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The largest number amongst the following that will perfectly [#permalink]
 02 Jun 2008, 22:19
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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, 22:51
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, 23:14
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, 23:21
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, 23:29
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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, 23:49
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, 07:09
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, 07:22, edited 1 time in total.
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Re: PS..750+ level [#permalink]
03 Jun 2008, 07:49
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
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Re: PS..750+ level [#permalink]
03 Jun 2008, 07:50
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, 07:53
101^100-1 = = 11574001359990451186283141070066493936597109898321600750632049572420389134812037149366326601286302964 9578890272024960082993285013905624357430362636553574489997143361936192994493701723942231042303500000  but I've not found my mistake...... 
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Re: PS..750+ level [#permalink]
03 Jun 2008, 07:56
Ahh... silly mistake...Kevincan, you are right! Should be 100 instead of 99...... 
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Re: PS..750+ level [#permalink]
03 Jun 2008, 07:56
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, 08:02
by the way, if I correctly calculated, D should be the answer....
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Re: PS..750+ level [#permalink]
03 Jun 2008, 08:18
Sorry, guys. It is 27048138294215260932671947108075308336779383827810027768902010491171015143067392794394560143467445909 7335651375483564268312519281766832427980496322329650055217977882315938008175933291885667484249510000
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Re: PS..750+ level [#permalink]
03 Jun 2008, 08:31
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, 08:44
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, 21:43
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, 21:59
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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Re: PS..750+ level [#permalink]
04 Jun 2008, 05:59
icognito has probably the best way to solve something like this..
There are such questions on gmat..
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Re: PS..750+ level
[#permalink]
04 Jun 2008, 05:59
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