问题11。求出最大数除以285和1249，分别剩下余数9和7。
解决方案：
The required number when divides 285 and 1249, should leave remainder 9 and 7 respectively,
285 – 9 = 276 and 1249 -7 = 1242 can divide them exactly.
The required number is equivalent to the H.C.F. of 276 and 1242.
Applying Euclid’s division lemma on 276 and 1242 , we get
1242 = 276 x 4 + 138
276 = 138 x 2 + 0.
Since, the remainder is now 0,
Therefore, the H.C.F.(req. number) = 138
问题12。找到最大数，该数精确地将280和1245相除，分别剩下余数4和3。
解决方案：
The required number when divides 280 and 1245, should leave remainder 4 and 3 respectively,
280 – 4 = 276 and 1245 – 3 = 1242 has to be exactly divisible by the number.
The required number is equivalent to the H.C.F. of 276 and 1242.
Applying Euclid’s division lemma on 276 and 1242 , we get
1242 = 276 x 4 + 138
276 = 138 x 2 + 0 (the remainder becomes 0 here)
Since, the remainder is now 0,
Therefore, the H.C.F.(req. number) = 138
问题13.将626、3127和15628除以余数分别为1、2和3的最大数是多少?
解决方案：
The required number when divides 626, 3127 and 15628, should leave remainder 1,2 and 3 respectively,
626 – 1 = 625, 3127 – 2 = 3125 and 15628 – 3 = 15625 has to be exactly divisible by the number.
The required number is equivalent to the H.C.F of 625, 3125 and 15625.
Considering 625 and 3125 first , we apply Euclid’s division lemma
3125 = 625 x 5 + 0
∴ H.C.F (625, 3125) = 625
Applying Euclid’s division lemma on 15625 and 625 , we get
15625 = 625 x 25 + 0
Now,
∴ H.C.F. (625, 3125, 15625) = 625
问题14。找到将除以445,572和699的最大数，分别剩下余数4、5和6。
解决方案：
The required number when divides 445,572 and 699, should leave remainder 4, 5 and 6 respectively,
445 – 4 = 441, 572 – 5 = 567 and 699 – 6 = 693 has to be exactly divisible by the number.
The required number is equivalent to the H.C.F of 441, 567 and 693.
Applying Euclid’s division lemma on 441 and 567 , we get
567 = 441 x 1 + 126
441 = 126 x 3 + 63
126 = 63 x 2 + 0.
∴ H.C.F (441 and 567) = 63
Applying Euclid’s division lemma on 63 and 693 , we get
693 = 63 x 11 + 0.
Now,
∴ H.C.F. (441, 567 and 693) = 63
问题15：找到除以2011和2623的最大数，分别剩下9和5。
解决方案：
The required number when divides 2011 and 2623 should leave remainder 9 and 5 respectively,
2011 – 9 = 2002 and 2623 – 5 = 2618 has to be exactly divisible by the number.
The required number is equivalent to the H.C.F. of 2002 and 2618
Applying Euclid’s division lemma, we get,
2618 = 2002 x 1 + 616
2002 = 616 x 3 + 154
616 = 154 x 4 + 0.
The remainder becomes 0,
Therefore, the H.C.F. (2002, 2618) = 154
问题16.使用Euclid的除法算法，找到最大数除以1251、9377和15628，分别剩下余数1、2和3。
解决方案：
The required number when divides 1251, 9377 and 15628 should leave remainder 1, 2 and 3 respectively,
1251 – 1 = 1250, 9377 – 2 = 9375 and 15628 – 3 = 15625 has to be exactly divisible by the number.
The required number is equivalent to the H.C.F of 1250, 9375 and 15625.
Applying Euclid’s division lemma on 1250, 9375 , we get,
9375 = 1250 x 7 + 625
1250 = 625 x 2 + 0
∴ H.C.F (1250, 9375) = 625
Applying Euclid’s division lemma on 625 and 15625 , we get,
15625 = 625 x 25 + 0
Since, the remainder is now 0,
∴ H.C.F. (1250, 9375, 15625) = 625
So, the required number is 625.
问题17.两种品牌的巧克力分别以24和15的包装提供。如果我需要购买相同数量的两种巧克力，那么我需要购买的最少每种盒子的数量是多少?
解决方案：
Number of chocolates of 1st brand in a pack = 24
Number of chocolates of 2nd brand in a pack = 15.
The least number of both brands of chocolates is equivalent to their LCM.
L.C.M. of 24 and 15 = 2 x 2 x 2 x 3 x 5 = 120
Now,
The number of packets of 1st brand to be bought = 120 / 24 = 5
The number of packets of 2nd brand to be bought = 120 / 15 = 8
问题18.石匠必须将最大尺寸的方形大理石瓷砖安装在浴室中。浴室的大小为10英尺。乘8英尺需要切割的瓷砖尺寸为多少英寸(英寸)?需要多少瓷砖?
解决方案：
Size of bathroom = 10 ft. by 8 ft.
Converting from ft. to inch.
= (10 x 12) inch by (8 x 12) inch
= 120 inch by 96 inch
The largest size of tile required is equal the HCF of the numbers 120 and 96.
Applying Euclid’s division lemma on 120 and 96 ,we get,
120 = 96 x 1 + 24
96 = 24 x 4 + 0
⇒ HCF = 24
Thus, the largest size of tile which required is 24 inches.
Also,
Number of tiles required = (area of bathroom) / (area of a tile)
= (120 x 96) / (24×24)
= 5 x 4
We obtain,
= 20 tiles
Therefore, 20 tiles each of size 24 inch by 24inch are required to be cut.
问题19：为学校宴餐捐赠了15个糕点和12个饼干小包。将它们包装在几个较小的相同的盒子中，每个盒子中有相同数量的糕点和饼干包装。每个盒子将容纳多少饼干包装和多少个糕点?
解决方案：
We have,
Number of pastries = 15
Number of biscuit packets = 12
The required number of boxes containing an equal number of both pastries and biscuits will be equal to the HCF of the numbers 15 and 12.
Applying Euclid’s division lemma on 15 and 12, we get
15 = 12 x 1 + 3
12 = 3 x 4 = 0
Therefore, the required boxes = 3
Now,
∴ Each box will contain 15/3 = 5 pastries and 12/3 = 4 biscuit packs.
问题20：必须把105只山羊，140只驴和175头母牛带到一条河上。只有一艘船才能这样做。懒惰的船夫有自己的运输条件。他坚持说，他每次旅行都将带走相同数量的动物，而且它们必须是同一种动物。他自然会喜欢每次都尽可能多地使用。您能说出每次旅行中有多少只动物吗?
解决方案：
We have,
Number of goats = 105
Number of donkeys = 140
Number of cows = 175
The largest number of animals in one trip is equivalent to the HCF (105, 140 and 175).
Applying Euclid’s division lemma on 105 and 140, we get
140 = 105 x 1 + 35
105 = 35 x 3 + 0
Therefore, the HCF (105 and 140) = 35
Applying Euclid’s division lemma on 35 and 175, we get
175 = 35 x 5 +0
Hence, the HCF (105, 140, 175) = 35.
Therefore, 35 animals went on each trip.
问题21.房间的长度，宽度和高度分别是8 m 25 cm，6 m 75 cm和4 m 50 cm。确定可以精确测量房间三个尺寸的最长杆。
解决方案：
We have,
Length of the room = 8m 25 cm = 825 cm (in cm)
Breadth of the room = 6m 75cm = 675 cm
Height of the room = 4m 50cm = 450 cm
The longest rod which can measure the room will be exactly equivalent to the HCF of given measurements 825, 675 and 450.
Applying Euclid’s division lemma on 675 and 450, we get
675 = 450 x 1 + 225
450 = 225 x 2 + 0
Therefore, the HCF (675, 450) = 225
Applying Euclid’s division lemma on 225 and 825, we get
825 = 225 x 3 + 150
225 = 150 x 1+75
Applying Euclid’s division lemma on 150 and 75, we get
150 = 75 x 2 + 0
Thus, HCF (225, 825) = 75.
Also,
HCF of 825, 675 and 450 is 75.
Therefore, the length of the longest required rod is 75 cm.
问题22.将468和222的HCF表示为468x + 222y，其中x，y以两种不同的方式表示整数。
解决方案：
We have the integers, 468 and 222, where 468 > 222
Applying Euclid’s division lemma on 468 and 222, we get
468 = 222 x 2 + 24……… (1)
Applying Euclid’s division lemma on 222 and remainder 24, we get
222 = 24 x 9 + 6………… (2)
Applying Euclid’s division lemma on 24 and remainder 6, we get
24 = 6 x 4 + 0……………. (3)
The remainder now is 0.
Therefore, the H.C.F. of 468 and 222
We can express the HCF as a linear combination of 468 and 222, by
6 = 222 – 24 x 9 [from (2)]
= 222 – (468 – 222 x 2) x 9 [from (1)]
= 222 – 468 x 9 + 222 x 18
6 = 222 x 19 – 468 x 9 = 468(-9) + 222(19)
∴ 6 = 468x + 222y, where x = -9 and y = 19.