问题1.定义两个正整数的HCF，并找到以下几对数字的HCF：
(i)32和54
解决方案：
Applying Euclid’s Division Lemma on 54 , we get,
54 = 32 x 1 + 22
Applying Euclid’s Division Lemma on 32 and remainder 22 , we get,
32 = 22 x 1 + 10
Now, remainder ≠ 0, apply division lemma on 22 and remainder 10, we get,
22 = 10 x 2 + 2
Applying Euclid’s Division Lemma on 10 and remainder 2 , we get,
10 = 2 x 5 + 0
Therefore, the H.C.F. of 32 and 54 is 2.
(ii)18和24
解决方案：
Applying Euclid’s Division Lemma on 24 and 18, we get,
24 = 18 x 1 + 6.
Applying Euclid’s Division Lemma on 18 and remainder 6., we get,
18 = 6 x 3 + 0.
Therefore, H.C.F. of 18 and 24 is 6
(iii)70和30
解决方案：
Applying Euclid’s Division Lemma on 70 and 30, we get,
70 = 30 x 2 + 10.
Applying Euclid’s Division Lemma on 30 and remainder 10, we get,
30 = 10 x 3 + 0.
Since, remainder is 0 now,
Therefore, H.C.F. of 70 and 30 is 10
(iv)56和88
解决方案：
Applying Euclid’s Division Lemma on 56 and remainder 88, we get,
88 = 56 x 1 + 32.
Applying Euclid’s Division Lemma on 56 and remainder 32, we get,
56 = 32 x 1 + 24.
Applying Euclid’s Division Lemma on 32 and remainder 24, we get,
32 = 24 x 1+ 8.
Applying Euclid’s Division Lemma on 24 and remainder 8, we get,
24 = 8 x 3 + 0.
Since, the remainder is now 0,
Therefore, H.C.F. of 56 and 88 is 8
(v)475和495
解决方案：
Applying Euclid’s Division Lemma on 475 and 495, we get,
495 = 475 x 1 + 20.
Applying Euclid’s Division Lemma on 475 and remainder 20, we get,
475 = 20 x 23 + 15.
Applying Euclid’s Division Lemma on 20 and remainder 15, we get,
20 = 15 x 1 + 5.
Applying Euclid’s Division Lemma on 15 and remainder 5, we get,
15 = 5 x 3+ 0.
Since, the remainder is now 0,
Therefore, H.C.F. of 475 and 495 is 5
(vi)75和243
解决方案：
Applying Euclid’s Division Lemma on 243 and 75, we get,
243 = 75 x 3 + 18.
Applying Euclid’s Division Lemma on 75 and remainder 18, we get,
75 = 18 x 4 + 3.
Applying Euclid’s Division Lemma on 18 and remainder 3, we get,
18 = 3 x 6+ 0.
Since, the remainder is now 0,
Therefore, H.C.F. of 75 and 243 is 3
(vii)240和6552
解决方案：
Applying Euclid’s Division Lemma on 6552 and 240, we get,
6552 = 240 x 27 + 72.
Applying Euclid’s Division Lemma on 240 and remainder 72, we get,
240 = 72 x 3+ 24.
Applying Euclid’s Division Lemma on 72 and remainder 24, we get,
72 = 24 x 3 + 0.
Since, the remainder is now 0,
Therefore, H.C.F. of 240 and 6552 is 24
(viii)155和1385
解决方案：
Applying Euclid’s Division Lemma on 1385 and 155, we get,
1385 = 155 x 8 + 145.
Applying Euclid’s Division Lemma on 155 and remainder 145, we get,
155 = 145 x 1 + 10.
Applying Euclid’s Division Lemma on 145 and remainder 10, we get,
145 = 10 x 14 + 5.
Applying Euclid’s Division Lemma on 10 and remainder 5, we get,
10 = 5 x 2 + 0.
Since, the remainder is now 0,
Therefore, H.C.F. of 155 and 1385 is 5
(ix)100和190
解决方案：
Applying Euclid’s Division Lemma on 190 and 100, we get,
190 = 100 x 1 + 90.
Applying Euclid’s Division Lemma on 100 and remainder 90, we get,
100 = 90 x 1 + 10.
Applying Euclid’s Division Lemma on 90 and remainder 10, we get,
90 = 10 x 9 + 0.
Therefore, H.C.F. of 100 and 190 is 10
(x)105和120
解决方案：
Applying Euclid’s Division Lemma on 120 and 105, we get,
120 = 105 x 1 + 15.
Applying Euclid’s Division Lemma on 105 and remainder 15, we get,
105 = 15 x 7 + 0.
Therefore, H.C.F. of 105 and 120 is 15
问题2。使用Euclid的除法算法来找到HCF的HCF
(i)135和225
解决方案：
On comparing both the integers, we find 225 > 135.
Applying Euclid’s Division Lemma on 225 and 135, we get,
225 = 135 x 1 + 90
Applying Euclid’s Division Lemma on 135 and remainder 90, we get,
⇒ 135 = 90 x 1 + 45
Applying Euclid’s Division Lemma on 90 and remainder 45, we get,
⇒ 90 = 45 x 2 + 0
Since, the remainder is now 0,
Therefore, the H.C.F of 225 and 135 is 45.
(ii)196和38220
解决方案：
On comparing both the integers, we find 38220 > 196.
Applying Euclid’s Division Lemma on 38220 and 196, we get,
38220 = 196 x 195 + 0
Since, the remainder is now 0,
Hence, the HCF of 38220 and 196 is 196
(iii)867及255
解决方案：
On comparing both the integers, we find 867 > 255.
Applying Euclid’s Division Lemma on 867 and 255 , we get,
867 = 225 x 3 + 192
Applying Euclid’s Division Lemma on 225 and remainder 192 , we get,
225 = 192 x 1 + 33
Applying Euclid’s Division Lemma on 192 and remainder 33 , we get,
192 = 33 x 5 + 27
Applying Euclid’s Division Lemma on 33 and remainder 27 , we get,
33 = 27 x 1 + 6
Applying Euclid’s Division Lemma on 27 and remainder 6 , we get,
27 = 6 x 4 + 3
Applying Euclid’s Division Lemma on 6 and remainder 3 , we get,
6 = 3 x 2 + 0
Since the remainder is now 0,
Hence, the HCF of 867 and 255 is 3.
(iv)184、230和276
解决方案：
We will first choose from 184 and 230 to find the HCF by using Euclid’s division lemma.
Thus, we obtain
230 = 184 x 1 + 46
Applying Euclid’s Division Lemma on 184 and remainder 46 , we get,
184 = 46 x 4 + 0
The HCF is therefore, 230.
Applying Euclid’s Division Lemma on 46 and 276 , we get,
276 = 46 x 6 + 0
Therefore, the HCF of the third number 276 and 46 is 46.
(v)136、170和255
解决方案：
We will first choose from 136 and 170 to find the HCF by using Euclid’s division lemma.
We get,
170 = 136 x 1 + 34
Applying Euclid’s Division Lemma on 136 and remainder 34 , we get,
136 = 34 x 4 + 0
Since, the remainder is now 0, the divisor will be the HCF i.e., 34 for 136 and 170.
Applying Euclid’s Division Lemma on 34 and 255 , we get,
255 = 34 x 7 + 17
Applying Euclid’s Division Lemma on 34 and remainder 17 , we get,
34 = 17 x 2 + 0
Since, the remainder is now 0,
Therefore, the HCF of 136, 170 and 255 is 17.
问题3.找到以下整数对的HCF并将其表示为它们的线性组合，
(i)963和657
解决方案：
Applying Euclid’s Division Lemma on 963 and 657 , we get,
963 = 657 x 1 + 306………. (1)
Applying Euclid’s Division Lemma on 657 and remainder 306 , we get,
657 = 306 x 2 + 45………… (2)
Applying Euclid’s Division Lemma on 306 and remainder 45 , we get,
306 = 45 x 6 + 36…………. (3)
Applying Euclid’s Division Lemma on 45 and remainder 36 , we get,
45 = 36 x 1 + 9…………… (4)
Applying Euclid’s Division Lemma on 36 and remainder 9 , we get,
36 = 9 x 4 + 0……………. (5)
Therefore, the HCF is 9.
We can express the HCF as a linear combination of 963 and 657, by
9 = 45 – 36 x 1 [from (4)]
= 45 – [306 – 45 x 6] x 1 = 45 – 306 x 1 + 45 x 6 [from (3)]
= 45 x 7 – 306 x 1 = [657 -306 x 2] x 7 – 306 x 1 [from (2)]
= 657 x 7 – 306 x 14 – 306 x 1
= 657 x 7 – 306 x 15
= 657 x 7 – [963 – 657 x 1] x 15 [from (1)]
= 657 x 7 – 963 x 15 + 657 x 15
= 657 x 22 – 963 x 15.
(ii)592和252
解决方案：
Applying Euclid’s Division Lemma on 592 and 252 , we get,
592 = 252 x 2 + 88……… (1)
Applying Euclid’s Division Lemma on 252 and remainder 88 , we get,
252 = 88 x 2 + 76………. (2)
Applying Euclid’s Division Lemma on 88 and remainder 76 , we get,
88 = 76 x 1 + 12………… (3)
Applying Euclid’s Division Lemma on 76 and remainder 12 , we get,
76 = 12 x 6 + 4………….. (4)
Applying Euclid’s Division Lemma on 12 and remainder 4 , we get,
12 = 4 x 3 + 0……………. (5)
Therefore, H.C.F. = 4.
We can express the HCF as a linear combination of 592 and 252, by
4 = 76 – 12 x 6 [from (4)]
= 76 – [88 – 76 x 1] x 6 [from (3)]
= 76 – 88 x 6 + 76 x 6
= 76 x 7 – 88 x 6
= [252 – 88 x 2] x 7 – 88 x 6 [from (2)]
= 252 x 7- 88 x 14- 88 x 6
= 252 x 7- 88 x 20
= 252 x 7 – [592 – 252 x 2] x 20 [from (1)]
= 252 x 7 – 592 x 20 + 252 x 40
= 252 x 47 – 592 x 20
= 252 x 47 + 592 x (-20)
(iii)506和1155
解决方案：
Applying Euclid’s Division Lemma on 506 and 1155 , we get,
1155 = 506 x 2 + 143…………. (1)
Applying Euclid’s Division Lemma on 506 and remainder 143 , we get,
506 = 143 x 3 + 77…………….. (2)
Applying Euclid’s Division Lemma on 143 and remainder 77 , we get,
143 = 77 x 1 + 66……………… (3)
Applying Euclid’s Division Lemma on 77 and remainder 66 , we get,
77 = 66 x 1 + 11……………….. (4)
Applying Euclid’s Division Lemma on 66 and remainder 11 , we get,
66 = 11 x 6 + 0………………… (5)
Therefore, H.C.F. = 11.
We can express the HCF as a linear combination of 506 and 1155 by,
11 = 77 – 66 x 1 [from (4)]
= 77 – [143 – 77 x 1] x 1 [from (3)]
= 77 – 143 x 1 + 77 x 1
= 77 x 2 – 143 x 1
= [506 – 143 x 3] x 2 – 143 x 1 [from (2)]
= 506 x 2 – 143 x 6 – 143 x 1
= 506 x 2 – 143 x 7
= 506 x 2 – [1155 – 506 x 2] x 7 [from (1)]
= 506 x 2 – 1155 x 7+ 506 x 14
= 506 x 16 – 1155 x 7
(iv)1288和575
解决方案：
Applying Euclid’s Division Lemma on 1288 and 575 , we get,
1288 = 575 x 2+ 138………… (1)
Applying Euclid’s Division Lemma on 575 and remainder 138 , we get,
575 = 138 x 4 + 23……………. (2)
Applying Euclid’s Division Lemma on 138 and remainder 23 , we get,
138 = 23 x 6 + 0……………….. (3)
Therefore, H.C.F. = 23.
We can express the found HCF as a linear combination of 1288 and 575, by
23 = 575 – 138 x 4 [from (2)]
= 575 – [1288 – 575 x 2] x 4 [from (1)]
= 575 – 1288 x 4 + 575 x 8
= 575 x 9 – 1288 x 4
问题4：找到将615和963相除的最大数，在每种情况下均剩下余数6。
解决方案：
Firstly, since 6 is required as the remainder, we subtract it from both the numbers.
So, the required numbers are 615 – 6 = 609 and 963 – 6 = 957.
The required number is the HCF of newly obtained numbers, 609 and 957.
Applying Euclid’s Division Lemma , we get,
957 = 609 x 1+ 348
609 = 348 x 1 + 261
348 = 261 x 1 + 87
261 = 87 x 3 + 0.
Since, the remainder is 0,
Therefore, the required number is 87
问题5.如果408和1032的HCF可以1032m – 408 x 5的形式表示，则找到m。
解决方案：
Firstly, the HCF of 408 and 1032 is to be found.
Applying Euclid’s Division Lemma on 408 and 1032 , we get,
1032 = 408x 2 + 216.
Applying Euclid’s Division Lemma on 408 and remainder 216 , we get,
408 = 216 x 1 + 192.
Applying Euclid’s Division Lemma on 216 and remainder 192 , we get,
216 = 192 x 1 + 24.
Applying Euclid’s Division Lemma on 192 and remainder 24 , we get,
192 = 24 x 8 + 0.
Since, the remainder is 0 ,
The H.C.F of 408 and 1032 i.e., 24
So, this HCF is expressed as a linear combination that is,
24 = 1032m – 408 x 5
1032m = 24 + 408 x 5
1032m = 24 + 2040
1032m = 2064
m = 2064/1032
We obtain,
∴ m = 2
问题6.如果657和963的HCF可以用657x + 963 x – 15的形式表示，则找到x。
解决方案：
Applying Euclid’s Division Lemma on 657 and 963 , we get,
963 = 657 x 1+ 306.
Applying Euclid’s Division Lemma on 657 and remainder 306 , we get,
657 = 306 x 2 + 45.
Applying Euclid’s Division Lemma on 306 and remainder 45 , we get,
306 = 45 x 6 + 36.
Applying Euclid’s Division Lemma on 45 and remainder 36 , we get,
45 = 36 x 1 + 9.
Applying Euclid’s Division Lemma on 36 and remainder 9 , we get,
36 = 9 x 4 + 0.
Now, the remainder = 0.
Hence, the last divisor is the H.C.F of 657 and 963 i.e., 9
by expressing the HCF as a linear combination, we get ,
9 = 657x + 963 (-15).
Finding the value of x, we get
9 = 657x —14445
9 + 14445 = 657x
14454 = 657x
⇒ x = 14454 / 657
∴ x = 22
问题7：一支由616名成员组成的部队队伍将在游行队伍中向一支由32名成员组成的军队乐队前进。两组将以相同的列数前进。他们可以前进的最大列数是多少?
解决方案：
We need to compute the max number of columns in which the army band can march, which can be done by finding the HCF of the given two numbers.
Now, this is equal to the H.C.F of 616 and 32.
Applying Euclid’s Division Lemma on 616 and 32 , we get,
616 = 32 x 19 + 8
32 = 8 x 4 + 0.
Therefore, H.C.F. = 8
∴ The maximum number of columns in which the army band can march is 8.
问题8.商人拥有120升一种油，180升另一种油和240升第三种油。他想通过将三种油装在容量相等的罐子中来出售油。这种锡的最大容量应该是多少?
解决方案：
The greatest capacity of the tin for filling three different types of oil is equivalent to the the H.C.F. of the three available quantities 120,180 and 240.
Applying Euclid’s Division Lemma on 180 and 120 , we get,
180 = 120 x 1 + 60
120 = 60 x 2 + 0
Since, the remainder is now 0 ,
The HCF = 60.
Computing the H.C.F of 60 and the third quantity 240.
Applying Euclid’s Division Lemma on 240 and 60 , we get,
240 = 60 x 4 + 0
Since the remainder is 0 ,
Therefore, the HCF = 60
Therefore, the tin should be of 60 litres.
问题9.在销售过程中，彩色铅笔的销售价格为每24支一包，蜡笔的零售价为每32支一包。如果您想同时购买一包和相同数量的铅笔和蜡笔，则需要购买几支?
解决方案：
We have,
Number of colour pencils in a pack = 24
Number of crayons in a pack = 32.
The least number of both colour pencils and crayons that needs to be purchased is equivalent to their LCM.
L.C.M of 24 and 32 = 2 x 2 x 2 x 2 x 2 x 3 = 96
Now,
The number of pencil packs to be bought = 96 / 24 = 4 packs
And, the number of crayon packs to be bought = 96 / 32 = 3 packs
问题10：将144箱可乐罐和90箱百事可乐罐堆叠在食堂中。如果每堆纸箱的高度相同并且要容纳相同饮料的纸箱，那么每堆纸箱最多可容纳多少个纸箱?
解决方案：
We have,
Number of coke can cartons = 144
Number of Pepsi can cartons = 90.
Therefore, the maximum number of cartons in a stack can be found by computing the H.C.F. of(144, 90).
Applying Euclid’s Division Lemma on 144 and 90 , we get,
144 = 90 x 1 + 54
90 = 54 x 1+ 36
54 = 36 x 1 + 18
36 = 18 x 2 + 0
∴ Since, the remainder is 0 ,
Therefore, the greatest number of cartons together in one stack is 18