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Minimum and 2015 Multiple Muse (Posted on 2023-06-16) Difficulty: 3 of 5
Determine the minimum value of a positive integer N such that:
•    3N - 2N is a positive integer multiple of 2015

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Solution computer solution | Comment 1 of 2
for n=sym(1):300
  v=3^n-2^n;
  if mod(v,2015)==0
    disp([n v v/2015])
  end
end

finds the minimum N is 60:

>> minimumAnd2015Multiple
[ N, formula value, value / 2015 ]

[60, 42391158274063282009687586225, 21037795669510313652450415]
[120, 1797010299914431210411850601513820123858571820477570761825, 891816526012124670179578462289737034172988496514923455]
[180, 76177348045866392339289727720614029254883935513536838376974415435773518603910854417425, 37805135506633445329672321449436242806394012661804882569218072176562540250079828495]
[240, 3229246017998554007515224836513361914702371285180037076635122988479942482774944911163345713522956067904578530553025, 1602603482877694296533610340701420305063211555920613933813956818104189817754315092388757177927025343873239965535]
[300, 136891479058588375991326027382088315966463695625337434434444213743882910909053388184422045104220547722314233804457544688432212969428989591168625, 67936217895081079896439715822376335467227640508852324781361892676864968193078604558025828835841462889485972111393322426020949364480888134575]
>> 

Edited on June 16, 2023, 10:54 am
  Posted by Charlie on 2023-06-16 10:52:21

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