Please solve the below...
Please solve the below reminder questions
1) find the remainder 7^(99) when divided by 2400
a 1
b 343
c 49
d 7
2) find the remainder when 83^(261) divided by 17
a 13
b 9
c 8
d 2
n/a
1)
7^99/(2^5 * 3 *5^2)
using chinease remainder the:
32x+23=3y+1=25z+18
solving the above eq
ans =343
2)
83^261/17 =15^5/17 using euler (16)
2 ans
n/a
find the remainder when 256256256..................300 terms
is divided with 37
Can you explain the procedure for solving the questions ?
1000/37 remainder = 1.
10^3 /37 rem = 1
10^6/37 rem = 1 (remainder theorem)
Consider the no,
256 256 256 = 256*10^6 + 256*10^3 + 256
when divided by 37,
applying remainder theorem
the remainder would be same as when { 256(1) + 256(1) + 256(1) } /37.
or 256*3/37
for 256256....300 terms there are 100 triplets of 256. So their sum would be
we just need to 25600/37
256/37 rem = -3 100/37 = -11 again applying remainder theorem
the remainder would be 33.
7^99= (7^4)^24 * 7^3
now (7^4)^24 will give remainder 1 when divided by 2400 (7^4 = 2401)
so by remainder theorem remainder is same as 7^3/2400 which is 343 :)
you can break it easily...
let say 7*7*7*7=2401=>7^4=2401
IF WE DIVIDE 2401 BY 2400 IT GIVES THE REMAINDER 1
WHEN YOU DIVIDE 7^4 24 TIMES THEN 7^3 WILL ONLY LEFT
NOW THE ANSWER TIME===7^3*1*1*1*1*.......=343.
wow... mr. abhishek i am really weak in maths .... will you mind helping me out ... genuine request...
my email id- its.saumyah@gmail.com
1) 7
2) 2
n/a
n/a