Pick a number that ends with 1, 3, 7, or 9. Will it evenly divide a number consisting entirely of ones (a repunit)? Answer: yes. Proof: Suppose 239 was chosen. Take the remainder of 239 dividing 10, 100, 1000, ..., . The chosen number will not divide evenly into any of those 239 powers of 10, so there are 238 possible remainders, 1 to 238. By the pigeonhole principle, two remainders must be the same, for some and . As it turns out, and both give remainder 44. Subtracting, 9,999,999,000 is the result, which yields 1,111,111 when divided by 9000. When , always returns an all-1 number multiplied by 9 and some power of 10, finishing the proof. The reciprocal of the chosen number has a repeating decimal of similar length. Consider: . . ….