perplexus.info :: Just Math : Smarandache Settlement

Smarandache Settlement (Posted on 2014-06-24)

Determine the remainder when 1234567...404142 is divided by 43.

See The Solution Submitted by K Sengupta

Rating: 3.0000 (2 votes)

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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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