IMO 1975 Problem B1 Let A be the sum of the decimal digits of 4444 4444 , and B be the sum of the decimal digits of A. Find the sum of the decimal digits of B.
Let X = 4444 4444 .Then X has less than 4.4444 = 17776 digits, so A is at most 9.17776 = 159984. Hence B is at most 6.9 = 54. But all these numbers are congruent mod 9. 4444 = -2 (mod 9), so X = (-2) 4444 (mod 9). But (-2) 3 = 1 (mod 9), and 4444 = 1 (mod 3), so X = -2 = 7 (mod 9). But any number less than 55 and congruent to 7 has digit sum 7 (possibilities are 7, 16, 25, 34, 43, 52). Hence the answer is 7. s are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48- X. 17th IMO 1975 © John Scholes jscholes@kalva.demon.co.uk 11 Oct 1998