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The largest number amongst the following that will perfectly

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Director
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The largest number amongst the following that will perfectly [#permalink] New post 11 Nov 2005, 20:48
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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 [#permalink] New post 11 Nov 2005, 21:18
C AND E...LOOK GOOD...NOT DOING ANY MATH HERE JUST APPROXIMATING

100^100/1000000=1

SO E LOOKS BEST
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 [#permalink] New post 11 Nov 2005, 21:18
A...

101 ^ 2 = 10101
101 ^ 3 = 1010101

It will keep on going for ever like this.

So: 10101....1010101 - 1 = 10101....1010100

Which is only perfectly divisble by 100
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 [#permalink] New post 11 Nov 2005, 21:33
nice one...totally bummed this one...

Blue agave wrote:
A...

101 ^ 2 = 10101
101 ^ 3 = 1010101

It will keep on going for ever like this.

So: 10101....1010101 - 1 = 10101....1010100

Which is only perfectly divisble by 100
Director
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 [#permalink] New post 11 Nov 2005, 21:46
Good one, Blue agave.
It is A
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 [#permalink] New post 11 Nov 2005, 21:48
It should be 100.

If you just multiply 101 * 101 * 101 -- you can find a trend, wherein the 10s digit is 0 and the units digit is obviously 1. So, a number with its units digit "1" and tens digit "0" minus "1" will give you a number with two 0's. Hence, such a number will divide by 100.
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 [#permalink] New post 11 Nov 2005, 21:49
OMG...did anyone notice...I became CEO in training...now where is my private jet and million dollar golden parachute..
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 [#permalink] New post 11 Nov 2005, 21:51
fresinha12 wrote:
OMG...did anyone notice...I became CEO in training...now where is my private jet and million dollar golden parachute..

You'll get it when you become CEO. First, complete your training well! :wink:
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 [#permalink] New post 11 Nov 2005, 22:22
Alright guys ...

OA is B

and OE

.

101^2 = 10201. 101^2 - 1 = 10200. This is divisible by 100. Similarly try for 101^3 - 1 = 1030301 - 1 = 1030300.

So you can safely conclude that (101^1 - 1) to (101^9 - 1) will be divisible by 100 & then (101^10 - 1) to (101^99 - 1) will be divisible by 1000 & therefore (101^100 - 1) will be divisible by 10,000.

:roll: :roll:
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 [#permalink] New post 11 Nov 2005, 22:43
I dont think, your assumption is correct.

101^11 will still have its last two digits as "01"....so the answer still seems to be 100.
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Re: Prefectly Divide [#permalink] New post 11 Nov 2005, 22:52
cool_jonny009 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 = (101-1) ( 101^99 +101^98 + .....+101^1+101^0 )
101-1= 100 divisible by 100
101^n ( n is positive integer) divided by 100 has remainder of 1. There're 99-0+1= 100 such 100^n inside the second ( ) , thus their sum is divisible by 100
----> 101^100-1 is divisible by 10,000 ( =100*100)
Re: Prefectly Divide   [#permalink] 11 Nov 2005, 22:52
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