Answer to Question #89325 in Combinatorics | Number Theory for Aravind

Question #89325
How do I find the answer? let n=9+99+999+…+9999…9;where the last number consist of 1999 digits of 9.How many times will the digit 1 appear in n ?
1
Expert's answer
2019-05-07T14:15:03-0400

Example:

Consider simpler case first 9+99+999 = (10-1) + (100-1) + (1000-1) = 1110 - 3 = 1107


Solution:

n = 9+99+999+...+9999...9 = (10-1) + (100-1) + (1000-1) +...+ (10000...0-1) =

10 + 100 + 1000 + 10000...0 - 1999 = {10000...0 consists 1999 digits of 0 }

1111...10 - 1999 = {1111...10 consists 1999 digits of 1 }

1111...100000 + 11110 - 1999 = {1111...100000 consists 1995 digits of 1 }

1111...100000 + 9111 {1111...100000 consists 1995 digits of 1 }

Number n consists of (1995 + 3) digits of 1.


Answer : Digit 1 appear in n 1998 times.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Assignment Expert
08.05.19, 18:29

Dear Aravind, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you!

Aravind
08.05.19, 18:08

Great work thanks a lot

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS