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number theory

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(1)
Consider
200
We have
12
1 1 1
1 1 ... 1
t
nn
p p p



where
12
, ,...
t
p p p
are positive integers.
The prime factors of
32
200 2 5
So
32
200 2 5n
32
200 2 5
11
200 1 1
25
80


Consider
210
The prime factors of
210 2 3 5 7  
So
210 2 3 5 7
1 1 1 1
210 1 1 1 1
2 3 5 7
48

(2)
Consider
12 28 mod20x
The congruence is
12 28 mod20x
Comparing with
ax b Mod m
, we have
12, 28, 20a b m
, 12,20 4d a m
which is not divides
28b
Therefore, the congruence
12 28 mod20x
has no solution.
(3)
Consider
3 mod9x
The congruence is
3 mod9x
Comparing with
ax b Mod m
, we have
1, 3, 9a b m
, 1,9 1d a m
which is divides
3b
And hence the solution of the given congruence exists and is unique.
Now
3 mod9x
…… (1)
Also
0 27 Mod 9
…… (2)
Adding congruence (1) and (2),
30 Mod 9x
Therefore,
30 Mod 9x
is a solution of the congruence
3 mod9x
Since
,1am
, the congruence
Mod max b
has a unique solution.
(4)
Consider
5 mod7x
The congruence is
5 mod7x
Comparing with
ax b Mod m
, we have
1, 5, 7a b m
, 1,7 1d a m
which is divides
5b
And hence the solution of the given congruence exists and is unique.
Now
5 mod7x
…… (1)
Also
0 35 Mod 7
…… (2)
Adding congruence (1) and (2),
40 Mod 7x
Therefore,
40 Mod 7x
is a solution of the congruence
5 mod7x
Since
,1am
, the congruence
Mod max b
has a unique solution.
(5)
Consider
2 mod5x
The congruence is
2 mod5x
Comparing with
ax b Mod m
, we have
1, 2, 5a b m
, 1,5 1d a m
which is divides
2b
And hence the solution of the given congruence exists and is unique.
Now
2 mod5x
…… (1)
Also
0 10 mod5
…… (2)
Adding congruence (1) and (2),
12 mod5x
Therefore,
12 mod5x
is a solution of the congruence
2 mod5x
Since
,1am
, the congruence
Mod max b
has a unique solution.
(6)
Compute the order of 5 modulo 36
That is
5 0 mod 36x
That implies
0 mod 36x
That implies
36x

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Anonymous
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