问题1.找到以下整数对的LCM和HCF，并验证LCM×HCF =整数的乘积：
(i)26和91
解决方案：
As we know that LCM is “Lowest Common Multiple” and HCF is “Highest Common Factor”. Here we have to
Verify that LCM x HCF = Product of the integers.
Given integers in questions are: 26 and 91
First, we will find the prime factors of 26 and 91.
26 = 2 × 13
91 = 7 × 13
⇒ L.C.M (26, 91) = 2 × 7 × 13 = 182
H.C.F (26, 91) = 13
According to question we need to verify the condition:
L.C.M × H.C.F = 182 × 13= 2366
Product of the integers = 26 × 91 = 2366
⇒ L.C.M × H.C.F = product of the integers
(ii)510和92
解决方案：
As we know that LCM is “Lowest Common Multiple” and HCF is “Highest Common Factor”. Here we have to
Verify that LCM x HCF = Product of the integers.
Given integers in question are: 510 and 92
We will find the prime factors of 510 and 92 to find LCM and HCF.
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
⇒ L.C.M (510, 92) = 2 × 2 × 3 × 5 × 23 × 17 = 23460
H.C.F (510, 92) = 2
According to question we need to verify the condition:
L.C.M × H.C.F = 23460 x 2 = 46920
Product of the integers = 510 x 92 = 46920
⇒ L.C.M × H.C.F = product of the integers.
(iii)336及54
解决方案：
As we know that LCM is “Lowest Common Multiple” and HCF is “Highest Common Factor”. Here we have to
Verify that LCM x HCF = Product of the integers.
Given integers in question are: 336 and 54
We will find the prime factors of 336 and 54 to find LCM and HCF.
336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 × 3
⇒L.C.M (336, 54) = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024
H.C.F (336, 54) = 2 × 3 = 6
According to question we need to verify the condition:
L.C.M × H.C.F = 3024 × 6 = 18144
Product of the integers = 336 × 54 = 18144
⇒L.C.M × H.C.F = product of the integers
问题2.通过应用素因数分解法，找到以下整数的LCM和HCF：
(i)12、15和21
解决方案：
We will find the prime factors of the given integers: 12, 15 and 21
Prime factor will help us to find LCM and HCF
12 = 2 × 2 × 3
15 = 3 × 5
21 = 3 × 7
Now for L.C.M we will find common multiples from each terms.
L.C.M of 12, 15 and 21 = 2 × 2 × 3 × 5 × 7
So L.C.M (12, 15, 21) = 420
And, for H.C.F we will find the highest common factors from each term.
H.C.F (12, 15 and 21) = 3
(ii)17、23和29
解决方案：
We will find the prime factors of the given integers: 17, 23 and 29
Prime factor will help us to find LCM and HCF
17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
Now for L.C.M we will find common multiples from each terms.
L.C.M of 17, 23 and 29 = 1 × 17 × 23 × 29
So L.C.M (17, 23, 29) = 11339
And, for H.C.F we will find highest common factors from each term.
H.C.F (17, 23 and 29) = 1
(iii)8、9和25
解决方案：
We will the find the prime factors of the given integers: 8, 9 and 25
Prime factor will help us to find LCM and HCF
8 = 2 × 2 × 2
9 = 3 × 3
25 = 5 × 5
Now for L.C.M we will find common multiples from each terms.
L.C.M of 8, 9 and 25 = 23 × 32 × 52
So L.C.M (8, 9, 25) = 1800
And, for H.C.F we will find the highest common factors from each term.
H.C.F (8, 9 and 25) = 1
(iv)40、36和126
解决方案：
We will find the prime factors of the given integers: 40, 36 and 126
Prime factor will help us to find LCM and HCF
40 = 2 × 2 × 2 × 5
36 = 2 × 2 × 3 × 3
126 = 2 × 3 × 3 × 7
Now for L.C.M we will find common multiples from each terms.
L.C.M of 40, 36 and 126 = 23 × 32 × 5 × 7
So L.C.M (40, 36, 126) = 2520
And, for H.C.F we will find highest common factors from each term.
H.C.F (40, 36 and 126) = 2
(v)84、90和120
解决方案：
We will find the prime factors of the given integers: 84, 90 and 120
Prime factor will help us to find LCM and HCF
84 = 2 × 2 × 3 × 7
90 = 2 × 3 × 3 × 5
120 = 2 × 2 × 2 × 3 × 5
Now for L.C.M we will find common multiples from each terms.
L.C.M of 84, 90 and 120 = 23 × 32 × 5 × 7
SO L.C.M (84, 90, 120) = 2520
And, for H.C.F we will find highest common factors from each term.
H.C.F (84, 90 and 120) = 6
(vi)24、15和36
解决方案：
We will find the prime factors of the given integers: 24, 15 and 36
Prime factor will help us to find LCM and HCF
24 = 2 × 2 × 2 × 3
15 = 3 × 5
36 = 2 × 2 × 3 × 3
Now for L.C.M we will find common multiples from each terms.
LCM of 24, 15 and 36 = 2 × 2 × 2 × 3 × 3 × 5 = 23 x 32 x 5
So LCM (24, 15, 36) = 360
And, for H.C.F we will find highest common factors from each term.
HCF (24, 15 and 36) = 3
问题3.鉴于HCF(306，657)= 9，找到LCM(306，657)
解决方案：
Given integers in questions are: 306 and 657
As we know that,
LCM × HCF = Product of the two integers
So here
⇒ LCM = Product of the two integers / HCF (putting values in formula)
= (306 × 657) / 9 = 22338
问题4.两个数字的HCF和16的LCM是否可以分别为16和380?给出理由。
解决方案：
If we divide 380 by 16 we will get, quotient as 23 and remainder as 12.
As we know that,
LCM × HCF = Product of the two integers
Now, the L.C.M is not exactly divisible by the HCF it’s can be said that two numbers cannot have HCF and LCM as 16 and 380 respectively.
问题5.两个数字的HCF为145，而其LCM为2175。如果一个数字为725，则查找另一个。
解决方案：
Here LCM and HCF of two numbers are 145 and 2175 given respectively.
One of the numbers is 725
As we know that,
LCM × HCF = First number × Second number
2175 × 145 = 725 × Second number (putting values in formula)
⇒ Second number = (2175 × 145)/ 725 = 435
Then the Second number is 435.
问题6.两个数字的HCF为16，其乘积为3072。找到他们的LCM。
解决方案：
Here we have, as given
HCF of two numbers = 16
Product of two numbwe = 3072
As we know that,
LCM × HCF = Product of the two numbers
LCM × 16 = 3072 (putting values in formula)
⇒ LCM = 3072/ 16 = 192
So the LCM of the two numbers is 192.
问题7.两个数字的LCM和HCF分别为180和6。如果其中一个数字为30，请查找另一个数字。
解决方案：
Here we have,
L.C.M = 180
H.C.F = 6
One of the numbers is 30
As we know that,
LCM × HCF = first number × second number
180 × 6 = 30 × second number
(putting values in formula)
Second number = (180 × 6)/ 30 = 36
Then the Second number is 36.
问题8。找到最小的数字，该数字增加17时就可以被520和468整除。
解决方案：
Firstly we will find LCM of both numbers which will be smallest number which is exactly divisible by both 520 and 468.
We will do prime factorization of given number, to find LCM:
520 = 23 × 5 × 13
468 = 22 × 32 × 13
LCM (520, 468) = 23 × 32 × 5 × 13 = 4680
Here 4680 is the smallest number which is exactly divisible by both 520 and 468.
Now, we will get a remainder of Zero in each case. But, we need to find the smallest number which when increased by 17 is exactly divided by 520 and 468.
So it can found by, subtracting,
4680 – 17 = 4663
4663 should be the smallest number which when increased by 17 is exactly divisible by both
520 and 468.
问题9.求出分别除以28和32得出余数8和12的最小数。
解决方案：
First, let’s find the smallest number which is exactly divisible by both 28 and 32.
Which is simply just the LCM of the two numbers.
By prime factorization, we get
28 = 2 × 2 × 7
32 = 25
L.C.M (28, 32) = 25
× 7 = 224
We found that 224 is the smallest number which is divisible by 28 and 32 both So, we will get a remainder of 0 in each case. But, we need the smallest number which will leave remainder of 8 and 12
When divided by 28 and 32 respectively.
So it can found by, subtracting 8 and 12 from 224
224 – 8 – 12 = 204
Here 204 must be the smallest number which will leave remainder of 8 and 12 when divided by both 28 and
32.
问题10.分别除以35、56和91得出的余数分别为7的最小数是多少?
解决方案：
Firstly we will find LCM of all three given numbers which will be smallest number which is exactly divisible by given three numbers 35, 56 and 91.
Now we will do prime factorization, to get LCM
35 = 5 × 7
56 = 23 × 7
91 = 13 × 7
So L.C.M (35, 56 and 91) = 23 × 7 × 5 × 13 = 3640
Here, we found that 3640 is the smallest number that can be divided by 35, 56 and 91 and gives remainder of 0 in each case. We require to find a number that leaves a remainder 7 in each case.
So it can be done by adding 7 in 3640,
3640 + 7 = 3647
So 3647 should be the smallest number that when divided by 35, 56 and 91 and gives the remainder
of 7 in each case.
问题11：一个矩形庭院长18m，长72cm，宽13m，长20cm。应使用相同尺寸的正方形瓷砖铺成。找到最少数量的此类图块。
解决方案：
Given data in the question are
Length of courtyard = 18 m 72 cm = 1800 cm + 72 cm = 1872 cm (We know that 1 m = 100 cm)
Breadth of courtyard = 13 m 20 cm = 1300 cm + 20 cm = 1320 cm
Here we need to the size of the square tile paved on the rectangular yard which will equal to the HCF of the length and breadth of the rectangular courtyard.
Now, we will find the prime factorization of 1872 and 1320
1872 = 24 × 32 × 13
1320 = 23 × 3 × 5 × 11
So the HCF (1872 and 1320) = 23 × 3 = 24
Here the length of side of the square tile is 24 cm.
Here We need to find number of tiles required. So, the number of tiles required = (area of the courtyard) / (area of a square tile)
As we know that, area of the courtyard = Length × Breadth
= 1872 cm × 1320 cm
(putting values in formula)
Area of a square tile = (side of square)2 = (24cm)2
The number of tiles required = (1872 x 1320) / (24)2 = 4290
(putting values in formula)
Here, we got the least possible number of tiles required is 4290.
问题12。找到最多可被24、15和36整除的6位数字。
解决方案：
The greatest 6 digit number is 999999.
If we assume that, 24,15 and 36 divides 999999 without leaving any remainder.
Then we need to find LCM (24, 15 and 36) because LCM will also divide 999999 exactly.
Now the prime factorization of 24, 15, and 36, to get LCM
24 = 2 × 2 × 2 × 3
15 = 3 × 5
36 = 2 × 2 × 3 × 3
Then L.C.M of 24, 15 and 36 = 360
If we divide (999999)/ 360 = 2777 × 360 + 279
Then remainder is 279.
Now to find greatest 6 digit number which is divisible by all three should be, we need to
We need to subtract remainder from 6 digit greatest number.
So, 999999 – 279 = 999720
Here 999720 is the greatest 6 digit number which is exactly divisible by 24, 15 and 36.
问题13.确定最接近110000但大于100000的数字，该数字可以被8、15和21中的每一个完全整除。
解决方案：
Firstly we will find the L.C.M of 8, 15 and 21.
By prime factorization, we will get LCM
8 = 2 × 2 × 2
15 = 3 × 5
21 = 3 × 7
Now L.C.M (8, 15 and 21) = 2
3 × 3 × 5 × 7 = 840
If we divide the 110000 by 840, then the remainder will be 800.
Now, If we substract 800 from the 110000, it will be divisible by each of 8, 15 and 21.
So, 110000 – 800 = 109200.
If we add 40, it will be also divisible by each of 8, 15 and 21
Now we have 110000 + 40 = 110040
We got two result, 109200 and 110040 both are greater than 100000 but 110040 is greater than 110000.
So, 109200 is the number nearest to 110000 and greater than 100000 which is exactly divisible by each of 8, 15 and 21.
问题14。找到可被1到10(包括两端)之间的所有数字整除的最小数字。
解决方案：
According to the question we need to find the least number that is divisible by all the numbers between 1 and 10
So, the L.C.M of 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 will be the least number that will be divisible by all the numbers between 1 and 10.
Hence, the prime factorization of all these numbers are, to get LCM:
1 = 1
2 = 2
3 = 3
4 = 2 × 2
5 = 5
6 = 2 × 3
7 = 7
8 = 2 × 2 × 2
9 = 3 × 3
10 = 2 × 5
L.C.M will be = 23 × 32 × 5 × 7 = 2520
Here we got 2520 as the least number that is divisible by all the numbers between 1 and 10 (both inclusive)
问题15.圆形场的周长为360 km。三个骑自行车的人一起开始，可以围绕场地每天骑48、60和72公里。他们什么时候会再见面?
解决方案：
Here we need to calculate the time taken before they meet again , so we need to find the individual time taken by each cyclist in covering the total distance.
Now, we will find Number of days a cyclist takes to cover the circular field by:
Number of days a cyclist takes to cover the circular field = (Total distance of the circular field) /
(distance covered in 1 day by a cyclist)
1st cyclist, number of days = 360 / 48 = 7.5 which will = 180 hours [we know that,1 day = 24 hours]
2nd cyclist, number of days = 360 / 60 = 6 which will = 144 hours
3rd cyclist, number of days = 360 / 72 = 5 which will 120 hours
Now, the LCM (180, 144 and 120) will help to know after how many hours the three cyclists meet again.
By prime factorization, we get LCM
180 = 22 x 32 x 5
144 = 24 x 33
120 = 23 x 3 x 5
L.C.M (180, 144 and 120) = 24 x 32 x 5 = 720
So, the after 720 hours the three cyclists meet again.
720 hours = 720 / 24 = 30 days [we know that, 1 day = 24 hours]
So, we can say that all the three cyclists will meet again after 30 days.
问题16：在一个早晨的步行中，三个人走在一起，他们的步长分别为80厘米，85厘米和90厘米。每个人应该走的最小距离是多少，以便他可以完整地覆盖该距离?
解决方案：
As per given in the question, that the required distance each should walk would be the
LCM of the measures of their steps that is 80 cm, 85 cm, and 90 cm.
So, L.C.M (80, 85 and 90) with the help of prime factorization,
80 = 24 × 5
85 = 17 × 5
90 = 2 × 3 × 3 × 5
L.C.M (80, 85 and 90) = 24 × 32 × 5 × 17 = 12240 cm = 122m 40cm [we know that, 1 m = 100cm]
We got, 122m 40cm as the minimum distance that each should walk so that all can cover the same distance in complete steps.