11no i dunno the answer i only know that it is not 1
but method please that would help me a lot
find the remainder when 3100 is divided by 100
(post ur entire method please )
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11
1http://targetiit.com/iit_jee_forum/posts/finding_last_two_digits_3747.html
1if the power had a 2 digit prime factor we can find last 2 digits easily .....but the other one is tough.......
1hey ankit wats the problem with prime number......hey yaar........solve 3100 u'll get the answer....
1easy way::
Fermat's little theorem states that if p is a prime number, then for any integer a, ap − a will be evenly divisible by p..
ap≡a(mod p)
A variant of this theorem is stated in the following form: if p is a prime and a is an integer coprime to p(eg 4 is coprime to 3), then ap − 1 − 1 will be evenly divisible by p.
ap-1≡1(mod p)
13100 = (80+1)25 = 25C0.8025 + 25C1.8024 + ... + 25C24.80 + 1
=> last two digits of 3100 are 0,1 respectively...
so when 3100 is divided by 100, the remainder will be 1... hence statement I is true...
d last digit of 3100 is 1... hence statement 2 is also true...
but statement II partly explains statement I ... even though d last digit of d expansion is 1, for d remainder to b 1 when divided by 100, the last but one digit (or d digit in ten's place should b 0)... hence d given statement is necessary but not sufficient condition to declare statement I...
well, the answer must b A... :)
13100=(32)50
(32)50=950
(92)25=8125
=(81)24.81
81 raised to any no /100 remainder is 1........because as we go simplifying we will get a 1 at the first place..........Remainder is always 1..........
33Deleted that...
[5]
actually did write 3.(33)33
33=9 [5] and then edited it... so that become wrong...(33 was odd naa so just changed 33 to 50 everywhere.. ) :P
:P
21priy : ye kaise ???
(10-1)^50 = 9^50
=-(1-10)^50 = -(-9)^50 [5] [11]
11actually the question asked was different
Statement 1: when 3100 is divided by 100 remainder is 1
Statement 2: The unit digit in 3100 is 1
11@KS ur answer is wrong
answer to that assertion question is D ie stat1 false stat2 true
11@tapanmast thanks for the effort but i would like to see an application rather than the link :)
13^100=9^50=(10-1)^50
Expand it,
50C010^50-......-500+1
=100(m)+1
Hence unit digit is one
So,When 3^100 is divided by 100,remainder is 1
21http://mathworld.wolfram.com/Congruence.html
http://en.wikipedia.org/wiki/Modular_arithmetic
