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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 17 Nov 2006, 02:54
00:00
A
B
C
D
E

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The largest number amongst the following that will perfectly divide 101^100 - 1 is

(1) 100
(2) 10,000
(3) 100^100
(4) 100,000
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Re: PS - Divisor [#permalink] New post 17 Nov 2006, 05:12
(100+1)^100 -1 = (x+1)^100 - 1 where x=100

now f(x) is divisible by x-a if f(a) = 0

f(0) = (0+1)^100 - 1 = 0

sp 100 divides the above no.
should be ans A

what's the oa??



karlfurt wrote:
The largest number amongst the following that will perfectly divide 101^100 - 1 is

(1) 100
(2) 10,000
(3) 100^100
(4) 100,000
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 [#permalink] New post 17 Nov 2006, 09:21
101 ^ 2 = 101 * 101 = 10201 = XXX01
101 ^ 3 = 101^2 * 101 = XXX01 * 101 = XXXXXX01
101^n = .................XXXXXX01
So 101^101 – 1 = ..............XXXXXX00
Thus divisible by 100
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 [#permalink] New post 18 Nov 2006, 02:30
:no
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 [#permalink] New post 18 Nov 2006, 06:40
karlfurt wrote:
:no


OK my bad

for any positive int n

(1+1/n) ^n always lies between 2 and 3

Now here in the problem 101^100 - 1

can be written as

100^100*(1/100 + 1) ^100 - (1/100)^100 *100^100

So it shouldbe perfectly divisible by 100^100

I am in a hurry to go out..If you guys didnt get it I will explain again.. :)
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 [#permalink] New post 19 Nov 2006, 06:35
trivikram wrote:
karlfurt wrote:
:no


OK my bad

for any positive int n

(1+1/n) ^n always lies between 2 and 3

Now here in the problem 101^100 - 1

can be written as

100^100*(1/100 + 1) ^100 - (1/100)^100 *100^100

So it shouldbe perfectly divisible by 100^100

I am in a hurry to go out..If you guys didnt get it I will explain again.. :)


:no

If you divide your result 100^100*(1/100 + 1) ^100 - (1/100)^100, then you get (1/100 + 1) ^100 - (1/100)^100 which might not be an integer!
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 [#permalink] New post 19 Nov 2006, 07:47
:oops:

try this

say

y = 101^100 - 1

take logs on both sides

log y = 100*log101 - log1

since log1 =0

1/100*logy = log101

means y/100 = 10^101

y = 10^103 = 10^3*10^100

which means 100^100 cannot divide it

so it should be 100,000

Sorry I just missed 2 more options :wink:

But I feel ashamed :oops:
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 [#permalink] New post 19 Nov 2006, 09:04
IMO it is B
Consider this.
Using binominal formula we get (100+1)^100-1. now the first term is 100^100 and the last term is 1^100. the term before the last term is (100C99)*(100^1)*(1^99) and when 1 is substracted the number is divisible by 100*100
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 [#permalink] New post 19 Nov 2006, 16:58
good point by BG.

I forgot the binomial formula. it should be the way to solve it.

but is it in the scope of Gmat?
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 [#permalink] New post 20 Nov 2006, 01:45
Sorry trivikram, three attempts were not sufficient!
Nice done BG.

Here is the OE and another way to solve it, even if I don't quite understand the 'you can safely conclude':

Quote:
The easiest way to solve such problems for CAT puposes is trial and error or by back substituting answers in the choices given.

1012 = 10201. 1012 - 1 = 10200. This is divisible by 100. Similarly try for 1013 - 1 = 1030301 - 1 = 1030300.

So you can safely conclude that (1011 - 1) to (1019 - 1) will be divisible by 100 & then (10110 - 1) to (10199 - 1) will be divisible by 1000 & therefore (101100 - 1) will be divisible by 10,000.
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Perhaps I am mistaken [#permalink] New post 22 Nov 2006, 06:35
Binomial formula, logs, wtf!?

Are these methods not far too complex for the GMAT and this type of problem (i know you use binomial for probability).

When I look at this problem, I see that no matter what, the number will always have a hundred value or a thousand value. That means the largest number that will divide it evenly is 1000. And that is only 1 out of every 10 times you multiply the number by itself. The other 9 it will be 100. Since 1000 isn't an option, then the answer is 100. Think of it like this.
Every power of 101 will end with the number 1. Every time you multiply one of the numbers by 101, it will there for always have a hundreds value except once every 10 times. Look at the first two powers of 101:

| 101
|X101
| 101
| 000
|+10100
|10,201
|
| 10,201
|X101
|10,201
| 000
|+1020100
|1030301


Now start thinking, since we are always multiplying the total by 101, and there is always a 1 in the unit digit of the total, then there will always be a hundred in our new total, except when the two lines of our addition from the multiplication add up to an even 1000. Which will be the case at 101^10. In fact, now that I think about it, you can even take this logic up to 101^101.

Everytime you multiply 101 X 101, the value of the hundreds place moves up by 1. Meaning 101^2 = 10,201, 101^3 = 1030301 and so on. Until it reaches 10, then the hundreds value will be a zero. Meaning for every power of 101, the largest number it is divisible by is 100, except when the power is a multiple of 10. Then it is divisible by 1000 (after you subtract the 1 of course).

Since the problem is 101^101, then the answer is 100.

Had it been 101^100, then the answer would be 1000.

Phew! Man I really hope my logic is right or that was a long, pointless post :-).
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Re: Perhaps I am mistaken [#permalink] New post 22 Nov 2006, 07:26
andrewnorway wrote:
Binomial formula, logs, wtf!?

Are these methods not far too complex for the GMAT and this type of problem (i know you use binomial for probability).

When I look at this problem, I see that no matter what, the number will always have a hundred value or a thousand value. That means the largest number that will divide it evenly is 1000. And that is only 1 out of every 10 times you multiply the number by itself. The other 9 it will be 100. Since 1000 isn't an option, then the answer is 100. Think of it like this.
Every power of 101 will end with the number 1. Every time you multiply one of the numbers by 101, it will there for always have a hundreds value except once every 10 times. Look at the first two powers of 101:

| 101
|X101
| 101
| 000
|+10100
|10,201
|
| 10,201
|X101
|10,201
| 000
|+1020100
|1030301


Now start thinking, since we are always multiplying the total by 101, and there is always a 1 in the unit digit of the total, then there will always be a hundred in our new total, except when the two lines of our addition from the multiplication add up to an even 1000. Which will be the case at 101^10. In fact, now that I think about it, you can even take this logic up to 101^101.

Everytime you multiply 101 X 101, the value of the hundreds place moves up by 1. Meaning 101^2 = 10,201, 101^3 = 1030301 and so on. Until it reaches 10, then the hundreds value will be a zero. Meaning for every power of 101, the largest number it is divisible by is 100, except when the power is a multiple of 10. Then it is divisible by 1000 (after you subtract the 1 of course).

Since the problem is 101^101, then the answer is 100.

Had it been 101^100, then the answer would be 1000.

Phew! Man I really hope my logic is right or that was a long, pointless post :-).


What a long post! But I am sorry, I ve posted already the answer just above, and the answer is 10,000!
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 [#permalink] New post 22 Nov 2006, 07:41
Yeah, I see that now, I thought it was 101^101 - 1, not 101^100 - 1. If it were 101^101 - 1, then it would be 100, since its 101^100 - 1, then its 10,000.

Just to make sure I understand this right though, if it were 101^90-1, then the answer would be 1000, correct? And 101^200 -1, the answer would be 100,000?
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 [#permalink] New post 22 Nov 2006, 17:37
Just use the binomial formula.

it is actually in high school syllabus.
  [#permalink] 22 Nov 2006, 17:37
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