exponenets and devisibility - Hard and time consuming

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

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...

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

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...

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..

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

distributing through, I think you get...

1^100 + 100^100 - 1

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