RMO 2009

Let X denote the set of all 3-digit natural numbers; let O be those numbers in X having only odd digits; and E be those numbers in X having only even digits. Then X - (O ∪ E) is the set of all 3-digit natural numbers having at least one odd digit and at least one even digit.The desired sum is therefore $\displaystyle\sum_{x\in X}x$ - $\displaystyle\sum_{y\in O}y$ - $\displaystyle\sum_{z\in E}z$ .

It is easy to find the first sum:

$\displaystyle\sum_{x\in X}x$ = $\displaystyle\sum_{j=1}^{999} j$ - $\displaystyle\sum_{k=1}^{99} k$ = $\frac{999\times 1000}{2}-\frac{99\times 100}{2}$ = $50\times 9891$ =494550.

Consider the set O. Each number in O has its digits from the set {1, 3, 5, 7, 9}. Suppose the digit in unit’s place is 1. We can fill the digit in ten’s place in 5 ways and the digit in hundred’s place in 5 ways. Thus there are 25 numbers in the set O each of which has 1 in its unit’s place. Similarly, there are 25 numbers whose digit in unit’s place is 3; 25 having its digit in unit’s place as 5; 25 with 7 and 25 with 9. Thus the sum of the digits in unit’s place of all the numbers in O is 25(1 + 3 + 5 + 7 + 9) = 25 × 25 = 625. A similar argument shows that the sum of digits in ten’s place of all the numbers in O is 625 and that in hundred’s place is also 625. Thus the sum of all the numbers in O is 625(100 + 10 + 1) = 625 × 111 = 69375.

Consider the set E. The digits of numbers in E are from the set {0, 2, 4, 6, 8}, but the digit in hundred’s place is never 0. Suppose the digit in unit’s place is 0. There are 4 × 5 = 20 such numbers. Similarly, 20 numbers each having digits 2,4,6,8 in their unit’s place. Thus the sum of the digits in unit’s place of all the numbers in E is 20(0 + 2 + 4 + 6 + 8) = 20 × 20 = 400. A similar reasoning shows that the sum of the digits in ten’s place of all the numbers in E is 400, but the sum of the digits in hundred’s place of all the numbers in E is 25 × 20 = 500. Thus the sum of all the numbers in E is 500 × 102 + 400 × 10 + 400 = 54400. The required sum is 494550 − 69375 − 54400 = $\boxed{370775}$