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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)

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We're going to need more precision (spoiler) | Comment 1 of 6
let the digit be d and the length of N be m-2.

Then d*10^m + 10N + d = 99N

so d(1+10^m)/89 = N

since 89 is prime, it must divide (1+10^m)

In other word, 10^m = 88 (mod 89)

By using excel, and multiplying the previous remainder mod 89 by 10, we can see that 10^23 = 88 mod n.

I don't have enough precision to finish, and don't feel like doing it by hand, but calculate (1+10^23)/89 and multiply it by the smallest digit necessary to make it a length 21 number.

That should be the sought after N.  

Edited on July 11, 2015, 12:16 pm
  Posted by Steve Herman on 2015-07-11 08:03:18

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