Let
\(N = 1234567891011...20192020 \)
be made by combining the first 2020 positive integers. Find the thousandth digit in this sequence
1 - 9 = puts us at the 9th digit
10- 99 = two digits for each integer * 90 integers = 90*2 = 180 digits
This puts us at 189 digits
So....we need to solve this where n is the number of integers after 99
189 + 3n = 1000
3n = 1000 - 189
3n = 811
n = 270 + 1/3
So the integer we are looking for is the 1st digit of the 271st integer after 99 =
99 + 271 = 370
So....the 1000th digit is 3
Sorry Cphill: My short computer code says it is "3" of the beginning of 370:
n=1234567891011..........3603613623633643653663673683693;s=0;p=0;cycle: s=s+(n%10);p=p+1;n=int(n/10);if(n!=0, goto cycle,0);print"Total Sum =",s;print"Total Num =",p
OUTPUT:
1234567891011........603613623633643653663673683693
Total Sum = 3738
Total Num = 1000