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lets see 101^2, is 10201, 101^3 is 1030301 you will notice that the 3rd digit from the left will become 0, when we hit ^10, so based on that 100,000,00 will divide it E it is..

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...

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...

I got this reply from one Guru and i dont get it myslef

gmatclubot

Re: Largest number
[#permalink]
21 Oct 2008, 06:05