2nd ASU 1962 problems Problem 9 Given a number with 1998 digits which is divisible by 9. Let x be the sum of its digits, let y be the sum of the digits of x, and z the sum of the digits of y. Find z.
x ≤ 9·1998 = 17982. Hence y ≤ the greater of 1+7+9+9+9=35 and 9+9+9+9=36. But 9 divides the original number and hence also x, y and z. Hence z=9. 2nd ASU 1962 © John Scholes jscholes@kalva.demon.co.uk 15 Sep 1998