All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Beginning and End (Posted on 2015-07-11) Difficulty: 3 of 5
When the digit is placed at the beginning and end of a positive integer N, the new number is 99*N.

Find the smallest value of N.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution anal. solution..................... spoiler | Comment 3 of 6 |

Let as start by assuming that the digit  is 1

N= (10^k+1)/89

It means that 10^k should be 88 mod 89

100=11 mod 89 , 1000=110 mod 89=21 mod 89 , 1000=32 mod 89…  later, 53, 85,..etc....80,88

Continuing this process with a simple calculator we arrive at the correct value, i.e,  k=22, meaning that with digit 1 our 89*N
 is 1000….00 (21 zeroes)0…  ..1, and N derived by dividing this number by 89 is

N=112,359,550,561,797,752,809

Checking with scientific calculator:

99*112,359,550,561,797,752,809 =11,123,595,505,617,977,528,091   ok

Since 89 is a prime number, there is clearly no way that digit other than 1 could bring a lesser result for N.


  Posted by Ady TZIDON on 2015-07-11 11:17:12
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information