# Contributions from Donald Eperson

We present six brief extracts from Donald Eperson's many contributions to

*Mathematics in School*and*The Mathematical Gazette*. We present them in chronological order.- Contained in a letter to
*The Mathematical Gazette***52**(381) (1968), 280-281:

May I invite the readers of the Gazette to let me know whether or not they consider that it is important for teachers to become aware of the beauty of mathematics, as distinct from its obvious utilitarian values (so that the children can find aesthetic satisfaction in their mathematical activities) and also to know by experience the satisfaction of "discovering" elementary mathematical relations for themselves (so that they are encouraged to provide similar experiences for those they are teaching)?

- Tests for Divisibility by 7,
*Mathematics in School***2**(2) (1973), 30-31.

To test a number, cross off the final figure, and subtract twice that figure from the remaining number: repeat the process as often as necessary, until you reach zero or a number that you recognize as a multiple of 7, in which case the original number was exactly divisible by 7. If however you reach a number that is not a multiple of 7, then the original number was not a multiple of 7.

17654 → 1765 - 8 = 1757 → 175 - 14 = 161 → 16 - 2 = 14.

- Puzzles, Pastimes and Problems,
*Mathematics in School***24**(3) (1995), 18, 32-33.

Proof by mathematical induction

I can still recall the pleasure I experienced when at school over 70 years ago I first met the method of proof by induction in Baker and Bourne's book of*Examples in Algebra*, where the first two problems occurred.

(a) Show that if $n$ is even, then $n^{2} + 2n$ is divisible by 8, and

(b) Show that $3^{4n} - 1$ is divisible by 80.

Can you prove these statements to be true?

Subsequently I discovered a number of similar relations about powers of integers that can be proved by induction or by simpler methods.

(c) $2^{n} + 2^{n+1}$ is always a multiple of 6.

(d) $4^{n} + 2$ is always a multiple of 6.

(e) $7^{n} + 5$ is always a multiple of 6.

(f) $5^{n} + 3$ is always a multiple of 4.

(g) $3^{n} + 3^{n+1} + 3^{n+2}$ is always a multiple of 13.

(i) $6^{n} + 4$ is always a multiple of 10.

(j) $13^{n} - 2 \times 7^{n} + 1$ is always a multiple of 12.

Can you prove these to be true? Can you create further, similar examples of your own?

- Memories of a mathematician,
*Mathematics in School***25**(5) (1996), 6-7.

In 1961, I enquired into the mathematical courses provided at over 120 Training Colleges for Teachers. The result of this research was that very few colleges appeared to make an effort in assessing the capability of students to teach mathematics. In consequence, many obtained the Teacher's Certificate, but dreaded the prospect of having to teach arithmetic in Primary schools. The head teachers of 101 Primary and Secondary were invited to express their opinions of the ability to teach mathematics shown by teachers in their probationary year. The majority expressed deep concern at the poor quality of teaching mathematics. Heads of Secondary Schools looked for better-trained all-subject teachers, and reported a dearth of 'specialists'. Many girls were allowed to drop mathematical studies as soon as they showed symptoms of 'mathophobia' when teenagers.

- Puzzles, Pastimes and Problems,
*Mathematics in School***25**(2) (1996), 19, 45.

Can you explain why some of these statements are true and at least one is false?

(i) The difference between the squares of any pair of consecutive odd numbers is always a multiple of 8.

(ii) The difference between the cubes of any pair of consecutive numbers is always a prime number.

(iii) All cubic numbers that are not multiples of 4 are either one more or one less than a multiple of 4.

(iv) All cubic numbers that are not multiples of 7 are either one more or one less than a multiple of 7.

(v) The sum of $n$ terms of the sequence of cubic numbers, 1, 8, 27, 64, 125 .... is always a square number.

- Puzzles, Pastimes and Problems,
*Mathematics in School***27**(3) (1998), 25.

Ask a friend to choose a number with two digits, which you will endeavour to discover if he/she tells you the answer to some simple calculations.

(1) Add together the two digits, and subtract 1 from their sum.

(2) Multiply the result by 11.

(3) Subtract the chosen number.

To find the chosen number, you add 11 to the answer, which will provide the chosen number's digits in reverse order. For example, if the answer is 47, add 11 to get 58, and the chosen number was 85. If the answer is a negative number, say -7, then -7 + 11 = 04, and the chosen number was 40. Can you explain how this prestidigitation works?

Last Updated May 2017