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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]  02 Jun 2008, 21:19
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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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Re: PS..750+ level [#permalink]  02 Jun 2008, 21:51
fresinha12 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

probably can also be solved by factorization.

but my logic is:
101^1 - 1 = 100
101^2 - 1 = 10201 - 1 = 10200
101^3 - 1 = 1030301 - 1 = 1030300

all are divided by 100.
so A.

any body has more succint way?
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Re: PS..750+ level [#permalink]  02 Jun 2008, 22:14
Expert's post
I'm between A and C. Don't know how to exclude C....
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Re: PS..750+ level [#permalink]  02 Jun 2008, 22:21
fresinha12 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

I'm going to take a stab at this and go with (B)

101^1 = 101
101^2 = 10201
101^3 = 1030301
.
.
101^10 = would end with 1001

Similarly, 101^100 would end with 10001

It follows that 101^100 - 1 (equivalent to N.....10,001 - 1) would be divisible by (b)
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Re: PS..750+ level [#permalink]  02 Jun 2008, 22:29
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fresinha12 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

(100 + 1)^100 - 1 = 100^100 + ... + 100((100)^1)((1)^99) + 1^100 - 1

This is a multiple of 10,000
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Re: PS..750+ level [#permalink]  02 Jun 2008, 22:49
Expert's post
kevincan wrote:
(100 + 1)^100 - 1 = 100^100 + ... + 100((100)^1)((1)^99) + 1^100 - 1
This is a multiple of 10,000

My reasons are the same, but

$$(100 + 1)^{100} - 1 = 1^{100}+99*1^{99}*100^1+\frac{99*98}{2}*1^{98}*100^2+....+99*1^1*100^{99}+100^{100} -1=$$

$$=99*100+\frac{99*98}{2}*1*100^2+....+99*1*100^{99}+100^{100}=$$

$$=99*100+100^2*(\frac{99*98}{2}+....+99*100^{97}+100^{98})$$

So, (100 + 1)^{100} - 1 is not divisible by 10000
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Re: PS..750+ level [#permalink]  03 Jun 2008, 06:09
OA is B..

here is how..

101^2 -1 =10201-1 =10200 this is divisible by 100..correct..

101^10 -1=102000-1= which is divisible 1000

101^100 - 1 will be divisible by 10,000.

Last edited by FN on 03 Jun 2008, 06:22, edited 1 time in total.
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Re: PS..750+ level [#permalink]  03 Jun 2008, 06:49
fresinha12 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

Why not A as well?

If it must divide by B then surely it will divide by A
Current Student
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Re: PS..750+ level [#permalink]  03 Jun 2008, 06:50
GMATBLACKBELT wrote:
fresinha12 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

Why not A as well?

If it must divide by B then surely it will divide by A

its asking for the largest number...

comeon no +1 for me???
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Re: PS..750+ level [#permalink]  03 Jun 2008, 06:53
Expert's post
101^100-1 =
= 11574001359990451186283141070066493936597109898321600750632049572420389134812037149366326601286302964
9578890272024960082993285013905624357430362636553574489997143361936192994493701723942231042303500000

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Re: PS..750+ level [#permalink]  03 Jun 2008, 06:56
Expert's post
Ahh... silly mistake...Kevincan, you are right! Should be 100 instead of 99......
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Current Student
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Re: PS..750+ level [#permalink]  03 Jun 2008, 06:56
incognito1 wrote:
fresinha12 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

I'm going to take a stab at this and go with (B)

101^1 = 101
101^2 = 10201
101^3 = 1030301
.
.
101^10 = would end with 1001

Similarly, 101^100 would end with 10001

It follows that 101^100 - 1 (equivalent to N.....10,001 - 1) would be divisible by (b)

this is the OE...excellent work ico
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Re: PS..750+ level [#permalink]  03 Jun 2008, 07:02
Expert's post
by the way, if I correctly calculated, D should be the answer....
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Joined: 17 Nov 2007
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Re: PS..750+ level [#permalink]  03 Jun 2008, 07:18
Expert's post
Sorry, guys. It is
27048138294215260932671947108075308336779383827810027768902010491171015143067392794394560143467445909
7335651375483564268312519281766832427980496322329650055217977882315938008175933291885667484249510000
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Re: PS..750+ level [#permalink]  03 Jun 2008, 07:31
walker wrote:
Sorry, guys. It is
27048138294215260932671947108075308336779383827810027768902010491171015143067392794394560143467445909
7335651375483564268312519281766832427980496322329650055217977882315938008175933291885667484249510000

how did you get that? my computer does not give the correct calculation.

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Re: PS..750+ level [#permalink]  03 Jun 2008, 07:44
Expert's post
GMAT TIGER wrote:
how did you get that? my computer does not give the correct calculation.

I found a program that can calculate with 5000 digits
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Re: PS..750+ level [#permalink]  03 Jun 2008, 20:43
Okay this is the type of problem I look at and go eenie meenie mineee mo .... guess and move on ...
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Re: PS..750+ level [#permalink]  03 Jun 2008, 20:59
Expert's post
bsd_lover wrote:
Okay this is the type of problem I look at and go eenie meenie mineee mo .... guess and move on ...

By the way, how two prove that 101^1-1 is not divisible by 100100?
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Current Student
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Re: PS..750+ level [#permalink]  04 Jun 2008, 04:59
icognito has probably the best way to solve something like this..

There are such questions on gmat..
Re: PS..750+ level   [#permalink] 04 Jun 2008, 04:59
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