最大公因数与最小公倍数基本概念
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choose
the
corre
ct meani
ng;
(4) to correct the
typos;
(5) so the child wr
ite wor
ds (ABA
B, a
nd A
ABB);
(6) in accorda
nce
with
written w
ords;
(7
) the compl
ete wor
d, a
nd ex
plain t
he mea
ning
of
the w
ord;
(8) collocati
on;
(9) make sentence
s wit
h the
word;
(10) the written lang
uage as r
equired. (C) t
he main
sente
nce types (1) compl
ete se
ntences; (2) write down t
he meaning
of a se
ntence or ex
pression of thoug
hts a
nd feeli
ngs; (3) write
sente
nce
s as re
quire
d; (4) finish malalig
nment
of the sentence; (5) modified
sente
nce
s. 2, knowle
dge cla
ssificati
on (1) the
common conjunctions
coordi
nate: ... ...
一
面
……
... 1, to examine the t
opi
c, i
dentify pr
oblem
s associate
d wit
h two 2, a
nalysis, alternative
question two is i
n dire
ct p
oportion to the amount of t
he associated re
lationshi
p is i
nversely
proporti
onal relati
onshi
p. 3, and set
unk
now
n,
colum
n pr
oportion type 4, and soluti
ons pr
oporti
on ty
pe 5, and test, wr
ote a
nswer la
nguage
ple
nary, and subje
ct:
appli
cation problem (1)--simple a
ppli
cation problem and composite applicati
on
probl
em review content sim
ple application
problem composite applicati
on pr
obl
em answer
s applicati
on pr
obl
em of general ste
ps 1, a
nd figure
out meaning--thr
oug
h examines t
he, find k
nown conditions a
nd by
seeki
ng problem 2, and analysis
number relationshi
p--analysis k
now
n conditions Zhiji
an, a
nd
conditions a
nd
probl
em Zhijia
n of relationshi
p,
determine pr
obl
em- solving method a
nd
problem-solving
steps.
3, and column ty
pe cal
culati
on--li
sts formula, i
s out subdivisions 4, a
nd test, a
nd wr
ote answer--che
ck, and checki
ng, a
nd wr
ote answ
ers typi
cal applicati
on
probl
em 13, and subje
ct: a
ppli
cation problem (3)--col
umn e
quati
on soluti
ons
appl
ication pr
oblem review
content overvi
ew pr
oblem
-solvi
ng steps 1, a
nd figure
out meani
ng, find by se
eking
of unk
now
n and x said 2, and accordi
ng to mea
ning find equivalent relati
onship, lists Equation 3, a
nd soluti
ons
equation
4, and test, and wrote a
nsw
ers accor
ding
to meaning find equivalent relati
onship
of common method 1 , And accor
ding to
common of num
ber relationship type, establi
she
d equivale
nt relationshi
p 2, and accor
ding to
has lear
n ha
d of cal
culati
on formula
, 3, and according to pr
obl
em in the
of focus
describe
d se
ntence from overall Sha
ng determine basi
c of equivale
nt relationshi
p 4, and using segme
nt figure, and list method, method a
nalysis
num
ber
基本概念：
1
、公约数和最大公约数
几个数公有的约数
，叫做 这几个数的公约数
；其中最大的一个
，叫做这几个数的最大公
．．．．．．．．
．．．．．．．．．．
．．．．．．．
．．．．．．．．．．
约数
。
．．
例如：
12
的约数有
1
，
2
，
3
，
4
，
6
，
12
；
30
的约数有
1
，
2
，
3
，
5
，
6
，
10
，
15
，
30
。
12
和
30
的公约数有
1
，
2
，
3
，
6
，其中
6
是
12
和
30
的最大公约数。
一般地我们用
（
a,b
）
表示
a ,b
这两个自然数的最大公约数，
如
（
12
，
30
）
=6
。
如果
（
a,b
）
=1,
则
a,b
两个数是互质数。
2
、公倍数和最小公倍数
几个数公有的倍数，叫做这几个数的公倍数；其中最小的一个，叫做这几个数的最小公
倍数。< br>
例如：
12
的倍数有
12
，
24< br>，
36
，
48
，
60
，
72
，…< br>
18
的倍数有
18
，
36
，
72< br>，
90
，…
12
和
18
的公倍数有 ：
36
，
72
…其中
36
是
12
和
18
的最小公倍数。
一般地，我们用
[a,b ]
表示自然数，
a,b
的最小公倍数，如
[12
，
18]= 36
。
3
、最大公约数与最小公倍数的求法
A
．最大公约数
求两个数的最大公约数一般有以下几种方法
（
1
）分解质因数法
（
2
）短除法
（
3
）辗转相除法
（
4
）小数缩倍法
（
5
）公式法
前两种方法在数学课本中已经学过，在这里我们主要介绍辗转相除法。
当两个整数不容易看出公约数时（一般是数字比较大）
，我们可以合用辗转相除法。
B
．最小公倍数
求几个数的最小公倍数的方法也有以下几种方法：
（
1
）分解质因数法
（
2
）短除法
（
3
）大数翻倍法
（
4
）a×b
=
（
a,b
）×[a,b]
上面的公式表示：两个数的乘积等于这两个数的最大公约数和最小公倍数的乘积。
choose
the
corre
ct meani
ng;
(4) to correct the
typos;
(5) so the child wr
ite wor
ds (ABA
B, a
nd A
ABB);
(6) in accorda
nce
with
written w
ords;
(7
) the compl
ete wor
d, a
nd ex
plain t
he mea
ning
of the w
ord;
(8) collocati
on;
(9) make sentence
s wit
h the
word;
(10) the written lang
uage as r
equired. (C) t
he main
sente
nce types (1) compl
ete se
ntences; (2) write down t
he meaning
of a se
ntence or ex
pression of thoug
hts a
nd feeli
ngs; (3) write
sente
nce
s as re
quire
d; (4) finish malalig
nment
of the sentence; (5) modified
sente
nce
s. 2, knowle
dge cla
ssificati
on (1) the
common conjunctions
coordi
nate: ... ...
一
面
……
... 1, to examine the t
opi
c, i
den
tify pr
oblem
s associate
d wit
h two 2, a
nalysis, alternative
question two is i
n dire
ct pr
oportion to the amount of t
he associated re
lationshi
p is i
nversely
proporti
onal relati
onshi
p. 3, and set
unk
now
n,
colum
n pr
oportion type 4, and soluti
ons pr
oporti
on ty
pe 5, and test, wr
ote a
nswer la
nguage
ple
nary, and subje
ct:
appli
cation problem (1)--simple a
ppli
cation problem and composite applicati
on
probl
em review content sim
ple application
problem composite applicati
on pr
obl
em answer
s applicati
on pr
obl
em of general ste
ps 1, a
nd figure
out meaning--thr
oug
h examines t
he, find k
nown conditions a
nd by
seeki
ng problem 2, and analysis
number relationshi
p--analysis k
now
n conditions Zhiji
an, a
nd
conditions a
nd
probl
em Zhijia
n of relationshi
p,
determine pr
obl
em- solving method a
nd
problem-solving
steps.
3, and column ty
pe cal
culati
on--li
sts formula, i
s out subdivisions 4, a
nd test, a
nd wr
ote answer--che
ck, and checki
ng, a
nd wr
ote answ
ers typi
cal applicati
on
probl
em 13, and subje
ct: a
ppli
cation problem (3)
--col
umn e
quati
on soluti
ons
appl
ication pr
oblem review
content overvi
ew pr
oblem
-solvi
ng steps 1, a
nd figure
out meani
ng, find by se
eking
of unk
now
n and x said 2, and accordi
ng to mea
ning find equivalent relati
onship, lists Equation 3, a
nd soluti
ons
equation
4, and test, and wrote a
nsw
ers accor
ding
to meaning find equivalent relati
onship
of common method 1 , And accor
ding to
common of num
ber relationship type, establi
she
d equivale
nt relationshi
p 2, and accor
ding to
has lear
n ha
d of cal
culati
on formula
, 3, and according to pr
obl
em in the
of focus
describe
d se
ntence from overall Sha
ng determine basi
c of equivale
nt relationshi
p 4, and using segme
nt figure, and list method, method a
nalysis
num
ber
例
1
、
437
与
323
的最大公约数是多少？
例
2
、
24871
和
3468
的最小公倍数是多少？
choose
the
corre
ct meani
ng;
(4) to correct the
typos;
(5) so the child wr
ite wor
ds (ABA
B, a
nd A
ABB);
(6) in accorda
nce
with
written w
ords;
(7
) the compl
ete wor
d, a
nd ex
plain t
he mea
ning
of the w
ord;
(8) collocati
on;
(9) make sentence
s wit
h the
word;
(10) the written lang
uage as r
equired. (C) t
he main
sente
nce types
(1) compl
ete se
ntences; (2) write down t
he meaning
of a se
ntence or ex
pression of thoug
hts a
nd feeli
ngs; (3) write
sente
nce
s as re
quire
d; (4) finish malalig
nment
of the sentence; (5) modified
sente
nce
s. 2, knowle
dge cla
ssificati
on (1) the
common conjunctions
coordi
nate: ... ...
一
面
……
... 1, to examine the t
opi
c, i
dentify pr
oblem
s associate
d wit
h two 2, a
nalysis, alternative
question two is i
n dire
ct p
oportion to the amount of t
he associated re
lationshi
p is i
nversely
proporti
onal relati
onshi
p. 3, and set
unk
now
n,
colum
n pr
oportion type 4, and soluti
ons pr
oporti
on ty
pe 5, and test, wr
ote a
nswer la
nguage
ple
nary, and subje
ct:
appli
cation problem (1)--simple a
ppli
cation problem and composite applicati
on
probl
em review content sim
ple application
problem composite applicati
on pr
obl
em answer
s applicati
on pr
obl
em of general ste
ps 1, a
nd figure
out meaning--thr
oug
h examines t
he, find k
nown conditions a
nd by
seeki
ng problem 2, and analysis
number relationshi
p--analysis k
now
n conditions Zhiji
an, a
nd
conditions a
nd
probl
em Zhijia
n of relationshi
p,
determine pr
obl
em- solving method a
nd
problem-solving
steps.
3, and column ty
pe cal
culati
on--li
sts formula, i
s out subdivisions 4, a
nd test, a
nd wr
ote answer--che
ck, and checki
ng, a
nd wr
ote answ
ers typi
cal applicati
on
probl
em 13, and subje
ct: a
ppli
cation problem (3)--col
umn e
quati
on soluti
ons
appl
ication pr
oblem review
content overvi
ew pr
oblem
-solvi
ng steps 1, a
nd figure
out meani
ng, find by se
eking
of unk
now
n and x said 2, and accordi
ng to mea
ning find equivalent relati
onship, lists Equation 3, a
nd soluti
ons
equation
4, and test, and wrote a
nsw
ers accor
ding
to meaning find equivalent relati
onship
of common method 1 , And accor
ding to
common of num
ber relationship type, establi
she
d equivale
nt relationshi
p 2, and accor
ding to
has lear
n ha
d of cal
culati
on formula
, 3, and according to pr
obl
em in the
of focus
describe
d se
ntence from overall Sha
ng determine
basi
c of equivale
nt relationshi
p 4, and using segme
nt figure, and list method, method a
nalysis
num
ber
例
3
、
把一块长
90
厘米，宽
42
厘米的长方形铁板剪成边长都是整厘米，面积都 相等的小正
方形铁板，恰无剩余。至少能剪
块。
（北京市第一届迎春杯数学竞赛刊赛试题）
【分析】
：根据题意，剪得的小 正形的边长必须是
90
和
42
的最大公约
6
。所以原长方形 的
长要分
90
÷
6
＝
15
段，宽要分
42
÷
6
＝７段，至少能剪
17
×７＝
105
（块）< br>
解：
（１）求
90
和
42
的最大公约数
2 90 42
3 45 21
15 7
（
90
，
42
）＝
60
（
2
）求至少剪多少块正方形铁板
90
÷
6
＝
15
45
÷
6
＝
7
15
×
7
＝
105
（块）
答：至少可以剪
105
块正方形铁板。
说明：用短除法求小数的最大公约数比较容易。
例
4
、
10
个自然数之和等于
1001
，求这十个自然数的最大公约数可能取的最大 值是多少？