CAT4MBA

  m gettin 888 

(2^999/1000) = 2^999/125 x 2^3 = 2^996/125
= 2^960 x 2^36/ 125

Now from Eulers theorem:
No of natural no's less than 125 & co-prime with it =125 x (1-1/5) x (1-1/5) x (1-1/5) = 64
& 64 x 15 =960, hence 2^960 is div by 125 as both 2 & 125 are co-prime.
Thus,
qn becomes: 2^36/125 = 2^35 x 2/125 = (2^7)^5 x 2/125 = 128^5 x 2 /125
= 3^5 x 2 /125 = 486/125 =Rem is 111
Bt as we cancelled 2^3 from the numerator in the beginning,so now multiplying it with remainder, we get

rem = 111 x 2^3 = 888

One more question: Rem of  40 ! / 83?

yaar euler no.of 1000 is also 64 so if proceede in that way the ans is sth else...pls explain

hi guys,
R(2^999/1000)

= 8 x R(2^996/125)----(i)

now 2 and 125 are co prime...

Euler no. of 125 is 125 x 4/5 = 100

hence R(2^100/125) = 1

hence {i} becomes,

= 8 x R(2^96/125)

2^10/125 gives remainder 24.

= 8 x R(24^9 x 2^6/125)

24^2/125 gives rem as 76

= 8 x R(76^4 x 24 x 64/125)

76^2/125 gives remainder as 51 and (24 x 64)/125 gives rem as 36

= 8 x R(51 x 36/125)

= 8 x 86

= 688

I will be happy if somebody comes up with a shorter method.... ;-)

regards,
shaheen

Thanx  ya for the solutn n clearing my doubt regarding  EULER'S theorem...

hi neel,
thx buddy....can u give the ans n if possible the soln for the qn."Rem of 40 ! / 83?"i tried solving it...but couldnt...

regards,
shaheen

hey the ans to the qn 40 ! / 83 is given as 81.....

n try 19^92/92 ?? m gettin "9" as remainder using Euler theorem bt the ans is 49..plz chk it once4me...

eluer no of 92 = 88 = 92/2/23*22

thus 19^92 = 19^4 mod 92

19 ^ 2 = -7 mod 92
thus 19^4 = (-7)^2 = 49 mod 92

eluer no of 92 = 88 = 92/2/23*22????????? HOW its cuming out to be 92 x 1/2 x 22/23 = 44