9th Iberoamerican 1994

Show that there is a number 1 < b < 1993 such that if 1994 is written in base b then all its digits are the same. Show that there is no number 1 < b < 1992 such that if 1993 is written in base b then all its digits are the same.

Solution

Any even number 2n can be written as 22 in base n-1. In particular 1994 = 22₉₉₆.

We have to show that we cannot write 1993 = aaa ... a_b. If the number has n digits, then 1993 = a(1 + b + ... + b^n-1) = a(bⁿ - 1)/(b - 1). But 1993 is prime, so a must be 1. Hence b^n-1 + ... + b - 1992 = 0. So b must divide 1992 = 2³3.83. We cannot have n = 2, for then b = 1992 and we require b < 1992. So n > 2. But 83² = 6889 > 1993, so b must divide 24. Hence b = 2, 3, 4, 6, 8, 12, or 24. But we can easily check that none of these work:

1 + 2 + 22 + ... + 29 = 1023, 1 + ... + 210 = 2047.

1 + 3 + ... + 36 = 1093, 1 + ... + 3^7 = 3280

1 + 4 + ... + 45 = 1365, 1 + ... + 46 = 5461

1 + 6 + ... 64 = 1555, 1 + ... + 65 = 9331

1 + 8 + 82 + 83 = 585, 1 + ... + 84 = 4681

1 + 12 + 122 + 123 = 1885, 1 + ... + 124 = 22621

1 + 24 + 242 = 601, 1 + ... + 243 = 14425



 









 


9th Ibero 1994


© John Scholes


jscholes@kalva.demon.co.uk


11 July 2002