5. All six digits of three 2-digit numbers are different. What is the largest possible sum of three such numbers?
A 237
B 246
C 255
D 264
E 273
Solution: $\mathbf{C}$
Suppose that the six digits that are used are $a, b, c, d, e$ and $f$ and that we use them to make the three 2-digit numbers ' $a d$ ', ' $b e$ ' and ' $c f$ '. The sum of these numbers is $(10 a+d)+(10 b+e)+(10 c+f)$, that is $10(a+b+c)+(d+e+f)$. To make this as large as possible, we need to choose $a, b$ and $c$ so that $a+b+c$ is as large as possible, and then $d, e$ and $f$ from the remaining digits so that $d+e+f$ is large as possible. Clearly, this is achieved by choosing $a, b, c$ to be $7,8,9$, in some order, and then $d$, $e, f$ to be $4,5,6$ in some order. So the largest possible sum is $10(7+8+9)+(4+5+6)=255$. [For example, the three 2-digit numbers could be 74,85 and 96.]