math 785 hw review 1
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Homework Review
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Notes 1, Problem 1
Let a and b be positive integers, and writea
b= m.d1d2 . . . dkdk+1dk+2 . . . dk+r
where m is a positive integer, the dj are digits, and r ischosen as small as possible. Prove that r divides (b)where is Eulers function.
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m = pe11 p
e22 perr
(m) =
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m = pe11 p
e22 perr
(m) = pe111 per1r
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m = pe11 p
e22 perr
(m) = pe111 per1r (p1 1) (pr 1)
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m = pe11 p
e22 perr
(m) = pe111 per1r (p1 1) (pr 1)
Remark: If b|b, then (b)|(b).
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ab= m.d1d2 . . . dkdk+1dk+2 . . . dk+r
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ab= m.d1d2 . . . dkdk+1dk+2 . . . dk+r
WLOG, k = 0. Why?
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ab= m.d1d2 . . . dkdk+1dk+2 . . . dk+r
WLOG, k = 0. Why?
Otherwise consider10ka
b.
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ab= m.d1d2 . . . dr
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ab= m.d1d2 . . . dr
(10r 1) a
b= md1d2 . . . dr m
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ab= m.d1d2 . . . dr
(10r 1) a
b= md1d2 . . . dr m
(10r 1 ) a
b Z
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(10r 1 ) a
b Z
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(10r 1 ) a
b Z
Supposea
b=a
bwhere gcd(a, b) = 1.
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(10r 1 ) a
b Z
Supposea
b=a
bwhere gcd(a, b) = 1.
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(10r 1 ) a
b Z
Supposea
b=a
bwhere gcd(a, b) = 1. Then
b|
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(10r 1 ) a
b Z
Supposea
b=a
bwhere gcd(a, b) = 1. Then
b|( 10r 1 ).
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(10r 1 ) a
b Z
Supposea
b=a
bwhere gcd(a, b) = 1. Then
b|( 10r 1 ).IDEA: r is the order of 10 modulo b
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(10r 1 ) a
b Z
Supposea
b=a
bwhere gcd(a, b) = 1. Then
b|( 10r 1 ).IDEA: r is the order of 10 modulo b
= r|(b)
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(10r 1 ) a
b Z
Supposea
b=a
bwhere gcd(a, b) = 1. Then
b|( 10r 1 ).IDEA: r is the order of 10 modulo b
= r|(b)= r|(b)
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(10r 1 ) a
b Z
Supposea
b=a
bwhere gcd(a, b) = 1. Then
b|( 10r 1 ).IDEA: r is the order of 10 modulo b
= r|(b)= r|(b)
Whats wrong with this IDEA?
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Finish the proof by showing that if t is such thatb|( 10t 1 ),
thena
b= m.d1d2 . . . dt.
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Finish the proof by showing that if t is such thatb|( 10t 1 ),
thena
b= m.d1d2 . . . dt.
So t r.
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Notes 1, Problem 2a
b= m.d1d2 . . . dkdk+1dk+2 . . . dk+r
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Notes 1, Problem 2a
b= m.d1d2 . . . dkdk+1dk+2 . . . dk+r
r = b 1 =
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Notes 1, Problem 2a
b= m.d1d2 . . . dkdk+1dk+2 . . . dk+r
r = b 1 = (b 1)|(b) =
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Notes 1, Problem 2a
b= m.d1d2 . . . dkdk+1dk+2 . . . dk+r
r = b 1 = (b 1)|(b) = b is prime
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Notes 1, Problem 2a
b= m.d1d2 . . . dkdk+1dk+2 . . . dk+r
r = b 1 = (b 1)|(b) = b is prime
b 6= 2
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Notes 1, Problem 2a
b= m.d1d2 . . . dkdk+1dk+2 . . . dk+r
r = b 1 = (b 1)|(b) = b is prime
b 6= 2 and b 6= 5
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Notes 1, Problem 2a
b= m.d1d2 . . . dkdk+1dk+2 . . . dk+r
r = b 1 = (b 1)|(b) = b is prime
b 6= 2 and b 6= 5
gcd(a, b) = 1
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Notes 1, Problem 2a
b= m.d1d2 . . . dkdk+1dk+2 . . . dk+r
r = b 1 = (b 1)|(b) = b is prime(10b1 1) a
b Z
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Notes 1, Problem 2a
b= m.d1d2 . . . dkdk+1dk+2 . . . dk+r
r = b 1 = (b 1)|(b) = b is prime(10b1 1) a
b Z
= k = 0
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Notes 1, Problem 2
Examples:
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Notes 1, Problem 2
Examples:
1
11= 0.090909 . . .
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Notes 1, Problem 2
Examples:
1
11= 0.090909 . . .
1
61= 0.01639344262295081967213114754098
3606557377049180327868852459 . . .
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Notes 1, Problem 2
Examples:
1
11= 0.090909 . . .
1
61= 0.01639344262295081967213114754098
3606557377049180327868852459 . . .
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Notes 1, Problem 2
Examples:
1
11= 0.090909 . . .
1
61= 0.01639344262295081967213114754098
3606557377049180327868852459 . . .
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Notes 1, Problem 2
Examples:
1
11= 0.090909 . . .
1
61= 0.01639344262295081967213114754098
3606557377049180327868852459 . . .
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Notes 1, Problem 2
Examples:
1
11= 0.090909 . . .
1
61= 0.01639344262295081967213114754098
3606557377049180327868852459 . . .
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Notes 1, Problem 2
Examples:
1
11= 0.090909 . . .
1
61= 0.01639344262295081967213114754098
3606557377049180327868852459 . . .
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Notes 1, Problem 2
Examples:
1
11= 0.090909 . . .
1
61= 0.01639344262295081967213114754098
3606557377049180327868852459 . . .
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Notes 1, Problem 2
Examples:
1
11= 0.090909 . . .
1
61= 0.01639344262295081967213114754098
3606557377049180327868852459 . . .
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Notes 1, Problem 2
Examples:
1
11= 0.090909 . . .
1
61= 0.01639344262295081967213114754098
3606557377049180327868852459 . . .
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Notes 1, Problem 2
Examples:
1
11= 0.090909 . . .
1
61= 0.01639344262295081967213114754098
3606557377049180327868852459 . . .
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Notes 1, Problem 2
Examples:
1
11= 0.090909 . . .
1
61= 0.01639344262295081967213114754098
3606557377049180327868852459 . . .
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Notes 1, Problem 2
Suppose r = b 1.(i) Prove that each of the digits 0, 1, . . . , 9 occurs among
the digits d1, d2, . . . , dr either[(b 1)/10] or [(b 1)/10] + 1
times.(ii) Prove that 0 occurs [(b1)/10] times among the digits
d1, d2, . . . , dr.
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b = p, a prime
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b = p, a prime
Note: r = p1 = the order of 10 modulo p is p 1
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b = p, a prime
Note: r = p1 = the order of 10 modulo p is p 1The values of 10ja modulo p are distinct for 0 j p 2
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b = p, a prime
Note: r = p1 = the order of 10 modulo p is p 1The values of 10ja modulo p are distinct for 0 j p2 and are congruent to 1, 2, . . . , p1 in some order.
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b = p, a prime
Note: r = p1 = the order of 10 modulo p is p 1The values of 10ja modulo p are distinct for 0 j p2 and are congruent to 1, 2, . . . , p1 in some order.
10ja = pqj + rj
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b = p, a prime
Note: r = p1 = the order of 10 modulo p is p 1The values of 10ja modulo p are distinct for 0 j p2 and are congruent to 1, 2, . . . , p1 in some order.
10ja = pqj + rj =10ja
p= qj +
rj
p
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b = p, a prime
Note: r = p1 = the order of 10 modulo p is p 1The values of 10ja modulo p are distinct for 0 j p2 and are congruent to 1, 2, . . . , p1 in some order.
10ja = pqj + rj =10ja
p= qj +
rj
p
10ja
p= d1d2 . . . dj .dj+1 . . . drd1d2 . . .
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b = p, a prime
Note: r = p1 = the order of 10 modulo p is p 1The values of 10ja modulo p are distinct for 0 j p2 and are congruent to 1, 2, . . . , p1 in some order.
10ja = pqj + rj =10ja
p= qj +
rj
p
10ja
p= d1d2 . . . dj
qj
.dj+1 . . . drd1d2 . . .
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b = p, a prime
Note: r = p1 = the order of 10 modulo p is p 1The values of 10ja modulo p are distinct for 0 j p2 and are congruent to 1, 2, . . . , p1 in some order.
10ja = pqj + rj =10ja
p= qj +
rj
p
10ja
p= d1d2 . . . dj
qj
.dj+1 . . . drd1d2 . . . rj/p
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10ja
p= d1d2 . . . dj
qj
.dj+1 . . . drd1d2 . . . rj/p
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10ja
p= d1d2 . . . dj
qj
.dj+1 . . . drd1d2 . . . rj/p
rj varies from 1 to p 1
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10ja
p= d1d2 . . . dj
qj
.dj+1 . . . drd1d2 . . . rj/p
0 rjp
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