1st Centromerican 1999 Problem A2 Find a positive integer n with 1000 digits, none 0, such that we can group the digits into 500 pairs so that the sum of the products of the numbers in each pair divides n.
11...1 2112 2112 ... 2112 (960 1s followed by 10 2112s) Solution Suppose we take 980 digits to be 1 and 20 digits to be 2. Then we can take 8 pairs (2,2), 4 pairs (2,1) and 488 pairs (1,1) giving a total of 528 = 16·3·11. The sum of the digits is 1020 which is divisible by 3, so n is certainly divisible by 3. We can arrange that half the 1s and half the 2s are in odd positions, which will ensure that n is divisible by 11. Finally, n will be divisible by 16 if the number formed by its last 4 digits is divisible by 16, so we take the last 4 digits to be 2112 (=16·132). So, for example, we can take n to be 11...1 2112 2112 ... 2112, where we have 960 1s followed by 10 2112s. 1st OMCC 1999 © John Scholes jscholes@kalva.demon.co.uk 26 November 2003 Last corrected/updated 26 Nov 03