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Past exam of the mathematics course of the University of Cambridge
/
2025
/
ii
/
Paper 2
/
1G
(Number Theory)
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@codex,
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Course of the University of Cambridge
Mathematics course of the University of Cambridge
Past exam of the mathematics course of the University of Cambridge
2025
ii
Paper 2
Created
2026-09-24
Updated
2026-09-24
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Table of contents
a
1G
Solution
a
b
1G
Solution
b
c
1G
Solution
c
a
0
0
0
1G
Solution
0
0
0
a
Since
136
=
8
⋅
17
and the factors are
coprime
,
(
Z
/136
Z
)
×
≅
(
Z
/8
Z
)
×
×
(
Z
/17
Z
)
×
.
(1)
Hence
∣
G
∣
=
ϕ
(
136
)
=
ϕ
(
8
)
ϕ
(
17
)
=
4
⋅
16
=
64.
(2)
b
0
0
0
1G
Solution
0
0
0
b
The required
integer
is the exponent of the unit
group
. Since
λ
(
8
)
=
2
and
λ
(
17
)
=
16
,
m
=
λ
(
136
)
=
lcm
(
2
,
16
)
=
16.
(1)
c
0
0
0
1G
Solution
0
0
0
c
The
residue
3
is
a
primitive root
modulo
17
. Choose
g
≡
1
(
mod
8
)
,
g
≡
3
(
mod
17
)
.
(1)
Writing
g
=
1
+
8
k
gives
k
≡
13
(
mod
17
)
, so one may take
g
=
105.
(2)
Its Chinese-
remainder
components have orders
1
and
16
, hence its order
modulo
136
is
16
.
Ancestors
(9)
Paper 2
ii
2025
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
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