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# The largest number amongst the following that will perfectly

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The largest number amongst the following that will perfectly [#permalink]

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17 Oct 2008, 08:28
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
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19 Oct 2008, 16:14
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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
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19 Oct 2008, 16:26
Thanks.., but that is not the OA....anyone ?
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19 Oct 2008, 17:45
its C
pls post the OA.

thanks!
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19 Oct 2008, 17:49
Sorry its not C either...I also went for C...We are running out of choices...
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19 Oct 2008, 19:20
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..
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19 Oct 2008, 19:42
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 ???
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19 Oct 2008, 19:54
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..
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19 Oct 2008, 21:09
Fresinha, is it necessary that 101^100 will result in XXXXX00001?
I am not doubting you answer but just want to confirm the rule.
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19 Oct 2008, 22:04
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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.
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20 Oct 2008, 11:48
Could you please explain how did you factor 101^100 -1.. I could not get it.

Thanks
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20 Oct 2008, 16:15
Hi scthakur

Could you explain how you broke down the equation please
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20 Oct 2008, 22:23
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)
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20 Oct 2008, 22:51
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.
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21 Oct 2008, 06:03
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...
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21 Oct 2008, 06:05
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, 06:05
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