# Puzzlist vs. Mathematician:A Practical Arithmetic Problem

Very much the same problem has been discussed in two disparate books. Morris Kline in Mathematics and the Physical World (Dover, 1981) cites the problem as the first example in which "mathematical reasoning can yield desirable knowledge." In Mathematical Puzzles of Sam Loyd (M. Gardner, Dover, 1959) it comes as the puzzle #84. The problem gets different treatment in the two books.

### Problem (Kline)

Suppose that a young man has a choice between two jobs. Each offers a starting salary of \$1800 per year, but the first one would lead to an annual raise of \$200 whereas the second would lead to a semiannual raise of \$50. Which job is preferable?

Before you read further, try to tackle the problem by yourself.

One would think that the answer is obvious. A raise of \$200 per year seems better than one that apparently would amount to only \$100 per year. But let us do a little arithmetic and put down what each job offers during successive six-month periods.

The first job will pay

900, 900, 1000, 1000, 1100, 1100, 1200, 1200 ...

The second job, which bears a semiannual increase of \$50, will pay

900, 950, 1000, 1050, 1100, 1150, 1200, 1250 ...

It is clear from a comparison of these two sets of salaries that the second job brings a better return during the second half of each year and does as well as the first job during the first half. The second job is the better one. With the arithmetic before us it is possible to see more readily why the second job is better. The semiannual increase of \$50 means that the salary will be higher at the rate of \$50 for six months or at the rate of \$100 for the year because the recipient will get \$50 more for each of the six-month periods. Hence two such increases per year amount to an increase at the rate of \$200 per year. Thus far the two jobs seem to be equally good. But on the second job the increases start after the first six months, whereas on the first job they do not start until one year has elapsed. Hence the second job will pay more during the latter six months of each year.

### Problem (Loyd)

Here is a problem from the ordinary affairs of life which is as interesting as it is puzzling to all who tackle it. The "Boss" was feeling pretty good the other day, so he said to his stenographer:

"Now, Mary, in view of the fact that you never indulge in useless vacations, I have determined to raise your salary \$100 every year. Beginning from today, for the ensuing year you will be paid weekly at the rate of \$600 a year; next year at the rate of \$700, the next at \$800, and so on, always increasing \$100 per year."

"On account of my weak heart," replied the grateful young woman, "I suggest that it would be safer to make the change less abrupt. Start the salary from today on the basis of \$600 a year, as suggested, but at the end of six months raise the yearly salary \$25, and continue to give me a \$25 yearly raise every six months, so long as my services are satisfactory."

The boss smiled benignly upon his faithful employee as he accepted the amendment, but a twinkle in his eye set some of the boys to figuring whether or not the boss made a wise move by accepting her proposition. Can you tell?

Loyd, the great puzzlist and entrepreneur, excelled in catching the public fancy (M.Gardner, The Second Scientific American Book of Mathematical puzzles and Diversions, Dover, 1987, p. 33.) He dressed up the problem as a nice story about 4 times longer than Kline's dry formulation. But styles aside, have you noticed a more significant difference between their presentations?

Here's the solution from Loyd's book:

In the puzzle of the young stenographer's salary, she gains \$12.50 the first year, but after that loses steadily. Some puzzlists fall into the error of adding the whole of each raise in a lump sum at the end of every six months, whereas the salary was raised each time to a yearly basis of \$25 better, which is only an improvement of \$12.50 every six months. Of course a raise of \$100 per year would give the clerk in five years, \$600 plus \$700 plus \$800 plus \$900 plus \$1,000, equaling \$4,000. Instead of which the clerk loses \$437.50 by her own plan, as follows:

 Yearly basis First six months \$300.00 \$600 Second six months 312.50 625 Third six months 325.00 650 Fourth six months 337.50 675 Fifth six months 350.00 700 Sixth six months 362.50 725 Seventh six months 375.00 750 Eighth six months 387.50 775 Ninth six months 400.00 800 Tenth six months 412.50 825

The second sentence (Some puzzlists fall into the error...) of Loyd's solution gives a clear indication that he was fully aware of Kline's variant. Why did he chose the other variant - much less entertaining, nay, plain insipid in my view?

The difference is that Kline offered a nice simple problem with an unexpected answer. Loyd's formulation is more like a trap based on possible misreading of the problem. I can't answer for Loyd. However, I feel on a pretty firm foundation answering for Kline. I am confident that neither Kline nor any other mathematician would take interest in Loyd's formulation. For the mathematics involved is on the level of early grades. There is nothing particularly interesting in adding the same number a few times over. The gist of Kline's quote is that, in mathematics, even with so trivial means it is possible to achieve a measure of surprise.

Kline's formulation carries an implicit message. Although it's not open-ended in the sense used in math education nowadays, it prompts one to wonder, Where did I go wrong?, Where has my intuition misled me? The result may be a desire to investigate other quantities or other proportions between suggested annual and semiannual raises. What is the border line proportion that would equate two accounting practices? And even in that case, would I rather have an annual or a semiannual raise and why?

Since writing this piece I have discovered a 3 page discussion in Mathematical Fallacies, Flaws, and Flimflam by Edward J. Barbeau (MAA, 2000). The problem has a respectable bibliography. It appeared in

1. March 15, 1992 column Ask Marilyn in Parade magazine,
2. The Chronicle of Higher Education, "In Box", p. 15, July 27, 1983
3. The Chronicle of Higher Education, "Letters to the editor", September 14, 1983
4. W.W. Sawyer, Mathematician's Delight, 1964, pp 68-69,
5. W.W. Rouse Ball, Mathematical Recreations and Essays, 1896, pp 26-27,
6. S. Loyd, Cyclopedia of Puzzles, 1914, pp 312 & 381,
7. R. Ripley, Believe It or Not: Book 2, Simon & Schuster, 1931, p 123.
8. The Tutorial Arithmetic, 1902, p425 (A straightforward exercise #16).

Needless to say, aside from Loyd's Cyclopedia, all other references follow Kline's variant. E.g., Ripley invites readers to compare a \$1/day with a \$35/week raise.

The earliest reference here is that by Rouse Ball - 1896. However, Loyd's Cyclopedia was published by his son - S. Loyd Jr. - three years after Loyd's death. It was a huge collection of puzzles Loyd produced during his long and distinguished career. So it is quite possible that the problem has originated with Loyd - an extra surprise.

### Other Puzzles by Sam Loyd

1. The Gordian Knot
2. Fifteen
3. Farmer and Wife To Catch Rooster and Hen
4. Puzzlist vs. Mathematician: A Practical Arithmetic Problem
5. Can you better Sam Loyd?
6. Sam Loyd's Geometric Puzzle

Copyright © 1996-2018 Alexander Bogomolny

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