1000, ?, 100, 31, 24, 22, 20, 17, 16, 15, 14, 13, 12, 11, 10

The solution to this problem is that all the numbers are the decimal number 16 in bases ranging from base 2 (binary) to base 16 (hexadecimal).

The first number, 10000, is in base 2, corresponding to 16 in decimal or base ten: with components one 16 and no 8's, 4's, 2's or 1's,

Therefore the second number should be in base 3, with components one 9, two 3's and one unit. The missing number is 121.

Solutions were received from Alex Skilton, age 15, The King's School, Canterbury, UK, Soh Yong Sheng, and Ling Xiang Ning of Raffles Institution, Singapore, and David Lowe, age 15, Trinity School, Carlisle, UK.