Question 9

By observation, you can see that the average of all three digits that can be used with the digits $n-1,$ $n,$ and $n+1$ for $n$ an integer between $2$ and $8,$ inclusive is $\overline{nnn}.$ Therefore, the average of the six numbers Sarah lists is $222,$ and the average of the six numbers Chloe lists is $666.$ These two averages are weighted equally, so the average of all twelve numbers is $\dfrac{222+666}{2}=\dfrac{888}{2}=\boxed{444}$

Using 3 digits once only numbers formed are 6

and sum of these 6 numbers = 2 × (sum of digits) × 111

sum of 6 numbers using 1, 2, 3 = 2 × 6 × 111 = 12 × 111

sum of 6 numbers using 5, 6, 7 = 2 × 18 × 111 = 36 × 111

Average of 12 numbers = (12 + 36) × 111/12 = 4 × 111 = 444

( 1+ 2+ 3+ 5+6+7) 111 4/(6*6) = 444