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# Number Problem

+2
47
3

Let

\(N = 1234567891011...20192020 \)

be made by combining the first 2020 positive integers. Find the thousandth digit in this sequence

Jan 2, 2020
edited by hellospeedmind  Jan 3, 2020

#1
+1

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   Jan 2, 2020
edited by CPhill  Jan 2, 2020
#2
+1

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

Jan 2, 2020
#3
0

You are correct guest....I miscalculated the  number of integers from 10-99  in my first answer

My corrected answer agrees with yours....!!!!   CPhill  Jan 2, 2020