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The largest number amongst the following that will perfectly

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New post 03 Aug 2008, 19:13
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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New post 03 Aug 2008, 19:20
I am not sure but I think I read in school days that
a^x - b^x can always be expressed as (a-b)(a^x-1+b^x-1+...)
So if I apply that thing
101 - 1 = 100 should surely be the factor.
Please confirm the OA.
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New post 03 Aug 2008, 19:31
Not sure, but judging by the pattern of multiplying the first 3 or 4 101s, I would say E
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New post 03 Aug 2008, 20:41
gmatnub wrote:
Not sure, but judging by the pattern of multiplying the first 3 or 4 101s, I would say E

gmatnub, lets try ur mathod. I belive finding pattern is the best method on GMAT. lets see.
101^2= 10201 and 10201 - 1 = 10200. ( can't be divided by (E)100,000,00)
101^3= 1030301 and 1030301 - 1 = 1030300 ( again E is not the correct option )

there is a rule i.e. if (a-b) is divided by n then (a^k - b^k) too is divided by n. for all k>=1
However I am not sure if this rules holds true for finding the greatest divisor.
so lets try
if the highest divisor that will divide (101 - 1) is 100 then 100 will also be the highest divisor for (101^100 - 1).....Option A

even if we try the pattern mathod, as we tried above for 101^2 and 101^3 then too 100 will be the highest divisor.

So, answer IMO is A

Last edited by rishi2377 on 04 Aug 2008, 12:46, edited 1 time in total.
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New post 04 Aug 2008, 00:40
(101)^100 - 1
= (100 + 1)^100 - 1
= 100^100 + 100C1 * 100^99 * 1^1 + 100C2 * 100^98 * 1^2 + ..................... + 100C98 * 100^2 * 1 + 100C99 * 100^1 * 1^99 + 1 -1

last two 1s will get cancelled...
second last term (bold) is 10000... and everry other term has a factor of 10000,
so 10000 should be the answer...
option B
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New post 04 Aug 2008, 08:16
durgesh79 wrote:
(101)^100 - 1
= (100 + 1)^100 - 1
= 100^100 + 100C1 * 100^99 * 1^1 + 100C2 * 100^98 * 1^2 + ..................... + 100C98 * 100^2 * 1 + 100C99 * 100^1 * 1^99 + 1 -1

last two 1s will get cancelled...
second last term (bold) is 10000... and everry other term has a factor of 10000,
so 10000 should be the answer...
option B


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New post 05 Aug 2008, 18:21
I will go for 100
IMO A
last two digits are 0 in the resulting number obtained from (101)^100 - 1

OA=?????
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Re: largest no.   [#permalink] 05 Aug 2008, 18:21
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