1. 3^302 mod 5
Solution:
Fast Modular Exponentiation
`3^302" mod "5`
Comparing with `A^B" mod "C`
We get `A=3,B=302,C=5`
Step 1: Divide B into powers of 2 by writing it in binary
`302=100101110` in binary
`302=2^1+2^2+2^3+2^5+2^8`
`302=2+4+8+32+256`
`3^302" mod "5=3^((2+4+8+32+256))" mod "5`
`3^302" mod "5=(3^2*3^4*3^8*3^32*3^256)" mod "5`
Step 2: Calculate mod C of the powers of two <= B
`3^1" mod "5=3" mod "5=3`
`3^2" mod "5=(3^1*3^1)" mod "5=(3^1" mod "5*3^1" mod "5)" mod "5=(3*3)" mod "5=9" mod "5=4`
`3^4" mod "5=(3^2*3^2)" mod "5=(3^2" mod "5*3^2" mod "5)" mod "5=(4*4)" mod "5=16" mod "5=1`
`3^8" mod "5=(3^4*3^4)" mod "5=(3^4" mod "5*3^4" mod "5)" mod "5=(1*1)" mod "5=1" mod "5=1`
`3^16" mod "5=(3^8*3^8)" mod "5=(3^8" mod "5*3^8" mod "5)" mod "5=(1*1)" mod "5=1" mod "5=1`
`3^32" mod "5=(3^16*3^16)" mod "5=(3^16" mod "5*3^16" mod "5)" mod "5=(1*1)" mod "5=1" mod "5=1`
`3^64" mod "5=(3^32*3^32)" mod "5=(3^32" mod "5*3^32" mod "5)" mod "5=(1*1)" mod "5=1" mod "5=1`
`3^128" mod "5=(3^64*3^64)" mod "5=(3^64" mod "5*3^64" mod "5)" mod "5=(1*1)" mod "5=1" mod "5=1`
`3^256" mod "5=(3^128*3^128)" mod "5=(3^128" mod "5*3^128" mod "5)" mod "5=(1*1)" mod "5=1" mod "5=1`
Step 3: Use modular multiplication properties to combine the calculated mod C values
`3^302" mod "5`
`=(3^2*3^4*3^8*3^32*3^256)" mod "5`
`=(3^2" mod "5*3^4" mod "5*3^8" mod "5*3^32" mod "5*3^256" mod "5)" mod "5`
`=(4*1*1*1*1)" mod "5`
`=4" mod "5`
`=4`
`:.3^302" mod "5=4`
This material is intended as a summary. Use your textbook for detail explanation.
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