The Merchant's Puzzle
(
CANTERBURY PUZZLES)
Of the Merchant the poet writes, "Forsooth he was a worthy man withal." He was thoughtful, full of schemes, and a good manipulator of figures. "His reasons spake he eke full solemnly. Sounding away the increase of his winning." One morning, when they were on the road, the Knight and the Squire, who were riding beside him, reminded the Merchant that he had not yet propounded the puzzle that he owed the company. He thereupon said, "Be it so? Here then is a riddle in numbers that I will set before this merry company when next we do make a halt. There be thirty of us in all riding over the common this morn. Truly we may ride one and one, in what they do call the single file, or two and two, or three and three, or five and five, or six and six, or ten and ten, or fifteen and fifteen, or all thirty in a row. In no other way may we ride so that there be no lack of equal numbers in the rows. Now, a party of pilgrims were able thus to ride in as many as sixty-four different ways. Prithee tell me how many there must perforce have been in the company." The Merchant clearly required the smallest number of persons that could so ride in the sixty-four ways.
Answer:
This puzzle amounts to finding the smallest possible number that has exactly sixty-four divisors, counting 1 and the number itself as divisors. The least number is 7,560. The pilgrims might, therefore, have ridden in single file, two and two, three and three, four and four, and so on, in exactly sixty-four different ways, the last manner being in a single row of 7,560.
The Merchant was careful to say that they were going over a common, and not to mention its size, for it certainly would not be possible along an ordinary road!
To find how many different numbers will divide a given number, N, let N = ap bq cr ..., where a, b, c ... are prime numbers. Then the number of divisors will be (p + 1) (q + 1) (r + 1) ..., which includes as divisors 1 and N itself. Thus in the case of my puzzle—
7,560 = |
23 |
× |
33 |
× |
5 |
× |
7 |
Powers = |
3 |
|
3 |
|
1 |
|
1 |
|
Therefore |
4 |
× |
4 |
× |
2 |
× |
2 |
= 64 divisors. |
To find the smallest number that has a given number of divisors we must proceed by trial. But it is important sometimes to note whether or not the condition is that there shall be a given number of divisors and no more. For example, the smallest number that has seven divisors and no more is 64, while 24 has eight divisors, and might equally fulfil the conditions. The stipulation as to "no more" was not necessary in the case of my puzzle, for no smaller number has more than sixty-four divisors.
Informational
