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exponenets and devisibility - Hard and time consuming

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by yezz » Tue Oct 21, 2008 2:12 am
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 reached an answer ( right one) however it took me more than 5 minutes - any shorter and practical answers would be appreciated
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Re: exponenets and devisibility - Hard and time consuming

by sudhir3127 » Tue Oct 21, 2008 4:42 am
yezz 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 reached an answer ( right one) however it took me more than 5 minutes - any shorter and practical answers would be appreciated
Ithink its not a Gmat question becasue it will test u with Eulers law. which i guess is not in GMAT.

Anyways .. i cant explain u the rule here...

Eulers rule says..If M and N are 2 co primes to each other.. HCF (M,N) = 1
and N= a^p*b^q.......remainder ( M^#(N)/N) = 1. where #N = N( 1-1/a)(1-1/b)..are Euler Totient function ..

here 101 and 100 are co primes

101 ( 1-1/101) = 100

hence the expression

101^100 will be divisble by 10000 will give an remainder of 1
101^100 - 1 is +1-1 = 0.

hence the whole expression is divisble by 10000.

I am sure its tough to understand.. but please let me know if u have any doubts...
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Re: exponenets and devisibility - Hard and time consuming

by stop@800 » Tue Oct 21, 2008 6:07 am
sudhir3127 wrote:
yezz 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 reached an answer ( right one) however it took me more than 5 minutes - any shorter and practical answers would be appreciated
Ithink its not a Gmat question becasue it will test u with Eulers law. which i guess is not in GMAT.

Anyways .. i cant explain u the rule here...

Eulers rule says..If M and N are 2 co primes to each other.. HCF (M,N) = 1
and N= a^p*b^q.......remainder ( M^#(N)/N) = 1. where #N = N( 1-1/a)(1-1/b)..are Euler Totient function ..

here 101 and 100 are co primes

101 ( 1-1/101) = 100

hence the expression

101^100 will be divisble by 10000 will give an remainder of 1
101^100 - 1 is +1-1 = 0.

hence the whole expression is divisble by 10000.

I am sure its tough to understand.. but please let me know if u have any doubts...
Thanks Sudhir.

More details at
https://en.wikipedia.org/wiki/Euler's_theorem
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Re: exponenets and devisibility - Hard and time consuming

by jsl » Tue Oct 21, 2008 6:35 am
yezz 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 reached an answer ( right one) however it took me more than 5 minutes - any shorter and practical answers would be appreciated
I think this question may be asking us to simplify exponents and thus, I think it's a valid GMAT-style question. Can someone correct or help push through my logic as I think I'm halfway there....

101^100 - 1 = (1+100)^100 - 1

distributing through, I think you get...

1^100 + 100^100 - 1

1^100 = 1 so....

1 + 100^100 - 1 = 100^100

100^100 = 10^200....

...but I'm not sure where to go after this step...
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Re: exponenets and devisibility - Hard and time consuming

by sudhir3127 » Tue Oct 21, 2008 6:45 am
jsl wrote:
yezz 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 reached an answer ( right one) however it took me more than 5 minutes - any shorter and practical answers would be appreciated
I think this question may be asking us to simplify exponents and thus, I think it's a valid GMAT-style question. Can someone correct or help push through my logic as I think I'm halfway there....

101^100 - 1 = (1+100)^100 - 1

distributing through, I think you get...

1^100 + 100^100 - 1

1^100 = 1 so....

1 + 100^100 - 1 = 100^100

100^100 = 10^200....

...but I'm not sure where to go after this step...
10^200 can be written as

(10^2)100
(100^2)50
(10000)^50 which is definitely divisible by10000

hope that helps..
Last edited by sudhir3127 on Tue Oct 21, 2008 7:27 am, edited 2 times in total.
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Re: exponenets and devisibility - Hard and time consuming

by jsl » Tue Oct 21, 2008 6:52 am
sudhir3127 wrote:
jsl wrote:
yezz 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 reached an answer ( right one) however it took me more than 5 minutes - any shorter and practical answers would be appreciated
I think this question may be asking us to simplify exponents and thus, I think it's a valid GMAT-style question. Can someone correct or help push through my logic as I think I'm halfway there....

101^100 - 1 = (1+100)^100 - 1

distributing through, I think you get...

1^100 + 100^100 - 1

1^100 = 1 so....

1 + 100^100 - 1 = 100^100

100^100 = 10^200....

...but I'm not sure where to go after this step...
10^200 can be written as

(100^2)50
(10000)^50 which is definitely divisible by10000

hope that helps..
Ah of course! Thanks - great teamwork there! ;-) I get 90% there and then mess up on the last mile!
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by bluementor » Tue Oct 21, 2008 8:03 am
Hi jsl,
101^100 - 1 = (1+100)^100 - 1

distributing through, I think you get...

1^100 + 100^100 - 1
could you explain this further please?.. I'm not sure how you arrived at 1^100 + 100^100 - 1 ?

BM
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by jsl » Tue Oct 21, 2008 8:09 am
bluementor wrote:Hi jsl,
101^100 - 1 = (1+100)^100 - 1

distributing through, I think you get...

1^100 + 100^100 - 1
could you explain this further please?.. I'm not sure how you arrived at 1^100 + 100^100 - 1 ?

BM
Sure yep...

101^100 = (101)^100 agreed?

You can manipulate anything in the bracket and thus you can split out 101 into multiple numbers in order to make them more useful to you... thus I logically thought lets make 101 into 100 and 1...

thus...

(100 + 1)^100

However, now I'm confused again because I don't think you can distribute the 100 to both numbers since you're adding exponents....

(100 + 1)^100 = 100^100 + 1^100 ????????

hmmmm - think I'm wrong here.....
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by Ian Stewart » Tue Oct 21, 2008 12:19 pm
jsl wrote:
(100 + 1)^100

However, now I'm confused again because I don't think you can distribute the 100 to both numbers since you're adding exponents....

(100 + 1)^100 = 100^100 + 1^100 ????????

hmmmm - think I'm wrong here.....
No, that's not quite right, though in some cases, it can be a clever little trick to start the problem the way you did- by writing 101 as 100+1. Try smaller numbers:

101^2 = (100 + 1)^2

Since you know (x+y)^2 = x^2 + 2xy + y^2 (and *not* just x^2 + y^2), then

101^2 = (100 + 1)^2 = 100^2 + 2*100*1 + 1^2 = 10,201

If you wanted to work out what (100 + 1)^100 is, you'd want to use the binomial theorem. It's not something you'll ever need to do on the GMAT, but you'd get something like this:

(100 + 1)^100 = 100^100 + (100C1) * 100^99 * 1 + (100C2) * 100^98 * 1^2 ... + (100C99) * 100^1 * 1^99 + (100C100) * 1^100

Notice it's a lot more complicated than just 100^100 + 1.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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