 zeros)
Hence Answer 10000 |
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shaji
Serious about CAT
Joined: 22 May 2005
Posts: 71
Location: UK
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| Posted: Sat Mar 18, 2006 9:27 pm Post subject: Re: Solution |
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| CZ@@@H@@@R wrote: | (101^100)-1 = (1+100)^100 - 1
Using Binomial Theorem
(1+100)^100 = 1 + 100 x 100 + (Terms containing > 4 zeros)
Hence (1+100)^100 - 1 = 10000 + (Terms containing > 4 zeros)
Hence Answer 10000 |
Hi;
The expansion(101^100)-1does not entail more than 4 trailing zeros.
Regards
Shaji |
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CZ@@@H@@@R
Serious about CAT
Joined: 14 Jan 2006
Posts: 64
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| Posted: Sun Mar 19, 2006 10:22 am Post subject: solution |
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Using Binomial Expansion
(1+100)^100 = 1 + (10000) + (495x10^5) + ( 1617x10^8 ) + --- (100^100)
Hence [ (1+100)^100 - 1 ] = 10000 (1 + 4950 + 16170000 + ---- 100^96)
Therefore (1+100)^100 - 1 = (10000 N)
where N is an integer with last digit 1.
101^100 - 1 = ......10000 (4 Zeros)
Number of zeros trailing in (101^100)-1 = 4
Hence Answer 10000 |
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