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Lecture 25

Math1xx3 Lecture 25: Lecture 25


Department
Mathematics
Course Code
MATH 1XX3
Professor
Adam Van Tuyl
Lecture
25

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Moreonfnierge
Example :Find the remainder when 2""
is divided by 7
210--1024=103 )Baise everything
"
Ma to "Imuttiplyby 24=16
2'"= 1049
Recall :a=b( modm)Trick :2'=2
723=1
(mod -7 )
a'±b' (modm)22=4/123
"taYCm°d7 )
a±a'±b±b' (modm
)[23--8] 2300=1 (mod 7)
aatbbfnodm )24=16 2300 .l=7q
antb "(modm)25=32 2300=79+1
For our Example :
154=3.5 ,+,
-
hexofodnoentestreagrhenthe
that we have 23 ?
2¥/(mod7)| raisetothepowerofsl
(
235¥15 'lmod
z'53I|( mod 7)
2=21 money }H¥g¥Ihkr :z'"=l .2(mod7)
2154 =2(mod7 )youneedto
find the right
iothenumeratorof 2'"divided by 7,2/7 power"
Test Question :4537 "=?( mod 5)
,
045 '±4lmod5)0
4÷4(
mod 5)
@y3±g(mds)¥o'
settle
't 42=-1 (mod Dlraiseto 's
H2p8÷P8(mod5 )
457.73=4 .2(mod5)456=1 (mod 5)
457.73=8 (mod 5) 457=4 (mod 5)
b
8=31mod 5) 2
7=7
b72=49 -=t(mod5)
49.7 "=3( mots)74=2401 -=l(mod5)|raiseto5
7
'2±1( mod 's)
7E2(mod5 )
7'3÷2( mod 5)
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