July 2001 solution

Solution to July 2001 Problem

We present two solutions; the first is based answer of Juan Mir Pieras (Spain). (His entire solution is on our Spanish solutions page.)

Ripley could not have been correct -- there are no persistent numbers, big or small! In fact, every integer divides some number whose digits consist of 9's and 0's (in the form 99...900...0). We shall see how this number arises from the standard method for converting a repeating decimal to a fraction: We start with an arbitrary integer n and find a multiple consisting of only 9's and 0's which, of course, implies that n could not be persistent.

We use the fact that 1/n is a repeating decimal for any whole number n. Suppose the repeating part starts after the kth place and has period p; in mathematical notation this amounts to

where m is a whole number. When n = 220 for example, 1/220 = 0.00454545..., so here k = 2 and p = 2:

a whole number as promised. Multiplying both sides by 220 we get on the left 10⁴ - 10² = 10000 - 100 = 9900; that equals (on the right) 45*220 - thus, a multiple of 220 is in the desired form. Returning to the general case we see that (10^k+p - 10^k)(1/n) = m, so that mn = 999...9*10^k, where the 9 is repeated p times.

Remarks on N.
In fact, it is easy to show directly that Ripley's number is not persistent. It is sufficient to multiply it by only 19:

19*526315789473684210 = 9999999999999999990. Mir deduced this from noting that 1/19 = 0.052631578947368421052631578947368421... The 18 digits of N form the repeating part of 1/19. More precisely, N = (1020 - 101) / (20 - 1). Perhaps what Ripley had in mind is that any multiple of N that is relatively prime to 19 will contain a cyclic permutation of these digits! That property follows from the fact that for any prime p, when 1/p repeats with period pÐ1 (which is the maximum possible period), then the repeating part of n/p is a cyclic permutation of 1/p. (You might wish to try this first for 1/7 = 0.1428571428571..., whose period is 7 - 1 = 6.) Here are the multiples of 1/19:   1/19 = 0. 052631578947368421052631578947368421...
  2/19 = 0. 105263157894736842105263157894736842...
  3/19 = 0. 157894736842105263157894736842105263...
  4/19 = 0. 210526315789473684210526315789473684...
  5/19 = 0. 263157894736842105263157894736842105...
  6/19 = 0. 315789473684210526315789473684210526...
  7/19 = 0. 368421052631578947368421052631578947...
  8/19 = 0. 421052631578947368421052631578947368...
  9/19 = 0. 473684210526315789473684210526315789...
10/19 = 0. 526315789473684210526315789473684210...
11/19 = 0. 578947368421052631578947368421052631...
12/19 = 0. 631578947368421052631578947368421052...
13/19 = 0. 684210526315789473684210526315789473...
14/19 = 0. 736842105263157894736842105263157894...
15/19 = 0. 789473684210526315789473684210526315...
16/19 = 0. 842105263157894736842105263157894736...
17/19 = 0. 894736842105263157894736842105263157...
18/19 = 0. 947368421052631578947368421052631578...
19/19 = 1. 000000000000000000000000000000000000 Note that there is a nice relationship between the multiplier of 1/19 and the number of decimal places that the repeating part gets displaced. (For example, 2/19 is displaced by 1 position, while 3/19 moves over 13 places. For details and further observations, see the Spanish solution.)

Alternative Solution.
For any number n, consider the remainders obtained by dividing it into the following n numbers: 1, 11, 111, ... , 111...1, where the last number consists of n 1's. Since at most nÐ1 nonzero remainders can result, we either have one of the above numbers divisible by n, in which case n is not persistent, or at least two of them, say R = 111...1 (r ones) and S = 111...1 (s ones), s > r, give the same remainder. In the latter case, their difference S - R = 11...100...0 (s-r ones and r zeros) is divisible by n and, therefore, is not persistent.

Further thoughts.
Mir wondered about persistent irrational numbers. We could define an irrational number to be persistent if the decimal representation of each of its integer multiples contains all ten decimal digits. As a consequence of the original problem, every digit would have to appear infinitely often. This question is closely connected with the so-called normal numbers Ñ irrational numbers whose decimal expansion contains every finite block of integers infinitely often. Such a decimal number can be constructed by a random process: simply chose the next digit by rolling a 10-sided die (or an icosahedron whose faces are labeled with the ten digits so that opposite faces have the same digit). It is still unknown whether naturally occurring irrationals such as e, pi, or the square root of 2 are normal or not.