The number #7^(1995)# when divided by 100 leaves remainder?( Solve using Binomial Theorem)

2 Answers
Mar 5, 2018

The remainder is #=43#

Explanation:

Perform this with modular arithmetic

#7^1 ≡7[ mod100]#

#7^2≡49 [mod100]#

#7^3≡43 [mod100]#

#7^4≡1 [mod100]#

But,

#1995=4xx498+3#

Therefore,

#7^1995≡ 7^(4*498+3)≡(7^4)^498*7^3≡1^484*43≡43[mod100]#

Mar 5, 2018

#43#

Explanation:

#7^4 = 2401=24 xx 100 + 1 = 100n+1#

#1995=4 xx 498+3 rArr 7^1995 = 7^3 xx (100n+1)^498#

then

#7^1995 equiv 7^3 xx (100n+1)^498 equiv 7^3 equiv 43 mod 100 #

NOTE

the binomial expansion theorem tell us that

#(100n+1)^498 = 100m + 1#