|
Author |
Message |
|
TAGS:
|
|
|
Intern
Joined: 23 Sep 2007
Posts: 37
Followers: 0
Kudos [?]:
1
[0], given: 0
|
The largest number amongst the following that will perfectly [#permalink]
 17 Oct 2008, 09:28
The largest number amongst the following that will perfectly divide
101^100 - 1 is
(a) 100
(b) 10,000
(c) 100100
(d) 100,000
(e) 100,000,00
|
|
|
|
|
|
|
Manager
Joined: 14 Jan 2006
Posts: 93
Schools: HKUST
Followers: 1
Kudos [?]:
26
[1] , given: 0
|
Re: Largest number [#permalink]
19 Oct 2008, 17:14
1
This post received
KUDOS
singaks wrote: The largest number amongst the following that will perfectly divide
101^100 - 1 is
(a) 100
(b) 10,000
(c) 100100
(d) 100,000
(e) 100,000,00 101^100 - 1^100 =(101-1) (101^99 + .... + 1) =100 (.....) Hence, A
|
|
|
|
|
|
Intern
Joined: 23 Sep 2007
Posts: 37
Followers: 0
Kudos [?]:
1
[0], given: 0
|
Re: Largest number [#permalink]
19 Oct 2008, 17:26
Thanks.., but that is not the OA....anyone ?
|
|
|
|
|
|
Intern
Joined: 14 Jul 2008
Posts: 12
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Re: Largest number [#permalink]
19 Oct 2008, 18:45
its C
pls post the OA.
thanks!
|
|
|
|
|
|
Intern
Joined: 23 Sep 2007
Posts: 37
Followers: 0
Kudos [?]:
1
[0], given: 0
|
Re: Largest number [#permalink]
19 Oct 2008, 18:49
Sorry its not C either...I also went for C...We are running out of choices...
|
|
|
|
|
|
Current Student
Joined: 28 Dec 2004
Posts: 3440
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 11
Kudos [?]:
134
[0], given: 2
|
Re: Largest number [#permalink]
19 Oct 2008, 20:20
singaks wrote: The largest number amongst the following that will perfectly divide
101^100 - 1 is
(a) 100
(b) 10,000
(c) 100100
(d) 100,000
(e) 100,000,00 lets see 101^2, is 10201, 101^3 is 1030301 you will notice that the 3rd digit from the left will become 0, when we hit ^10, so based on that 100,000,00 will divide it E it is..
|
|
|
|
|
|
Intern
Joined: 23 Sep 2007
Posts: 37
Followers: 0
Kudos [?]:
1
[0], given: 0
|
Re: Largest number [#permalink]
19 Oct 2008, 20:42
Sorry Guys....The OA is not even E....Since we are down to only 2 options...I will give the answer...The OA is B...can anybody wrestle is one ???
|
|
|
|
|
|
Current Student
Joined: 28 Dec 2004
Posts: 3440
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 11
Kudos [?]:
134
[0], given: 2
|
Re: Largest number [#permalink]
19 Oct 2008, 20:54
singaks wrote: Sorry Guys....The OA is not even E....Since we are down to only 2 options...I will give the answer...The OA is B...can anybody wrestle is one ??? actually i just made a mistake posting my ans...the working is actually correct..ans should be b... 101*101=101^2 or 10201 101*101*101=101^3 or 1030301 101^10 will get you XXXXX001 101^100 will get XXX00001 which is divisible by 10,000 after you -1 from the unit digit..
|
|
|
|
|
|
Senior Manager
Joined: 18 Jun 2007
Posts: 300
Followers: 2
Kudos [?]:
12
[0], given: 0
|
Re: Largest number [#permalink]
19 Oct 2008, 22:09
Fresinha, is it necessary that 101^100 will result in XXXXX00001?
I am not doubting you answer but just want to confirm the rule.
|
|
|
|
|
|
SVP
Joined: 17 Jun 2008
Posts: 1593
Followers: 7
Kudos [?]:
131
[1] , given: 0
|
Re: Largest number [#permalink]
19 Oct 2008, 23:04
1
This post received
KUDOS
I would have surely answered this wrong in the real exam. However, after spending more than 5 mins, here is my explanation of the OA.
101^100-1
= (101-1)*(101^99 + 101^98 + 101^97 + .......+1)
= 100 * ((101^99 + 101^98 + 101^97 + .......+1)
= 100 * (101-1) * (101^98 + 101^97 + ........ + 1)
= 100 * 100 * (101^98 + 101^97 + ........ + 1)
Thus, the above expression will be completely divisible by 100, 10000, 1000000, ....etc.
Since the largests of the numbers in the answer choices from the series above is 10000, hence B should be the answer.
|
|
|
|
|
|
Manager
Joined: 28 Jul 2004
Posts: 142
Location: Melbourne
Schools: Yale SOM, Tuck, Ross, IESE, HEC, Johnson, Booth
Followers: 1
Kudos [?]:
5
[0], given: 2
|
Re: Largest number [#permalink]
20 Oct 2008, 12:48
Could you please explain how did you factor 101^100 -1.. I could not get it. Thanks
_________________
kris
|
|
|
|
|
|
Manager
Joined: 09 Jul 2008
Posts: 111
Location: Dallas, TX
Schools: McCombs 2011
Followers: 1
Kudos [?]:
5
[0], given: 1
|
Re: Largest number [#permalink]
20 Oct 2008, 17:15
Hi scthakur
Could you explain how you broke down the equation please
|
|
|
|
|
|
SVP
Joined: 17 Jun 2008
Posts: 1593
Followers: 7
Kudos [?]:
131
[0], given: 0
|
Re: Largest number [#permalink]
20 Oct 2008, 23:23
kman wrote: Hi scthakur
Could you explain how you broke down the equation please a^n - 1 can be written as (a-1)(a^n-1 + a^n-2 + a^n-3 + ......+1). For example, a^3-1 = (a-1)(a^2 + a + 1). a^4-1 = (a-1)(a^3 + a^2 + a + 1)
|
|
|
|
|
|
Director
Joined: 23 May 2008
Posts: 842
Followers: 3
Kudos [?]:
19
[0], given: 0
|
Re: Largest number [#permalink]
20 Oct 2008, 23:51
scthakur wrote: I would have surely answered this wrong in the real exam. However, after spending more than 5 mins, here is my explanation of the OA.
101^100-1
= (101-1)*(101^99 + 101^98 + 101^97 + .......+1)
= 100 * ((101^99 + 101^98 + 101^97 + .......+1)
= 100 * (101-1) * (101^98 + 101^97 + ........ + 1)
= 100 * 100 * (101^98 + 101^97 + ........ + 1)
Thus, the above expression will be completely divisible by 100, 10000, 1000000, ....etc.
Since the largests of the numbers in the answer choices from the series above is 10000, hence B should be the answer. doesnt this mean that 100,000,000 is a factor, which would be E? unless the comma is in the wrong place on answer choice E.
|
|
|
|
|
|
SVP
Joined: 05 Jul 2006
Posts: 1562
Followers: 4
Kudos [?]:
63
[0], given: 32
|
Re: Largest number [#permalink]
21 Oct 2008, 07:03
singaks wrote: The largest number amongst the following that will perfectly divide
101^100 - 1 is
(a) 100
(b) 10,000
(c) 100100
(d) 100,000
(e) 100,000,00 Ithink its not a Gmat question becasue it will test u with Eulers law. which i guess is not in GMAT. Anyways .. i cant explain u the rule here... Eulers rule says..If M and N are 2 co primes to each other.. HCF (M,N) = 1 and N= a^p*b^q.......remainder ( M^#(N)/N) = 1. where #N = N( 1-1/a)(1-1/b)..are Euler Totient function .. here 101 and 100 are co primes 101 ( 1-1/101) = 100 hence the expression 101^100 will be divisble by 10000 will give an remainder of 1 101^100 - 1 is +1-1 = 0. hence the whole expression is divisble by 10000. I am sure its tough to understand.. but please let me know if u have any doubts...
|
|
|
|
|
|
SVP
Joined: 05 Jul 2006
Posts: 1562
Followers: 4
Kudos [?]:
63
[0], given: 32
|
Re: Largest number [#permalink]
21 Oct 2008, 07:05
yezz wrote: singaks wrote: The largest number amongst the following that will perfectly divide
101^100 - 1 is
(a) 100
(b) 10,000
(c) 100100
(d) 100,000
(e) 100,000,00 Ithink its not a Gmat question becasue it will test u with Eulers law. which i guess is not in GMAT. Anyways .. i cant explain u the rule here... Eulers rule says..If M and N are 2 co primes to each other.. HCF (M,N) = 1 and N= a^p*b^q.......remainder ( M^#(N)/N) = 1. where #N = N( 1-1/a)(1-1/b)..are Euler Totient function .. here 101 and 100 are co primes 101 ( 1-1/101) = 100 hence the expression 101^100 will be divisble by 10000 will give an remainder of 1 101^100 - 1 is +1-1 = 0. hence the whole expression is divisble by 10000. I am sure its tough to understand.. but please let me know if u have any doubts... I got this reply from one Guru and i dont get it myslef 
|
|
|
|
|
|
|
Re: Largest number
[#permalink]
21 Oct 2008, 07:05
|
|
|
|
|
|
|
|
|
|
|
close
