12. (5 points) The five-digit number "12345" is repeated 403 times to form a 2015-digit number "123451234512345...". From left to right, first delete all the digits in the odd positions to get a new number, then delete all the digits in the odd positions of the new number; continue this process until only one digit remains. What is the last remaining digit? $\qquad$.

【Solution】Solution: According to the analysis, number the 2015-digit number from left to right, from 1 to 2015. In the first round of operations, the number of digits deleted $=(2015+1) \div 2=1008$,

remaining $=2015-1008=1007$, leaving the digits with even numbers;
In the second round of operations, the number of digits deleted $=(1007+1) \div 2=504$, remaining $=1007-504=503$, leaving the digits with multiples of 4;

In the third round of operations, the number of digits deleted $=(503+1) \div 2=252$, remaining $=503-252=251$, leaving the digits with multiples of 8;

In the fourth round of operations, the number of digits deleted $=(251+1) \div 2=126$, remaining $=251-126=125$, leaving the digits with multiples of 16;

In the fifth round of operations, the number of digits deleted $=(125+1) \div 2=63$, remaining $=125-63=62$, leaving the digits with multiples of 32;

In the sixth round of operations, the number of digits deleted $=(63+1) \div 2=63$, remaining $=63-32=31$, leaving the digits with multiples of 64;

In the seventh round of operations, the number of digits deleted $=(31+1) \div 2=16$, remaining $=31-16=15$, leaving the digits with multiples of 128;

In the eighth round of operations, the number of digits deleted $=(15+1) \div 2=8$, remaining $=15-8=7$, leaving the digits with multiples of 256;

In the ninth round of operations, the number of digits deleted $=(7+1) \div 2=4$, remaining $=7-4=3$, leaving the digits with multiples of 512;

In the tenth round of operations, the number of digits deleted $=(3+1) \div 2=2$, remaining $=3-2=1$, leaving the digits with multiples of 1024;

A total of 10 rounds of operations are required, and the original 2015-digit number is arranged according to the pattern of 5 digits "12345" repeated 403 times. $1024 \div 5=204 \cdots 4$, the extra 4 digits are "1234", therefore, the digit at position 1024 is $4$, hence the answer is: 4.
12. (5 points) Write the five-digit number “12345” 403 times to form a 2015-digit number “$123451234512345 \cdots$”. From left to right, first delete all the digits in the odd positions of this number to get a new number, then delete all the digits in the odd positions of the new number; continue deleting according to the above rule until only one digit remains. What is the last remaining digit? $\qquad$