: 12. A real number may have infinitely many digits. The strategy therefore has to be to exhibit some rational number which requires N to get a 2, and then to show that for any real number some special digit, such as the first, is a 2 for some multiple <= N. A little experimentation reveals that for most integers fairly small multiples suffice. We notice that 15 requires 8, but most seem to need less. If x has first digit 7, 8 or 9, then 3x has first digit 2. If x has first digit 5 or 6, then 4x has first digit 2. If x has first digit 4, then 5x has first digit 2. If x has first digit 3, then 7x has first digit 2. Obviously, if x has first digit 2, then 1x has first digit 2. So we need only consider numbers with first digit 1. If x has second digit 2, we are home. If x has second digit 0 (including of course the case x has no second digit), 1, 3 or 4, then 2x has first digit 2. If x has second digit 5, then 8x has second digit 2. It is convenient to notice that x and 10x behave the same way, so we can restrict x to be between 1 and 10. The analysis above shows that we need only consider 1.6 ≤ x < 1.9. If 5/3 ≤ x < 2, then 20 ≤ 12x < 30. We might also look at 5/3. Its multiples are 1.66... , 3.33... , 5, 6.66... , 8.33... , 10, 11.66... , 13.33... , 15, 16.66... , 18.33... , 20. So it requires N = 12. So we are left with 1.6 ≤ x < 5/3. The first digit is not 2 for N up to 12. But we notice that 2x has a 2 for 1.6 ≤ x < 1.65, and for 1.66 ≤ x < 1.665 and for 1.666 ≤ x ≤ 1.6665 and so on. Also 5x has a 2 for 1.64 ≤ x < 1.66 and for 1.664 ≤ x < 1.666 and so on. Thus 2x or 5x has a 2 for any x in the range 1.6 ≤ x < 5/3. Note that there are other x which require N = 12, such as 1.6795 and 1.6835. 27th BMO 1991 © John Scholes jscholes@kalva.demon.co.uk 15 Aug 2002
27th BMO 1991 Problem 4 Let N be the smallest positive integer such that at least one of the numbers x, 2x, 3x, ... , Nx has a digit 2 for every real number x. Find N. Failing that, find upper and lower bounds and show that the upper bound does not exceed 20. Solution