5. (10 points) On the blackboard, there is a sequence of numbers: $1,2,3, \cdots, 2011$. If every time the first 16 numbers are erased, and the sum of the erased 16 numbers is written at the end of the sequence, until only one number remains, then:
(1) What is the last remaining number?
(2) What is the total sum of all numbers that have appeared on the blackboard?
【Analysis】(1) Each operation does not affect the total sum of all numbers on the blackboard, so the final remaining sum $=1+2+3+\cdots$ +2011, which can be completed using the Gauss summation formula.
(2) Since the second-to-last operation, there are 16 numbers on the blackboard, with a total sum of 2023066. These 16 numbers come from $16 \times$ $16=256$ numbers, and the sum of these 256 numbers is the same as above. $2011-(16-1) x=256, x=117$ times, clearly, from the beginning, it only takes 117 operations for the blackboard to have 256 numbers left. According to the rules, we can analyze accordingly. (1) Originally, there are 2011 numbers, with a sum of 2023066.
(2) After 117 operations, 256 numbers remain on the blackboard: 1873 to 2011, and 117 new sums appear. These 117 sums $=2023066$ - $(1873+2011) * 139 / 2=1753128$
(3) After 16 operations, 16 new numbers remain on the blackboard, with a sum of $=2023066$
(4) After 1 operation, 1 number remains on the blackboard $=2023066$;
In summary, all the numbers that have appeared $=2023066+1753128+2023066+2023066=7822326$
【Solution】Solution: (1) $1+2+3+\cdots+2011$
$=(1+2011) \times 2011 \div 2$
$=2012 \times 2011 \div 2$
$=2023066$
Answer: The final remaining number is 2023066.
(2) Since the second-to-last operation, there are 16 numbers on the blackboard, with a total sum of 2023066. These 16 numbers come from $16 \times$ $16=256$ numbers, and the sum of these 256 numbers is the same as above.
2011 - (16 - 1) $x=256, x=117$ times,
Clearly, from the beginning, it only takes 117 operations for the blackboard to have 256 numbers left.
(1) Originally, there are 2011 numbers, with a sum of 2023066.
(2) After 117 operations, 256 numbers remain on the blackboard: 1873 to 2011, and 117 new sums appear.
These 117 sums $=2023066-(1873+2011) \times 139 \div 2=1753128$
(3) After 16 operations, 16 new numbers remain on the blackboard, with a sum of $=2023066$
(4) After 1 operation, 1 number remains on the blackboard $=2023066$
In summary, all the numbers that have appeared $=2023066+1753128+2023066+2023066=7822326$.