Question

how can i find x without a the use of a calculator

Answer 1

\[\large 17^{133}\equiv x\mod 2257\]

Answer 2

interestig, binary exponentiation requires calculator too as 2257 and 133 are not small numbers to work in head

Answer 3

@ikram002p @mukushla may be having some neat methods

Answer 4

\[\phi(2257)=2160\] is given here,also
\[2257=37\times 61\]
so,by Euler
\[17^{2160}\equiv 1 \mod 2257\] is known also,hope it helps

Answer 5

mmm is 2257 prime ??

Answer 6

2257 = 37x61

Answer 7

then we can use the chinese remander

Answer 8

gcd(37,61)=1

Answer 9

17^133=a1mod(37)
17^133=a2 mod(61)

Answer 10

ok i got an idea to solve it but i dint got the same answer :/
2257 = 37x61
gcd(37,61)=1
17^133=a1mod(37) .......1
17^133=a2 mod(61) ......2
nw for equation 1
using fermat thm
if p prime , p do not devide a
a^(p-1)=1 mod p

17^36=1mod 37
17^108=1 mod 37#

17^2=30 mod 37
17^4=12 mod 37
17^8=33 mod 37
17^24=6 mod 37
17^25=4 mod 37 ##
from # & ##
17^133=(4*1) mod 37
***i know there is an errer in calculating but the method is correct
nw for equation 2
17^133=a2 mod(61)
using fermat thm

17^60=1 mod 61
17^120=1 mod 61#

17^4=22 mod 61
17^8=57 mod 61
17^12=20 mod 61
17^13=29 mod 61##

for #& ##
17^133=29 mod 1
*********
u just need to check my calculating error cuz the remainder in the two cases must be the same :)