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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] New post 25 Feb 2006, 10:47
The largest number amongst the following that will perfectly divide (101^100) - 1 is

a) 100
b) 10,000
c) 100^100
d) 100,000

:beat

Last edited by jlui4477 on 25 Feb 2006, 18:40, edited 2 times in total.
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Re: 100^100-1 [#permalink] New post 25 Feb 2006, 16:39
jlui4477 wrote:
The largest number amongst the following that will perfectly divide (100^100) - 1 is

a) 100
b) 10,000
c) 100^100
d) 100,000


There must be some typo in the question because each of the choices is a multiple of 10 whereas the units digit of (100^100) - 1 cannot be 0.
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 [#permalink] New post 25 Feb 2006, 16:48
I too believe question to be wrong for the same reason.
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 [#permalink] New post 25 Feb 2006, 18:33
100 - A it is.

What I did is

101 * 101 = 10201

10201 * 101 = 1030301

So I concluded that right most digit 101 ^ 100 will be a 1 second from right will be 0 and third from right will be a non-zero. That means max number will be 100.

I don't know if this is correct or not.

Please show your approach also.
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 [#permalink] New post 25 Feb 2006, 18:44
ps_dahiya wrote:
100 - A it is.

What I did is

101 * 101 = 10201

10201 * 101 = 1030301

So I concluded that right most digit 101 ^ 100 will be a 1 second from right will be 0 and third from right will be a non-zero. That means max number will be 100.

I don't know if this is correct or not.

Please show your approach also.

ps_dahiya my little note was in reference to the typo I made earlier. The equation is (101^100) - 1. Sorry for the confusion. :pc
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 [#permalink] New post 25 Feb 2006, 20:18
The correct answer should be B, 10,000.
101*101=10201
if we raise this to 10 we get the hundreds position as zero.
last four digits of (101)^10= 1001
If we see the trend,the last two digits remain the same --01
the digts from third onward are equal to the power,if it is 2 it will be 2 and if we make it 10 the third and fourth become 10.
So if we take this to 100, (101)^100 we get the last 5 digits as 10001
subtract 1 from this we get our answer 10,000
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Re: 101^100-1 [#permalink] New post 25 Feb 2006, 22:59
jlui4477 wrote:
The largest number amongst the following that will perfectly divide (101^100) - 1 is

a) 100
b) 10,000
c) 100^100
d) 100,000

:beat



Confirm that it's B!
The problem can be understood this way: find the largest divisor of 101^100 -1 among the following.
let's express 101^100 this way:
101^100= (100+1)^100 = 100^100*1^0+ A*100^99*1^1+ B*100^98*1^2 + ........+ Z*100^98*1^2 + Y*100^1*1^99+ T*100^0*1^100

The rule of expression is : the power of 100 reduces from 100 to 0 WHEREAS the power of 1 increases from 0 to 100

A, B, Z..... are coefficients which obeys some strict rule of the so-called Pascal triangle. From the Pascal triangle rule, Y (the last coefficient) is 100 and T=1

The remainder of this expression depends on the 2nd last term, (since the last term T*100^0*1^100 =1 . This last term is already terminated by -1 )
the second last term is Y*100^1*1^99= Y*100^1 ...Y=100 ---> Y*100= 10000.

Even the smallest term of the expression is perfectly divided by 10000 ---> the whole expression is the same.

The explanation looks lengthy but the logical processing is prompt if we know the generic expression of (a+b)^n
Re: 101^100-1   [#permalink] 25 Feb 2006, 22:59
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