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 Beginning and End (Posted on 2015-07-11)
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)

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 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

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