Problem 571

Consider three three-digit numbers with digits $RST$, $UVW$ and $XYZ$.

The integer with digits $RST$ equals $100R+10S+T$, the integer with digits $UVW$ equals $100U+10V+W$, and the integer with digits $XYZ$ equals $100X+10Y+Z$.

Therefore, $$\begin{aligned} RST+UVW+XYZ&=100R+10S+T+100U+10V+W+100X+10Y+Z\\ &=100(R+U+X)+10(S+V+Y)+(T+W+Z)\end{aligned}$$ We note that each of $R,S,T,U,V,W,X,Y,Z$ can be any digit from 0 to 9, except that $R$, $U$ and $X$ cannot be 0.

Max wants to make $100(R+U+X)+10(S+V+Y)+(T+W+Z)$ as large as possible. He does this by placing the largest digits (9, 8 and 7) as hundreds digits, the next largest digits (6, 5 and 4) as tens digits, and the next largest digits (3, 2 and 1) as units digits. We note that no digits can be repeated, and that the placement of the digits assigned to any of the place values among the three different three-digit numbers is irrelevant as it does not affect the actual sum.

Max’s sum is thus $100(9+8+7)+10(6+5+4)+(3+2+1)=2400+150+6=2556$.

Minnie wants to make $100(R+U+X)+10(S+V+Y)+(T+W+Z)$ as small as possible. She does this by placing the smallest allowable digits (1, 2 and 3) as hundreds digits, the next smallest remaining digits (0, 4 and 5) as tens digits, and the next smallest digits (6, 7 and 8) as units digits.

Minnie’s sum is thus $100(1+2+3)+10(0+4+5)+(6+7+8)=600+90+21=711$.

The difference between their sums is $2556-711=1845$.