Sam Lloyd, Master of Puzzles

Sam Lloyd was born in Philadelphia in 1841, but lived most of his life in New York. He was primarily a chess player and composer of chess problems, but he also delighted in mathematical puzzling. Actually, not just mathematical puzzles, but word puzzles; picture puzzles; tangrams; he composed thousands of them. In fact, it was one of his pictorial puzzles which grabbed my attention at a very early age, and possibly sparked my enjoyment of such puzzles. Have a look at it here: you will need to print off the picture and cut out the shapes! (You do need to try and solve the puzzle yourself before looking at the solution.)

So here are some of Sam Lloyd’s mathematical puzzles – no easier now than when he first composed them.

Puzzle 1:   A bicycle rider went a mile in three minutes with the wind, and returned in four minutes against the wind. How fast could he ride a mile if there was no wind?

Puzzle 2:   A gypsy queen who ekes out a precarious existence by gathering quarters from gullible victims who want the future revealed, laments the decline of the fortune-telling industry as follows: ‘The week before last I earned less than three dollars, last week only a third as much as the previous week and this week somewhat less than half as much as the week before.’

How much did she earn in three weeks? (Giveaway note: A quarter is a quarter of a dollar.)

Puzzle 3:  An Arab Sheik, finding himself about to die, called his sons about him and said: ‘Divide my camels among you in the proportion of one-half of the herd to the eldest son, the second son one-third, and to the youngest son one-ninth.’
Thereupon the oldest son cried: ‘O, my father, one-half, one-third, and one-ninth do not constitute a whole. To whom, therefore, shall the remainder of the herd be given?’

‘To any poor man who may be standing by when the division is made,’ replied the Sheik, who thereupon died.

When the herd was collected a new difficulty arose. The number of the camels could not be divided either by two or three or nine. While the brothers were disputing, a poor but crafty Bedouin, standing by with his camel, exclaimed, ‘Behold, I will sell you my beast for ten pieces of silver, so that you may then divide the herd.’

Seeing that the addition of one camel would solve the difficulty, the brothers jumped at the offer, and proceeded to divide the herd, but when each had received his allotted portion there yet remained one camel.

‘I am the poor man standing by.’ Said the crafty Bedouin, and gaily mounting the camel, he rode away, with the ten pieces of silver in his turban.

Now, how many camels were in the Sheik’s herd?

(Not actually a very hard puzzle, but such a surprising outcome!)



In Puzzle 1 the obvious, but wrong, answer is 3{\textstyle{1 \over 2}} minutes. We need to look at average speeds. Going into the wind his average speed is 1 \div {\textstyle{3 \over {60}}} = 20 mph, and going with the wind it is 1 \div {\textstyle{4 \over {60}}} = 15 mph. Without any wind his average speed will therefore be 17{\textstyle{1 \over 2}} mph, and his time is therefore 1 \div 17{\textstyle{1 \over 2}} = 0.0571 hours = 3.43 minutes, or about 3 mins 26 secs. The maths behind this is equivalent to the fact that {\textstyle{1 \over 3}} is not halfway between {\textstyle{1 \over 2}} and {\textstyle{1 \over 4}}; the reciprocal function is not a linear one.

The gypsy queen in puzzle 2 only took quarters, so in the first week the amount she took must be less than 3 dollars, it must be divisible by 3, and it has to be a whole number of quarters. The only possible sum is 2.25 dollars. So the second week she took 75c and the third week less than half of that which, still in whole quarters, must be 25c. Total? – $3.25.

The smallest number of camels in puzzle 3 which can be divided by 2, 3 and 9 is 18. Let’s suppose the brothers had 17 camels; when they added 1 more they could then divide the herd up. 18 ÷ 2 = 9, 18 ÷ 3 = 6, 18 ÷9 = 2. 9 + 6 + 2 = 17, leaving one camel over. Incidentally, there’s a bit of a feel in this puzzle of “perfect numbers” which are numbers whose proper divisors (all its factors except itself) sum to give the number itself. For example, the proper divisors of 6 are 1, 2 and 3; and 1 + 2 + 3 = 6. Can you find the next perfect number? It’s less than 30.

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