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Smarandache Settlement (Posted on 2014-06-24) Difficulty: 3 of 5
Determine the remainder when 1234567...404142 is divided by 43.

No Solution Yet Submitted by K Sengupta    
Rating: 1.0000 (1 votes)

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Solution Solution Comment 2 of 2 |
Start by expressing the number as (12345678*10^67) + (9*10^66 + 10*10^64 + 11*10^62 + ... + 41*10^2 + 42*10^0).

The large sum is a arithmetico–geometric series.  Its sum can be expressed as ((9*99+1)*10^68 - 99*43 - 1) / 99^2.  (Note this is an integer)

Now compute each part mod 43:
12345678 = 34 mod 43
10^67 = 24 mod 43
(9*99+1)*10^68 - 99*43 - 1 = 32*240 - 1 = 25 mod 43
99^2 = 40 mod 43

Then the AG sum is 25/40 mod 43, which is 6.
Finally 34*24+6 = 5 mod 43.  The remainder of the number is 5.

  Posted by Brian Smith on 2018-12-15 18:32:19
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