56
JAVOBLAR va KO’RSATMALAR I-BOB BUTUN SONLAR XALQASIDA BO’LINISH NAZARIYASI 1-§ 1. a )
q = 7; 8 va
r = 2; 6;
b )
q = 8; 9 va
r = 2; 6.
2. a )
Yechish: (2
n +1)
2
= 4
n (
n +1) + 1, bu yerda
n (
n + 1) 2 ga bo’linadi;
b )
Yechish: n 2
+ (
n + 1)
2
= 2
n (
n + 1) + 1, bu yerda
n (
n + 1) 2 ga bo’linadi.
3. Ko’rsatma. 15 = 7
⋅
2 + 1. Agar 15
n =7
q +1, u holda 15
n +1
= 5
n ⋅
15 = 7
Q + 1.
4. Yechish. Masala sharti bo’yicha,
−
=
−
+
t p m pq mn butun son.
(
) (
)
n q p m p m n p m q p m pq mn p m np mq t p m np mq −
=
−
−
−
−
=
−
+
−
−
+
=
−
−
+
.
Bundan
−
+
−
=
−
+
t n q p m np mq butun son. Demak,
mq +
np m –
p ga bo’linadi.
5. Yechish. Masala sharti bo’yicha,
ad – bc = nt va a – b = nt 1
. Ikkinchi teng-
likni
d ga ko’paytirib, birinchisidan ayiramiz:
b (
c-d ) =
n (
dt 1
-
t ). Bundan
b va
n ga qo’yilgan shartlarlarga asosan
s –
d ni
n ga
bo’linishi kelib chiqadi.
6. c) Yechish. m 5
–
m = (
m - 1)
m (
m + 1) (
m 2
+ 1) = (
m - 1)
m (
m + 1). [(
m 2
- 4)
+ 5] = (
m - 2) (
m - 1)
m (
m + 1) (
m + 2) + 5 (
m - 1) (
m + 1). Qo’shiluvchilarning har
biri 30 ga bo’linadi, chunki
k ta ketma-ket sonlar ko’paytmasi
k ! ga bo’linadi (bu
(
)(
) (
)
k k n n n n C k n ....
3
.
2
.
1
1
...
2
1
+
−
−
−
=
- butun son bo’lishidan kelib chiqadi). Bundan
yig’indi ham 30 ga bo’linadi, demak
m 5 – m 30 ga bo’linadi.
7. Yechish. 10
x + 5 – izlanayotgan sonl bo’lsin. 5 raqamni chap tomondan bi-
rinchi o’ringa qo’yib 5
⋅
10
5
+
x hosil qilamiz
. Berilgan shartlarga ko’ra 5
⋅
10
5
+
x = 4 (10
x + 5) tenglamaga kelamiz. Bundan
x = 112820 kelib chiqadi.
8. Yechish. Masala shartini quyidagicha yozib olamiz:
n (
n + 1) (2
n + 1) =
n (
n + 1) [(
n - 1) + (
n + 2)] = (
n -1)
n (
n + 1) +
n (
n + 1)(
n + 2). Har bir qo’shiluvchi 6 ga
bo’linishidan (6 masala yechimidan) yig’indini 6 ga bo’linishi kelib chiqadi.
9. Yechish. (
) (
)
(
) (
)
(
)(
)
(
)
[
]
n m n m n m n m n m n m +
+
+
−
+
+
=
+
+
+
+
−
+
2
2
2
2
2
2
2
2
1
4
1
2
1
2
1
2
1
2
hosil bo’lgan kasrni
faqat 2 ga qisqartirish mumkin.
10. Yechish. (
)
(
)
100
10
101
10
1
100
1000
2
+
+
=
+
+
+
+
=
y x y x y x N .
Bundan
(
)(
)
.
8181
,
91
,
101
10
10
10
2
=
=
−
+
=
+
N N N N y x
57
11. Yechish. (
n - 2)
2
+ (
n - 1)
2
+
n 2
+ (
n + 1)
2
+ (
n + 2)
2
= 5 (
n 2
+ 2) to’la kvad-
rat bo’lishi uchun
n 2
+ 25 ga karrali bo’lishi kerak yoki
n 2
ning oxirgi raqami 8
yoki 3 bo’lishi kerak, bu mumkin emas.
12. Yechish. Har qanday butun sonni quyidagilardan birortasi shaklida yozish
mumkin: 9
k , 9
k ±
1, 9
k ±
2, 9
k ±
3, 9
k ±
4. Bu sonlar kvadratlari:
(9
k )
2
= 9(9
k 2
); (9
k ±
1)
2
= 9 (9
k 2
±
2
k ) + 1; (9
k ±
2)
2
= 9(9
k 2
±
4
k ) + 4;
(9
k ±
3)
2
= 9 (9
k 2
±
6
k + 1); (9
k ±
4)
2
= 9 (9
k 2
±
8
k + 1) + 7. Natijada butun son kvad-
rati 9 ga bo’lganda qoldiq faqat 0, 1, 4, 7 bo’lishi mumkinligi kelib chiqadi.
13. Yechish. 9
1
10
...
9
1
10
9
1
10
9
1
10
(
7
)
1
...
111
...
111
11
1
(
7
3
2
−
+
+
−
+
−
+
−
=
+
+
+
+
=
n raqam ta n n S 3
2
1
=
).
10
9
10
(
81
7
1
−
−
=
+
n n 14. Yechish. .
1
3
...
333
1
3
1
10
3
2
10
6
9
1
10
10
5
10
9
1
10
56
...
555
1
...
111
2
2
1
2
1
1
1
+
=
+
−
=
+
=
+
−
⋅
⋅
+
⋅
−
=
+
+
+
+
3
2
1
4
3
42
1
3
2
1
raqam ta n n n n n n raqam ta n raqam ta n 15 . Yechish. m n (m 4 – n 4 ) = n (m 5 - m) – m (n 5 - n) 30 ga karrali
(1 misolga ko’ra).
16 . Yechish. y 2
= 3
x 2
+ 2 tenglama butun sonlarda yechimga ega emas. Haqi-
qatdan ham,
u ni
y =3
n yoki
y = 3
n ±
1 shakllardan birortasi ko’rinishida ifodalash
mumkin va bundan
y 2
ni 3 ga bo’lganda qoldiq faqat 0 yoki 1 bo’ladi masala shar-
tiga ko’ra qoldiq 2 bo’lishi kerak.
17 . Yechish. Matematik induksiya usulini qo’llaymiz:
ааа
son 3 ga bo’linadi,
chunki
a + a + a = 3
a; Agar
3
2
1
т
а
аа
3
... son 3
n ga bo’linsa, u holda
( )
01
...
0100
...
100
...
...
10
...
10
...
...
...
...
...
3
3
3
2
3
3
3
3
3
1
3
⋅
=
=
+
⋅
+
⋅
=
=
+
a aa a aa a aa a aa a aa a aa a aa а
аа
n n n n n n n n n 3
2
1
3
2
1
3
2
1
3
2
1
3
2
1
3
2
1
3
2
1
3
n +1
ga bo’linadi.
2-§ 18 . a ) 21,
b ) 13;
c ) 119;
d ) 3;
e ) 23.
19 .
a ) 2520;
b ) 138600;
c ) 99671;
d ) 881200.
20 . Yechish. (
a, b, c ) =
d bo’lsin, u holda
a =
cq +
r ,
b =
cq 1
+
r 1
dan
d|r va
d|r 1
kelib chiqadi.
d = (
c, r, r 1
) ni isbotlaymiz. (
c, r, r 1
) =
D bo’lsin.
a = cq + r va
b = cq 1
+
r 1
tengsizliklardan
D
a, D
b va shart bo’yicha
D|c . Bundan
D = (
a, b, c )
58
va demak,
D =
d .
n ta son uchun (
a 1
,
a 2
,…,
a n ) = (
a n ,
r 1
,
r 2
,.,
r n -1
) ni olamiz, bu
yerda
r 1
,
r 2
,…,
r n -1
–
a 1
,
a 2
,…,
a n -1
sonlarni
a n ga bo’lgandagi qoldiq.
21 .
a ) 23;
b ) 7;
c ) 21.
22 .
a ) 3776;
b ) 1116;
c ) 67818;
d ) 5382;
e ) 6409.
23 .
a )
Yechish. (
d,m ) = (
d , [
dx ,
dy ]) =
d (1, [
x ,
y ]) =
d . Agar
d = (
a 1
,
a 2
,…,
a n )
va
m = [
a 1
,
a 2
,…,
a n ] deb olsak, natija o’zgarmaydi;
b )
Yechish. Agar
p –
a + b va
a ⋅
b sonlarning umumiy bo’luvchichi bo’lsa,
u holda
a yoki
b sonlardan birortasi
p ga bo’linishi kerak.
a +
b ni
p ga
bo’linishidan
p son
a va
b larni umumiy bo’luvchisi ekanligi kelib chiqadi. Bu
masala shartiga zid, chunki (
a ,
b ) =1;
c )
Yechish . (
a, b ) =
d va
a =
dx ,
b =
dy bo’lsin, bunda (
x, y ) = 1. Bu holda
(
a +
b ,
m ) = (
d (
x +
y ),
dxy ) =
d (
x +
y ,
xy ) =
d . Demak,
(
a +
b , [
a,b ]) = (
a ,
b ).
24 .
Yechish .
x va
y – izlanayotgan sonlar bo’lsin va (
x, y ) =
d , bundan
x =
dm va
y =
dn va (
m, n ) = 1. Shartga ko’ra,
x +
y =
d (
m +
n ) =
= 667 = 23
⋅
29. Shart bo’yicha
,
120
)
,
(
]
,
[
=
y x y x bundan [
x,y ]=120
⋅
(x, y)=120
d, boshqa
tomondan
[ ]
.
120
,
120
,
2
d xy yoki d d xy d xy y x =
=
⇒
=
Bulardan
=
⋅
=
+
2
120
29
23
d xy y x sistemani hosil qilamiz.
d (
m +
n ) = 23
⋅
29 dan
d = 23 va
d = 29
(
d = 1 yoki
d = 23
⋅
29 - o’rinli bo’lmaydi) bo’lishi mumkin.
d = 23 bo’lganda,
x =
552,
y = 115.
d = 29 da
x = 435,
y = 232.
25 .
Yechish. x va
y – noma’lum sonlar va (
x ,
y ) =
d bo’lsin. U holda
n d y m d x =
=
,
, bunda
.
1
)
,
(
=
n m Shart bo’yicha
m + n = 18,
[ ]
.
13
5
3
975
,
2
⋅
⋅
=
=
=
⋅
=
=
mnd d dn dm d xy y x Bundan
⋅
⋅
=
=
+
13
5
3
18
2
mnd n m ni
hosil
qilamiz
va
uning
yechimi
15
,
13
,
5
=
=
=
d n m bo’ladi. Demak,
x = 75,
y = 195.
26. Yechish . Shart bo’yicha,
a = 899,
b = 493. Yevklid algoritmiga ko’ra:
a =
b ⋅
1 + 406,
b = 406
⋅
1 + 87, 406 = 87
⋅
4 + 58, 87 = 58
⋅
1 + 29, 58 = 29
⋅
2 bo’ladi. Oxiri-
dan ikkinchi tenglikdan boshlab: 29 = 87– 58 =87 – (406–87
⋅
4) = 87
⋅
5–406 = (
b -
406)
⋅
5–406 = 5b - 406
⋅
6 = 5b - (
a - b )6 =
a(-b) +
b ⋅
11 ni olamiz. 29 = 899
x + 493
y bilan solishtirsak,
x = - 6,
y = 11 kelib chiqadi.
27 .
a ) 17 =
a (-10) +
b ⋅
23 =
ax +
by ;
b ) 43 =
a ⋅
(-4) +
b ⋅
5 =
ax +
by ;
c ) 47 =
a ⋅
2 +
b (-5) =
ax + by .
59
28 .
a )
Yechish. x = 45
u va
y = 45
v , bu yerda (
u, v ) = 1,
7
11
=
v u dan
u = 11
va
v = 7, demak
x = 495 va
y = 315;
b )
Yechish .
x = 20
u va
y = 20
v , bu
yerda (
u,v ) = 1,
uv = 21 dan
u = 1; 3; 7; 21 va
x = 20; 60; 140; 420.
x y 8400
=
bo’lganligi sababli
y = 420, 140, 60, 20.
d )
x = 140,
y = 252.
29 .
Yechish. (
a, b, c ) =
d bo’lsin, u holda
a =
md ,
b =
nd ,
c =
kd .
2
2
;
2
2
;
2
2
d k n c b d k m c a d n m b a +
=
+
+
=
+
+
=
+
Bundan
d son
2
,
2
,
2
c b c a b a +
+
+
sonlarning umumiy bo’luvchisi bo’lishini ko’rsatadi.
Faraz qilamiz,
D c b c a b a =
+
+
+
2
,
2
,
2
bundan
d|D ,
.
2
;
2
,
2
1
1
1
D k c b D n c a D m b a =
+
=
+
=
+
Birinchit va ikkinchi tengliklar yig’indisidan
uchinchi tenglikni ayirib,
a = (
m 1
+
n 1
–
k 1
)
D ni hosil qilamiz. Shu usulda
b = (
m 1
–
n 1
+
k 1
)
D ,
c = (-
m 1
+
n 1
+
k 1
)
D larni hosil qilamiz. Bu tengliklardan
a ,
b ,
c larni
D ga bo’linishi kelib chiqadi va demak,
D|d . Natijada
(
)
.
2
,
2
,
2
,
,
,
+
+
+
=
=
c b c a b a c b a яъни
d D 30 .
a )
Yechish. (
a, b, c ) =
d bo’lsin, u holda (
a, b ) =
md , (
a, c ) =
nd , (
b,c ) =
kd , bu yerda (
m, n, k ) = 1. Bu tenglikdan
adm va
dn ga bo’linishi kelib chiqadi, de-
mak,
a =
dmn α
. Xuddi shunday:
b =
dmk β
;
c =
dnk γ
. Bu yerda (
α
,
β
,
γ
) = 1.
[
]
(
)(
)(
)
(
)
( )( )( )
;
,
,
,
,
,
,
,
2
2
2
2
4
c b c a b a c b a abc dk dn dm d dnk dmk bmn mnk d k n m dmnk c b a =
⋅
⋅
=
=
=
γ
β
α
αβγ
α
αβγ
b )
Ko’rsatma :
[ ]
( )
b a ab b a ,
,
=
dan foydalaning.
31 .
Yechish. qN = 100
q +
bq = 100
q +
a –
a +
bq =
am – (
a -
bq ), bundan tas-
diq to’g’riligi kelib chiqadi, chunki (
q, m ) = 1.