当然，在800年成书，远早于印刷术的出现，意味着这部著作是通过抄写员抄写流传下来的。正如这类传播方式通常发生的那样，流传给我们的各种版本存在差异，这些差异源于抄写错误，也源于为“改进”文本而有意做出的改动。一个差异是谜题的数量，有些文本版本包含53个谜题，另一些则包含56个。历史学家中的多数意见是，原始文本包含56个谜题，而只有53个的是后来的（有些缺陷的）版本。在一些手稿中谜题没有编号，而在另一些只有53个谜题的版本中则有标准编号。我们在谜题列表中遵循了标准编号，因此，例如有谜题11、11(a)和11(b)。
- 蜗牛问题。
A snail was invited by a swallow to lunch a league away. However, it could only crawl one inch per day. How many years and days did it take for the snail to crawl to that lunch?
[假设1里格=1500步，1步=5英尺，1英尺=12英寸，一年有365天。]
解答。
1500 × 5 × 12 = 90000。用90000除以365，得到246年零210天。 - 在街上行走的人问题。
A certain man walking in the street saw a group of men coming towards him, and he said to them: "Suppose there were as many more of you as you are now; and then to this half of half were added; and then half of this number were added. Then, together with myself, we would number 100." How many men were in the group that the man saw?
解答。
假设这群人中有x个男人。那么2x+2x+4x+1=100。所以x=36.阿尔琴当然不使用方程。他只是通过检查72 + 18 + 9 + 1 = 100来验证36满足给定条件。这类问题非常古老，出现在莱因德纸草书中。 - 两个人和鹳问题。
Two men were walking along the street when they saw some storks. They asked each other, "How many storks are there?" Their discussion went as follows: "Suppose the number of storks was doubled, then the original number added again, and then half of a third of this sum were added. Then, together with another two, they would number 100." How many storks did the men see?
解答。
假设有x只鹳。那么2x+x+63x+2=100.所以x=28.同样，阿尔琴给出了答案并验证了28满足条件。 - 人与马的谜题。
A certain man saw some horses grazing in a field and wanted them. He said: "Suppose that they were mine, and that the number were doubled, and then a half of half of this sum were added. In this case I would have 100 horses." How many horses did the man see grazing in the field?
解答。
假设有x匹马。那么2x+42x=100.所以x=40.阿尔琴给出了答案40并验证了它满足条件。 - 养猪农有100磅的谜题。
A pig farmer said: "I want to buy 100 pigs with 100 pounds. Now a boar costs ten pounds, a sow costs five pounds and two piglets can be bought for a pound." How many boars, sows, and piglets can the pig farmer buy so that he spends all his money?
解答。
假设农夫买了x头公猪、y头母猪和z对小猪。那么x+y+2z=100，以及10x+5y+z=100。由第二个方程可知z必能被5整除，于是记z=5t.，则x+y+10t=100，以及2x+y+t=20。第一个方程给出t<10，但将方程相减得到−x+9t=80，因此t≥9（因为9t>80）。所以t=9,x=1且y = 9。
养猪农必须买1头公猪、9头母猪和90头小猪。阿尔琴表明这个答案满足条件，但没有给出任何找到它的方法。人们推测本意是用试错法。这类问题已知的最早出现是在5世纪的中国. - 两个农夫有100磅的谜题。
There were two farmers who had 100 pounds between them, with which they bought some pigs. They bought groups of five pigs at two pounds for a group, and they intended to fatten them and to sell them at a profit. But when they saw that the time was not right to fatten the pigs, and being unable to keep them on their farms, they tried to make a profit by selling them. However, they were unsuccessful because they could only sell the pigs for what they had paid (i.e., five pigs for two pounds). When they realized this, they said to each other, "We shall divide the pigs." But by dividing and selling the pigs for as much as they had paid, they made a profit. How many pigs were there at first, and how did the men divide and sell for a profit that which they could not do together?
解答。
他们必定买了250头猪。如果他们能够通过不同的分配方式获利，那么5头一组中的猪不可能价值相等。一个农夫拿125头质量最好的猪，另一个拿125头质量最差的猪。第一个以2头1磅的价格卖出120头，而第二个农夫以3头1磅的价格卖出120头。因此他们收回了自己的100磅，并且每人还有5头猪可以出售获利。这可能是阿尔琴原创的谜题。当然，已知没有更早的这类谜题版本。 - 重30磅的盘子的谜题。
There is a plate weighing 30 pounds or 600 shillings. In it, there is gold, silver, brass and tin. It has three times are much silver as gold, three times as much brass as silver, and three times as much tin as brass. How much does each type of metal weigh?
解答。
假设有x先令的金子。那么有3x先令的银子、9x先令的黄铜和27x先令的锡。x+3x+9x+27x=40x=600所以x=15，其中有15先令的金，45先令的银，135先令的黄铜和405先令的锡。像往常一样，阿尔琴给出答案并检验它满足条件。 - 有三个裂缝的木桶的谜题。
There is a cask which has three cracks in it. It is filled with 7200 pints of water. A third plus a sixth part runs out through one crack. Through another crack a third part runs out. Only a sixth part runs out through the third crack. How many pints ran out through each crack.
解答。
一半从第一个裂缝流出，即3600品脱，31从第二个裂缝流出，即2400品脱，61从第三个裂缝流出，即1200品脱。 - 斗篷谜题。
I have material which is 100 feet long and 80 feet wide. From it, I wish to make cloaks from portions in such a way that each portion is five feet in length and four feet wide. How many cloaks can be made from the material?
解答。
从100英尺中，我们可以按5英尺裁出20块，从80英尺中，我们可以按4英尺裁出20块。因此，我们可以裁出400块5英尺乘4英尺的布料。 - 亚麻布谜题。
I have a single linen cloth which is 60 feet long and 40 feet wide. I wish to cut it into smaller portions, each being six feet in length, four feet in width, so that each piece is big enough to make a tunic. How many tunics can be made from the single linen cloth?
解答。
如前一个谜题，10×10即100件束腰外衣。 - 两个男人娶了对方姐妹的谜题。
If two men should marry one another's sister. What will be the sons' relations to each other?
解答。
这些儿子们是双重表亲，每人都有一位家长以两种方式与另一人的一位家长是兄弟姐妹。这个谜题以及接下来的两个“亲属关系”谜题可能是阿尔琴的原创。当然，已知没有更早的版本。
(a) 两个男人娶了对方母亲的谜题。
如果两个男人各自娶了对方的母亲为妻，他们的儿子之间会是什么关系？
解答。
设A和B为娶了对方母亲的两个男人。再假设这两桩婚姻的儿子是S和T。那么S的父亲是A，而B是他的同父异母兄弟。同样，T的父亲是B，而A是他的同父异母兄弟。因此，这两个儿子S和T彼此既是叔伯又是侄子。从现存的手稿来看，阿尔琴似乎没有给出这个谜题的解答。然而，此后这个谜题一直可见，并不时地重新出现。尽管阿尔琴没有问及在这种情况下出现的任何其他奇怪关系，但引发一个谜语和一首歌的关系如下。
B的母亲是A的母亲的婆婆，因此是A的祖母。由于A娶了B的母亲，他娶了自己的祖母，因此也是自己的祖父。(b) 父亲、儿子、寡妇及其女儿的谜题。
If a widow and her daughter take a father and son in marriage, so that the son marries the mother and the father the daughter, what is the relationship of their sons?
解答。
这导致了与前一个谜题类似的情况，两个儿子彼此既是叔伯又是侄子。同样，阿尔琴似乎没有给出解答。如前一个谜题，父亲的儿子是他自己的祖父。 - 父亲及其三个儿子的谜题。
A certain father died and left as an inheritance to his three sons 30 glass flasks, of which 10 were full of wine; another 10 were half full, while the last 10 were empty. Divide the wine and flasks so that an equal share of each should come down to the three sons, both of wine and glass.
解答。
每个儿子必须得到10个玻璃瓶和5瓶量的酒。因此，分配瓶子，使一个儿子得到10个半满的酒瓶，另外两个儿子各得到5个满瓶的酒和5个空瓶。 - 国王军队的谜题。
A king ordered his servant to collect an army from 30 villages as follows: He should bring back from each successive village as many men as he had taken there. The servant went to the first village alone; he went with one other man to the second village; he went with three other men to the third village. How many men were collected from the 30 villages.
解答。
经过第一个村庄后，军队中有2人，经过第二个村庄后有4人，经过第三个村庄后有8人，……。因此经过第n个村庄后，将为军队征集到2n人。经过第30个村庄后，军队中将有230=1,073,741,824人。
[最后几个村庄必须比地球上任何城市都大，才能提供那么多的人！] - 牛的谜题。
How many footprints are left in the last furrow by an ox which has been ploughing all day?
解答。
这是一个带陷阱的问题。犁沟里没有脚印，因为牛后面的犁会把它们抹掉。 - 犁夫的谜题。
How many furrows might a farmer have in his field if the ploughman shall have made three turns at each end of the field?
解答。
第一条犁沟不需要转弯，然后转一次弯后他将有2条犁沟，转2次弯后有3条犁沟，所以转6次弯后有7条犁沟。 - 两个牵牛人的谜题。
Two men were leading oxen along the road when one said to the other: "Give me two oxen, and I'll have as many oxen as you." Then the other said: "If after that you give me two oxen, I'll have twice as many as you." How many oxen there were, and how many did each man have?
解答。
假设第一个人有x头牛，第二个人有y头牛。那么x+2=y−2和2(x+2−2)=y−2+2.将2x=y代入第一个方程得到x=4，所以y=8.像通常一样，阿尔琴只给出答案，然后说明它满足条件。 - 三兄弟和他们的姐妹的谜题。
There were three men, each having a sister, who wanted to cross a river. They found only a small boat in which only two persons could cross at a time. How did they cross the river, so that none of the girls was ever left alone with a man other than her brother either in the boat or on the bank?
解答。
假设M1有姐妹S1，M2有姐妹S2，M3有姐妹S3。开始时所有人都在同一岸，每一后续行表示一次渡河后的位置，船用=标记：M1, M2, M3, S1, S2, S3 =-----
M2, M3, S2, S3 -----= M1, S1
M2, M3, S2, S3, M1 =----- S1
M2, M3, M1 -----= S2, S3, S1
M2, M3, M1, S1 =----- S2, S3
M1, S1 -----= M2, M3, S2, S3
M1, S1, M2, S2 =----- M3, S3
S1, S2 -----= M1, M2, M3, S3
S1, S2, S3 =----- M1, M2, M3
S2 -----= S1, S3, M1, M2, M3
S2, M2 =----- S1, S3, M1, M3
-----= S2, M2, S1, S3, M1, M3
这是阿尔琴的解答，当然他是用文字写出来的，而不是我们使用的符号。它需要11次渡河，但实际上可以用更少的次数完成。看看你能否找到只需要9次渡河的解答。这个谜题以及下面的“渡河”谜题可能是阿尔琴的原创谜题。已知没有更早的版本。 - 狼、山羊和卷心菜的谜题。
A man needed to take a wolf, a goat and a box of cabbage across a river. However, he could only find a boat which would carry two of these at a time. How did he get all of them across unharmed?
[除非他在场，否则他不能把狼和山羊留在一起，或者把山羊和卷心菜留在一起。]
解答。
分别用M、W、G、C表示人、狼、山羊和卷心菜。开始时所有人都在同一岸，每一后续行表示一次渡河后的位置：W, G, C, M =-----
W, C -----= M, G
W, C, M =----- G
C -----= M, W, G
C, G, M =----- W
G -----= M, C, W
G, M =----- C, W
-----= M, G, C, W
- 重人与女人的谜题。
A man and his wife, each the weight of a loaded cart, had two children each the weight of half a cart. They needed to cross a river but the boat they had could only carry the weight of one cart. Find a way of crossing so that the boat should not sink.
解答。
分别用 M、W、C1、C2 表示男人、女人和两个孩子。开始时所有人都在同一岸，每一后续行表示完成一次渡河后的位置：M, W, C1, C2 =-----
M, W -----= C1, C2
M, W, C1 =----- C2
M, C1 -----= W, C2
M, C1, C2 =----- W
M -----= C1, C2, W
M, C1 =----- C2, W
C1 -----= M, C2, W
C1, C2 =----- M, W
-----= C1, C2, M, W
- 渡河谜题。
A man and woman who wished to cross a river. They see two children with a boat on the bank but the boat can only carry one adult or two children. How do the adults cross so that the children end up on the original bank with their boat.
解答。
这本质上与前一题是同一个问题。分别用 M、W、C1、C2 表示男人、女人和两个孩子。开始时所有人都在同一岸，每一后续行表示完成一次渡河后的位置：M, W, C1, C2 =-----
M, W -----= C1, C2
M, W, C1 =----- C2
M, C1 -----= W, C2
M, C1, C2 =----- W
M -----= C1, C2, W
M, C1 =----- C2, W
C1 -----= M, C2, W
C1, C2 =----- M, W
- 田地里羊的谜题。
There is a field which is 200 feet long, 100 feet wide. I want to put sheep in it as follows: Each sheep should have an area five feet long and four feet wide. How many sheep can be put in such a place?
解答。
这是一个相当容易的谜题。我们可以在田地上划出网格，长边方向线间距离为 5 英尺，短边方向线间距离为 4 英尺。这样就把田地分成 40 × 25 = 1000 个所需大小的矩形。因此我们可以把 1000 只羊放进田里。 - 不规则田地的谜题。
There is an irregular field which is 100 yards on each side, 50 metres on one front, 60 yards in the middle, and 50 yards on the other front. How many square yards does this field enclose?
解答。
阿尔琴的田地究竟应该是什么形状并不完全清楚。让我们假设田地的上半部分是一个梯形，其平行边长度为 50 码和 60 码，另外两条边长度为 50 码。这样每边都是 100 码，当然，这些边不是直的。我们可以用毕达哥拉斯算出梯形中平行线之间的距离约为 49.75 米。现在把各部分的面积相加，一个矩形和 4 个直角三角形，得到总面积为 5472.5 平方米。阿尔琴使用近似方法，假设面积与一个31(50+50+60)码乘100码的矩形相同。这给出5533平方码。 - 四边形田地的谜题。
There is a field which is 30 yards on one side, 32 yards on another, 34 metres in the front, and 32 yards on the remaining side. How many square yards are contained in such a field?
解答。
这个问题无法求解，因为给出田地各边的长度并不能确定它。事实上，能够确定的是这块田地可能具有的最大面积。当形状为圆内接四边形时达到这个最大值，此时面积由海伦公式A=√(s−a)(s−b)(s−c)(s−d)给出，其中s=(a+b+c+d)/2和a,b,c,d是各边的长度。这给出（近似）1022 平方米。事实上，阿尔琴使用了一个不正确的面积公式，但它仍然给出了正确答案的良好近似，得到1023平方码。 - 三角形田地的谜题。
There is a field which is 30 yards on one side, 30 yards on another, and 18 metres in the front. How many square yards is the area of such a field?
解答。
这块田地是一个等腰三角形，所以我们可以用毕达哥拉斯算出高为 28.6 码。面积就是 9 乘以 28.6，即 257.4 平方码。阿尔琴计算 9 乘以 30，得到 270 平方码。除非谜题的表述已被讹误，否则阿尔琴的解答看来就是错的。 - 圆形田地的谜题。
There is a round field which contains 400 yards in its circumference. How many square yards will its area be?
解答。
半径为 400/2π = 200/π。于是面积约为 πr2=40000/π=12732 平方码。阿尔琴 得到 10000 平方码。你可能会问，他是怎么得到的？他使用了近似 π=4。这确实显得奇怪。如果他使用了 π=3，或许还能原谅他！ - 狗追兔子的谜题。
There is a field which is 150 feet long. At one end stood a dog, at the other, a hare. The dog chased the hare advancing nine feet per leap, while in the same time the hare made a seven foot leap. How many feet and how many leaps did the dog take in pursuing the fleeing hare until it was caught?
解答。
经过 x 次跳跃后，狗跑了 9x 英尺，兔子跑了 7x 英尺。因此它们之间的距离缩短了 2x 英尺。狗要追上兔子，需要 2x=150，所以 x=75。于是狗在兔子被追上之前跑了 675 英尺，兔子跑了 525 英尺，共 75 次跳跃。这类“追赶”问题在 阿尔琴 之前 2000 年就已在中国出现。 - 四边形城市的谜题。
There is a four-sided city which has one side of 1100 feet, another side of 1000 feet, a front of 600 feet, and a final side of 600 feet. I want to put some houses there so that each house is 40 feet long and 30 feet wide. How many houses ought the city to contain?
解答。
这是一个奇怪的问题。我们将假设城市为等腰梯形来计算其面积。这给出（近似）627808.7 平方英尺的面积。一座房子的面积为 1200 平方英尺，所以田地中有 523 个房子面积。阿尔琴 使用了错误的田地面积公式，得到 63000 平方英尺。他发现这能容纳 525 个房子面积。当然，除了田地面积的错误公式外，这个解答还有另一个问题，即你无法把那么多房子放进田地里。试试看你能放进去多少。 - 三角形城市的谜题。
There is a triangular city which has one side of 100 feet, another side of 100 feet, and a third of 90 feet. Inside of this city, I want to build houses each of which is 20 feet in length and 10 feet in width. How many houses can I build in the city?
解答。
这个问题与上一个类似，阿尔琴 的解答中包含了所有相同的错误。如果我们用 毕达哥拉斯 来计算三角形的高，则为 89.3 英尺，城市的面积算出来是 4018.63 平方英尺。阿尔琴 用他错误的公式得到 4500 平方英尺。现在，他没有像上一个谜题那样用 4500 除以一座房子的面积（200 平方英尺），而是用 4000 除以 200，得到 20 座房子。不清楚他是犯了错误，还是在以某种方式补偿房子放不进去这一事实。我看不出如何能把超过 15 座房子放进这座城市。 - 圆形城市的谜题。
There is a city which is 8000 feet in circumference. How many houses could the city contain if each house is 30 feet long and 20 feet wide?
解答。
这一次，阿尔琴 确实试图考虑房子必须放得进去这一事实。然而，他的论证似乎完全不正确。他得到 6400 座房子，与之前这类谜题不同，现在给出的估计过低了。至少可以轻松地把 8000 座房子放进这座城市，但我们留给读者去做得更好。 - 长方形教堂的谜题。
A basilica is 240 feet long and 120 feet wide. The basilica paved with tiles 23 inches long and 12 inches wide. How many tiles are needed to cover the basilica?
[1 英尺有 12 英寸。]
解答。
由于瓷砖宽 1 英尺，每行需要 120 块瓷砖。长方形教堂长 2880 英寸，2880/23 = 125.2。因此我们需要 126 行瓷砖，每行 120 块。所以总共需要 15120 块瓷砖。阿尔琴 得到了这个正确答案。 - 酒窖之谜。
A wine cellar is 100 feet long and 64 feet wide. How many casks can it hold, given that each cask is seven feet long and four feet wide, and given that there is an aisle four feet wide down the middle of the cellar?
解答。
如果“沿着酒窖的中间”是指两侧各留出30英尺宽的区域，那么我们可以这样做。我们将每侧分成一条14英尺宽的条带和一条16英尺宽的条带，在14英尺条带中按一种方式放置酒桶，在16英尺条带中按另一种方式放置，这样在通道每侧可以容纳2 × 25 + 4 × 14 = 106个：总共212个。
如果“沿着酒窖的中间”是指“在中间某处”有一条过道，并且允许我们一侧有28英尺、另一侧有32英尺，那么我们可以做得更好。
将28英尺的一侧放满4 × 25 = 100个酒桶。在32英尺的一侧放入8 × 12 = 96个酒桶，在一端留下16 × 32英尺的空隙。再放入4 × 4 = 16个酒桶，留下4 × 16英尺的空隙。这足够再放入2个酒桶，总计214个。由于只剩下4 × 2英尺，这已经是最优的了！阿尔琴得到210，显然没有意识到这个非常漂亮的谜题的微妙之处。 - 家长分配粮食之谜。
The head of a household had 20 servants. He ordered them to be given 20 measures of corn as follows. The men must receive three measures, the women must receive two measures, and the children half a measure each. How many men, women and children servants are there in the household?
解答。
假设有x个男人、y个女人和z个小孩。那么x+y+z=20和6x+4y+z=40.用第二个方程减去第一个方程得到5x+3y=20.。因此y能被5整除，又因为它必须小于7（因为3乘以7大于20），所以必须有y = 5。于是x = 1，z = 14，所以家中有1个男人、5个女人和14个孩子作为仆人。与这类问题通常的情况一样，阿尔琴只是给出答案并检验它满足所有条件。 - 另一位家长分配粮食之谜。
A head of household had 30 servants whom he ordered to be given 30 measures of corn as follows. The men should each receive three measures, the women should each receive two measures, and the children should receive a half measure each. How many men, women and children servants are there in the household?
解答。
假设有x个男人、y个女人和z个小孩。那么x+y+z=30和6x+4y+z=60。用第二个方程减去第一个方程得到5x+3y=30。因此y能被5整除，又因为它必须小于10（因为3乘以10等于30，这不允许有正的x），所以必须有y = 5。于是x = 3，z = 22，所以家中有3个男人、5个女人和22个孩子作为仆人。同样，对于这类问题，阿尔琴只是给出答案并检验它满足所有条件。(a) 另一个类似的分配粮食之谜。
A gentleman has a household of 90 persons and ordered that they be given 90 measures of grain. He directed that each man should receive three measures, each woman two measures, and each child half a measure. How many men, women, and children were there?
解答。
假设有x个男人、y个女人和z个小孩。那么x+y+z=90和6x+4y+z=180。用第二个方程减去第一个方程得到 5x+3y=90。因此 y 能被 5 整除，但我们现在有多种可能。若 y = 5，则 x = 15，z = 70。若 y = 10，则 x = 12，z = 68。若 y = 15，则 x = 9，z = 66。若 y = 20，则 x = 6，z = 64。若 y = 25，则 x = 3，z = 62。在 x,y,z 全为正数的情况下，这是仅有的 5 个可能解。因此家中的仆人数目为：3个男人、25个女人和62个小孩，或
6个男人、20个女人和64个小孩，或
9个男人、15个女人和66个小孩，或
12个男人、10个女人和68个小孩，或
15个男人、5个女人和70个小孩。阿尔琴 只给出了这 5 个答案中的一个，即 9 个男人、15 个女人和 66 个小孩。由于他没有给出任何方法，无法看出他是如何得到这个特定答案的，但这个答案恰好是前一个问题的答案的 3 倍，这几乎可以肯定地表明他是如何得到它的。 - 另一个关于分配粮食的谜题。
A head of a household had 100 servants. He ordered that they be given 100 measures of corn as follows. The men should receive three measures, the women should receive two measures, and the children should receive half a measure each. How many men, women, and children servants are there in the household?
解答。
这有点重复了！假设有x个男人、y个女人和z个小孩。那么x+y+z=100和6x+4y+z=200。用第二个方程减去第一个方程得到 5x+3y=100。因此 y 能被 5 整除，但我们现在有多种可能。若 y = 5，则 x = 17，z = 78。若 y = 10，则 x = 14，z = 76。若 y = 15，则 x = 11，z = 74。若 y = 20，则 x = 8，z = 72。若 y = 25，则 x = 5，z = 70。若 y = 30，则 x = 2，z = 68。在 x,y,z 全为正数的情况下，这是仅有的 6 个可能解。因此家中的仆人数目为：2个男人、30个女人和68个孩子，或
5个男人、25个女人和70个孩子，或
8个男人、20个女人和72个孩子，或
11个男人、15个女人和74个孩子，或
14个男人、10个女人和76个孩子，或
17个男人、5个女人和78个孩子。阿尔琴只给出了这6个答案中的一个，即11个男人、15个女人和74个孩子。 - 临终之人的遗嘱谜题。
A certain father died and left behind children, a pregnant wife, and 960 pounds from his estate. However, in his will, he stipulated that if a son should be born to his wife, then the son should receive three quarters of the inheritance. In this case, the mother should get a quarter of his estate. However, if a daughter were born, she should receive seven twelfths of his estate, and in this case the mother would receive five twelfths. But his wife gave birth to twins, one boy and one girl. How much did the mother, son and daughter each receive?
解答。
没有进一步的信息，这个问题无法解决。阿尔琴据以出这道谜题的原始问题可能基于伊斯兰法，或者阿尔琴可能假定读者会依据罗马法来解答。然而，他的解答假定每个人将得到两种可能性的平均值，即：母亲得到960磅的21(41+125)=31，即320磅，
儿子得到960磅的21(43+0)=83，即360磅，
女儿得到960磅的21(127+0)=247，即280磅。大多数历史学家认为阿尔琴不理解罗马继承法！ - 老人向男孩问好的谜题。
An old man greeted a boy as follows "May live for a long time - as long as you have already lived, and then an amount equal to your age at that time, and then three times as much. And if God will grant you one more year than that, and you shall live to be 100." How old was the boy at the time the old man greeted him?
解答。
假设老人向男孩问好时，男孩为x岁。那么2(x+x)乘以3得99。因此4x=33，且x = 8年3个月。阿尔琴给出了答案，并验证它满足谜题的条件。 - 男人建造房屋的谜题。
A man wanted to build a house. He employed six workmen, of whom five were master builders and one was an apprentice. It was agreed between the man who wanted to build the house and the workmen he employed, that a total of 25 pounds should be given to them per day as pay, and that the apprentice should receive half what the master builders receive. How much did each of them receive per day?
解答。
假设每位工匠师傅得到x磅。那么5x+2x=25.这给出11x=50，所以是x=4磅和116磅。这就是每个师傅得到的，而学徒得到2磅和113磅。阿尔琴先分22磅，其中每个师傅得到4磅，而学徒得到2磅。然后他分剩下的3磅，给每个人正确的分数。 - 男人购买100只动物的谜题。
A certain man bought 100 various animals for 100 pounds. The cost was three pounds per horse, one pound per cow, and one pound per 24 sheep. How many horses, cows and sheep were there?
解答。
假设有x匹马、y头牛和z只羊。那么x+y+z=100和3x+y+24z=100。用第二个减去第一个得到2x=(2423)z或48x=23z。因此x能被23整除，并且由第二个方程，不能大到46。因此x=23,z=48和y=29。所以这个人买了23匹马，29头牛，和48只羊。像往常一样，阿尔琴给出答案并检查它可行。 - 东方商人的谜题。
A certain merchant bought 100 assorted animals for 100 pounds on a trip to the Orient. He paid five pounds for each camel, one pound for each ass, and one pound for 20 sheep. How many camels, asses and sheep did the merchant buy?
解答。
假设有x头骆驼、y头驴和z只羊。那么x+y+z=100和5x+y+20z=100。用第二个减去第一个得到4x=(2019)z或80x=19z。因此x能被19整除，并且由第二个方程，不能大于19。因此x=19,z=80,y=1。所以这个人买了19头骆驼，1头驴，和80只羊。像往常一样，阿尔琴给出答案并检查它可行。 - 人与吃草的羊的谜题。
A certain man saw from sheep grazing on the mountainside and said, "I wish I had that number of sheep, and then just as many more, plus a half of half of this number, and then another half of the last amount added. Then if I took these sheep back to my home together with me there would be, counting myself, 100." How many sheep did the man see grazing?
解答。
假设他看到x只羊在山坡上吃草。那么我们得到(x+x)+2x+4x+1=100.由此得到(411)x=99，所以x=36。因此这个人看到山坡上有36只羊在吃草。阿尔琴给出答案36，并验证了它是正确的。 - 母猪与猪圈的谜题。
A certain farmer built a large square enclosure in which he placed a sow. The sow gave birth to seven piglets in the centre of the sty. The offspring, along with the mother, the eighth pig, each gave birth to another seven piglets in the first of the four corners of the sty. Next the sow and all the offspring each give birth to seven more piglets in the second corner. The same happens in the third corner, and then in the fourth corner. Finally the sow and all the offspring each give birth to seven more piglets in the centre of the sty. How many pigs, including the mother, were in the sty by this time?
解答。
我们必须忽略这个谜题中的生物学困难。所有小猪都是雌性的，并且在没有雄性存在的情况下就能生育后代！这个谜题涉及8的幂。母猪在猪圈中央生下7只小猪后，总共有8头猪。这些猪中的每一头都在第一个角落生下7只小猪，所以在这个阶段有64头猪。在每一头都在第二个角落生下一只小猪后，将有83 = 512头猪。然后在第三个角落之后有84头猪，在第四个角落之后有85头猪。同样，每一头都在猪圈中央生下一窝7只小猪，所以在这个最后阶段，猪圈中将有86 = 262144头猪。I·J·古德的工作，那是一个很大的猪圈！阿尔琴正确地解决了这个谜题，但在85阶段似乎犯了一个数值错误，得到32788而不是32768。这不是手稿中简单的抄写错误，因为32788被乘以8以得到最终数字。几份现存的手稿对这个谜题有不同的错误答案，似乎都是由于算术错误。 - 百级台阶的谜题。
There is a ladder which has 100 steps. One pigeon sat on the first step, two pigeons on the second, three pigeons on the third, four on the fourth, five on the fifth, and so on up to the hundredth step. How many pigeons were there in total on the ladder?
解答。
鸽子的总数是前100个自然数之和。前n个自然数之和是n(n+1)/2，所以我们得到答案5050。因此梯子上有5050只鸽子。阿尔琴给出了这个谜题的一个很好的解法。他指出，第1步和第99步总共有100只鸽子。同样，第2步和第98步总共有100只鸽子。继续下去，直到第49步和第51步，它们加起来也是100只鸽子。到这一步，我们已经数了4900只鸽子，但还需要加上第50步的50只鸽子和第100步的100只鸽子。这样总共有5050只鸽子。 - 猪的谜题。
A certain man had 300 pigs. He ordered all of them slaughtered in three days, but with an uneven number being killed each day. He wished the same thing to be done with 30 pigs. What odd number of pigs out of 300 or 30 were to be killed on each of the three days?
解答。
当然这个问题没有解，因为三个奇数之和永远不可能是偶数。阿尔琴知道这个问题无解。他建议这是一个很好的问题，可以给那些调皮的孩子做！ - 男孩问候父亲的谜题。
A certain boy addressed his father, saying, "Hello, father!" His father answered, "Hello, my son. May you live to twice your present age and then at that time three times the age you will then be. If I gave you one of my years to add to this, then you will live to be 100 years old." How old was the boy at the time?
解答。
假设男孩x岁。那么3(2x)+1=100。因此x = 16.5岁。所以男孩16岁零6个月大。阿尔琴给出了16岁6个月的答案，并验证了它是正确的。 - 鸽子的谜题。
A pigeon sitting in a tree saw some other pigeons flying by and said to them, "Suppose there were as many of you again and then the same number were added again. Then, along with me, you would number 100." How many pigeons were there flying by?
解答。
假设有x只鸽子飞过。那么x+x+x+1=100。由此得到x = 33，所以有33只鸽子飞过。阿尔琴给出答案33只鸽子，并验证它成立。 - 被抢劫者的谜题。
A certain man was walking along the street when he found a small bag containing 2 talents. A crowd of people noticed that he had found a purse and said to him: "Friend, give us a part of your discovery." But the man shook his head and said he didn't want to give them any of the money. Those in the crowd rushed at him, emptying the money out of the bag, and each grabbing 50 gold shillings. The man was left with 50 gold shillings after the crowd left. How many men were there in the crowd?
[一塔兰特价值75磅，每磅有72枚金币。]
解答。
一塔兰特是5400枚金币，所以袋中有10800枚金币。如果人群中的每个人拿走50枚金币后还剩50枚金币，那么被拿走的是10750枚金币。人群中必定有10650/50 = 215人。阿尔琴计算出有216份50枚金币。 - 有12个面包的主教的谜题。
A certain bishop ordered 12 loaves of bread to be divided amongst the clergy. He stipulated that each priest should receive two loaves, each deacon should receive half a loaf and each reader should receive a quarter of a loaf. It turned out that the number of clerics and the number of loaves were the same. How many priests, deacons and readers must there have been?
解答。
假设有x名祭司、y名执事和z名诵经士。那么x+y+z=12和2x+2y+4z=12。将第二个方程乘以4，再从中减去第一个方程的两倍，得到6x−z=24.。由于6x>24，x至少为5。但第二个方程表明x不可能大到6（因为那样y和z就必须为0或负数）。因此x=5,y=1且z=6.。所以有5名祭司、1名执事和6名诵经士。 - 遇到学生的男子的谜题。
A certain man met some students and asked them, "How many of you are there in your school?" One of the students replied: "I do not want to tell you directly but I'll tell you how to work it out. You double the number of students, then triple that number, then divide that number into four parts. If you add me to one of the quarters, there will be 100." How many students are in the school?
解答。
假设学校里有x名学生。那么3(2x)/4+1=100.因此x = 66，所以学校里有66名学生。阿尔琴写道：“33的两倍是66。这就是那个数。”然后他验证66满足条件。也许当他写下“33的两倍是66”时，这正是他如何算出答案的线索。 - 木匠的谜题。
Seven carpenters each made seven wheels. How many carts did they build?
解答。
这个谜题假定一辆大车需要4个轮子。那么，由于木匠们制造了49个轮子，他们有足够的轮子制造12辆大车。将会剩下一个轮子。这正是阿尔琴的解答。 - 酒瓶的谜题。
I ask a question to which anyone may reply. How many pints do 100 measures of wine contain, and how many cups do 100 measures contain?
[一量器包含48品脱，一品脱包含6杯。]
解答。
100量器中有4800品脱和28800杯。这个谜题似乎没有什么可算的。 - 分酒壶的谜题。
A certain dying man had four flasks of wine which he wanted to divide between his four sons. In the first flask there were 40 measures of wine; in the second there were 30 measures of wine, in the third there were 20 measures of wine, and in the fourth there were 10 measures of wine. Calling his servant, he said to him, "Divide these four flasks containing wine amongst my four sons in such a way that each son receives an equal portion of wine and flasks." The servant had no means of measuring wine and no container other than the four flasks. How did he carry out the dying man's wishes?
解答。
酒的总量为40 + 30 + 20 + 10 = 100份。因此每个儿子必须得到25份酒。所以仆人必须找到一种方法，在无法测量的情况下，将25份酒装入每个酒壶。他拿起两个酒壶，一个装有40份，另一个装有10份。他将酒从一个壶倒入另一个壶，直到两者酒量相等。现在这两个壶各装有25份。同样，他拿起装有30份和20份的两个酒壶，将酒从一个倒入另一个，直到两者酒量相等。现在这两个壶也各装有25份，问题得以解决。 - 户主之谜。
A certain head of household ordered that 90 measures of grain be taken from one of his houses to another 30 leagues away. This load of grain can be carried by a camel in three trips. Given that the camel eats one measure of grain for each league it goes (the camel only eats when carrying a load), how many measures were left after the grain was transported to the second house?
解答。
这是一个巧妙的谜题。乍一看似乎什么都不会剩下。骆驼每趟能驮30份谷物，而到达第二座房子时它会吃掉全部30份。但运输并非如此进行。你会怎么做呢？ - 修道院长与十二名修士的谜题。
A certain abbot of a monastery was in charge of 12 monks. He asked his treasurer to give an equal share of 204 eggs to each of the monks. The monks consisted of 5 priests, 4 deacons and 3 readers? How many eggs were given to priests? How may to deacons? How many to readers?
解答。
这个问题似乎有些不对劲，因为按照它的提法几乎没有什么可做的。每个修士将得到12204=17个鸡蛋。因此，85个鸡蛋给神父，68个给执事，51个给诵经员。也许我们能给读者出的最好的谜题，是要求一个更有趣的谜题，它能得出给定的答案。