116
. a) 200; b) 192; c) 432; d) 320; e) 400; f) 1152.
117
. a) 288; b) 24; c) 480; d) 388800.
118
. 88.
119
. Yechish.
Masala sharti bo’yicha,
a
= 3
α
5
β
7
ϒ
. Bu soning Eyler funksiyasi
ϕ
(
a
) = 3
α
-1
⋅
2
⋅
5
β
-1
4
⋅
7
ϒ
-1
⋅
6 = 2
4
3
α
-1
5
β
-1
7
γ
-1
. shart bo’yicha
ϕ
(
a
) = 3600 = =2
4
⋅
3
2
⋅
5
2
, demak 2
4
⋅
3
α
⋅
5
β
-1
⋅
7
γ
-1
= 2
4
⋅
3
2
⋅
5
2
, bundan
α
= 2,
β
= 3,
γ
=1 va
a
= 3
2
⋅
5
3
⋅
7 = 7875.
120
. Yechish.
Shart bo’yicha:
ϕ
(
a
) =
ϕ
(
pq
) =(
p
- 1) (
q
– 1) = 120 va
p
-
q
= 2. Natijada
(
)(
)
;
2
120
1
1
=
−
=
−
−
q
p
q
p
sistemani hosil qilamiz. Uning yechimi
p
= 13,
q
= 11. Demak,
a
=
p q
= 143.
121
. Yechish.
Shart bo’yicha
( )
(
)
(
) (
)
1
1
2
2
−
−
=
=
q
q
p
p
q
p
a
ϕ
ϕ
va
( )
.
17
7
3
2
11424
5
⋅
⋅
⋅
=
=
a
ϕ
Demak,
(
) (
)
,
17
7
3
2
1
1
5
⋅
⋅
⋅
=
−
−
q
q
p
p
yoki
(
) (
)
,
6
7
16
17
1
1
⋅
⋅
⋅
=
−
−
q
q
p
p
bundan
7
,
17
=
=
q
p
va
14161
7
17
2
2
=
⋅
=
a
.
122
.
Yechish.
Shart bo’yicha
( )
(
)
(
)
(
)
1
...
1
1
1
2
1
2
1
1
1
2
1
−
−
−
=
−
−
−
n
n
p
p
p
p
p
p
a
n
α
α
α
ϕ
va
( )
.
11
7
5
3
2
462000
3
4
⋅
⋅
⋅
⋅
=
=
a
ϕ
Demak,
(
)
(
)
(
)
11
7
5
3
2
1
...
1
1
3
4
1
2
1
2
1
1
1
2
1
⋅
⋅
⋅
⋅
=
−
−
−
−
−
−
n
n
p
p
p
p
p
p
n
α
α
α
.
O’ng tomondagi ko’paytuvchilarni chap tomondagi kabi ko’rinishda
ko’paytuvchilarni o’rnini almashtiramiz:
(
)
(
)
(
) (
)( )
(
)
,
4
5
6
7
10
11
1
...
1
1
2
1
2
1
2
2
1
1
1
1
⋅
⋅
⋅
=
−
−
−
−
−
−
n
n
n
p
p
p
p
p
p
α
α
α
bundan
11
1
=
p
va
7
;
2
2
1
=
=
p
α
va
5
;
2
3
2
=
=
p
α
va
;
3
3
=
α
.
741125
5
7
11
3
2
2
=
⋅
⋅
=
a
71
123
. Yechish.
(
a, m
)=1 shart bajarilganda (
a, m-a
) = 1 ni bajarilishini
ko’rsatamiz. Teskarisini faraz qilamiz, ya’ni (
a, m - a
) =
d
> 1, u holda
a
=
dk
,
m – a
=
dt
, bundan
m
=
d
(
t
+
k
) va (
a
,
m
) =
d
> 1, bu esa (
a, m
)= 1 shartga ziddir.
m
dan kichik va u bilan tub bo’lgan sonlarni tartib bilan yozamiz:
1,
a
1
,
a
2
, …,
m
–
a
2
,
m
–
a
1
,
m
– 1; bu qatorda
ϕ
(
m
) son bor. Har qanday
a
i
songa
m
– a
i
son mos keladi; ular yig’indisi
a
i
+ (
m
-
a
i
) =
m
, bu juftliklar soni
( )
m
ϕ
2
1
va demak
( )
.
2
1
m
m
S
ϕ
=
124
. a) 24; b) 54; c) 37500.
125
. a)
Yechish.
.
...
2
1
2
1
k
x
k
x
x
p
p
p
a
=
bo’lsin. U holda
( )
( )
.
1
1
...
1
1
1
1
1
1
...
1
1
1
1
1
2
1
1
2
1
a
a
p
p
p
a
a
p
p
p
a
a
k
k
ϕ
ϕ
α
α
α
α
−
−
=
=
−
−
−
=
−
−
−
=
126
. Birinchi hol (
a
, 2) = 1 bo’lganda o’rinli, ikkinchi hol esa (
a
, 2) = 2
bo’lganda o’rinli bo’ladi.
127
. a)
Yechish.
ϕ
(4
n
+ 2) =
ϕ
(2)
ϕ
(2
n
+ 1) =
ϕ
(2
n
+ 1);
b)
Yechish.
Agar (
n
, 2)=1 bo’lsa, u holda
ϕ
(4
n
) =
ϕ
(4)
ϕ
(
n
) = 2
ϕ
(
n
). Agar
n
= 2
α
⋅
k
bo’lib, (
k
, 2) = 1 bo’lsa, u holda
ϕ
(4
n
) =
ϕ
(2
α
+2
⋅
k
) = 2
α
+1
⋅ϕ
(
k
) = 2
⋅ϕ
(2
α
+1
⋅
k
) = 2
ϕ
(2
n
).
128
. a)
x
= 3; b)
x
= 3; c) tenglama
p
> 2 da yechimga ega emas.
p
= 2 da ixtiyoriy natural sonlar uchun o’rinli.
129
.
ϕ
(
b
).
130
. a) 4 kasr:
.
10
9
.
10
7
,
10
3
,
10
1
c) 12 kasr.
131
.
ϕ
(2) +
ϕ
(3) +…+
ϕ
(
n
).
132
. a) 9; b) 31; c) 71.
133
. Yechish.
Shart bo’yicha, (300,
x
) = 20 va barcha
x
lar 300 dan kichik, 20
ga qisqartirgandan so’ng (15,
y
) = 1, bu yerda barcha
y
Lar 15 dan kichik va 15 bilan
o’zaro tub; ular soni
ϕ
(15) = 8. Bu
y
= 1, 2, 4, 7, 8, 11, 13, 14 sonlar va bundan
x
=
20, 40, 80, 140, 160, 220, 260, 280.
134
.
ϕ
(45) = 24.
135
.
ϕ
(36) = 12.
136
. Ko’rsatma.
ϕ
(
a
) ning juftligi 123 masala yordamidan kelib chiqadi.
137
. Yechish.
Agar (
m
, 2) = 1 bo’lsa, u holda
ϕ
(
m
) =
ϕ
(2
m
).
138
. Yechish.
m
va
n
larning tub bo’luvchisi
p
uchun
ϕ
(
mn
) sonda
,
1
1
p
−
ko’paytuvchi bor, a
ϕ
(
m
)
ϕ
(
n
) – sonida esa
.
1
1
2
−
p
ko’paytuvchi bor.
,
1
1
1
<
−
p
bo’lganligi sababli
ϕ
(
m
)
ϕ
(
n
) <
ϕ
(
mn
). Xususiy holda
ϕ
2
(
m
)
≤
ϕ
(
m
2
), tenglik
m
= 1
bo’lganda bajariladi.
72
139
. Ko’rsatma.
q
1
,
q
2
,…,
q
t
–
m
ning kanonik yoyilmasidagi tub sonlar;
p
1
,
p
2
,…,
p
k
–
m
va
n
ning kanonik yoyilmasidagi tub sonlar va
−
S
l
l
l
,...,
,
2
1
faqat
n
ning kanonik yoyilmasidagi tub sonlar bo’lsin. U holda
( )
( ) ( ) ( )
.
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
d
d
n
m
p
d
d
p
n
p
q
m
p
q
mn
mn
k
i
i
k
i
S
i
i
i
k
i
i
t
i
i
k
i
S
i
i
i
t
i
i
ϕ
ϕ
ϕ
ϕ
=
∏
−
⋅
⋅
∏
∏
−
−
×
×
∏
−
∏
⋅
−
=
∏
∏
−
⋅
−
∏
⋅
−
=
=
=
=
=
=
=
=
=
l
l
1
-eslatma
. Shu usulda
m
va
n
sonlarning umumiy bo’luvchisi
A
uchun
( ) ( ) ( ) ( )
A
A
n
m
mn
ϕ
ϕ
ϕ
ϕ
=
munosabatni keltirib chiqarish mumkin.
2-
eslatma.
Chiqarilgan formula yordamida 138 masala yechimi juda osonlik
bilan topiladi:
( ) ( )
( )
,
mn
n
m
ϕ
ϕ
ϕ
≤
chunki
( )
1
≥
α
ϕ
d
. Tenglik o’rinli bo’lishi uchun
d
= 1 zarur va yetarlidir.
140
. Yechish.
( )
( )
( ) ( )
( )
.
µ
δϕ
γ
δ
µ
ϕ
δ
ϕ
δµ
ϕ
ϕ
τ
=
⋅
=
=
mn
141
.
p
α
.
142
.
m.
144
. a)
Yechish
. Gaussa formulasidan
x
= 2
y
3
z
5
u
(
y
≥
0,
z
= 0; 1 va
U
= 0; 1) kelib chiqadi.
x
= 2
y
; 2
y
⋅
3; 2
y
⋅
5; 2
y
⋅
15; 3; 5; 15 imkoniyatlarni tek-
shirishi
x
= 2
α+1
; 2
α
⋅
3; 2
α
-1
⋅
5; 2
α
-2
⋅
15 (
α
≥
2); 15 (
α
= 3)larni beradi;
b)
p
≠
3 da yechim yo’q.
r
= 3 da tenglamalarni ixtiyoriy butun
x
≥
2 qiymat-
lar qanoatlantiradi.
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