Answer

Any rational number can be expressed as a repeating decimal. The length of the repeating segment is called the period of the decimal. So 1/7 = 0.142857142857... has period 6, while 1/11 = 0.9090909... has period 2.

What prime numbers have reciprocals with period five or less? This problem can be tackled crudely with a computer, but a technology-free solution is also possible.

Answer:
We first note that 2 and 5 are the only prime numbers with terminating decimal expansions.

If p is prime and 1/p has period k, then

1/p = 10^-kA + 10^-2kA + ...

where A is a k-digit number. So:

1/p = 10^-kA (1 + 10^-k + (10^-k)² + ...)
= 10^-k A / (1-10^-k) (summing the geometric series)
= A / (10^k - 1)

Therefore pA = 10^k-1, and hence p is a prime divisor of 10^k-1.

• if k = 1, p is a divisor of 9 = 3²
• if k = 2, p is a divisor of 99 = 3²× 11
• if k = 3, p is a divisor of 999 = 3³ × 37
• if k = 4, p is a divisor of 9999 = 3² × 11 × 101
• if k = 5, p is a divisor of 99999 = 3² × 41 × 271

Thus the only possible values of p which give a reciprocal with period of five or less are 3, 11, 37, 41, 101 and 271.

• Rainer Hoft of Groote Schuur High School wins the R20 prime voucher.
  Sorry, prize voucher...