BMAD Posted February 16, 2015 Report Share Posted February 16, 2015 Among all the numbers representable as 36k - 5m (k and m are natural numbers) find the smallest. Prove that it is really the smallest. Quote Link to comment Share on other sites More sharing options...
0 Barcallica Posted February 16, 2015 Report Share Posted February 16, 2015 Among all the numbers representable as 36k - 5m (k and m are natural numbers) find the smallest. Prove that it is really the smallest. You mean smallest natural? Quote Link to comment Share on other sites More sharing options...
0 Barcallica Posted February 16, 2015 Report Share Posted February 16, 2015 k=1 m=2 smallest 11. Quote Link to comment Share on other sites More sharing options...
0 BMAD Posted February 16, 2015 Author Report Share Posted February 16, 2015 it does not need to be natural, only k and M have to. Quote Link to comment Share on other sites More sharing options...
0 gavinksong Posted February 16, 2015 Report Share Posted February 16, 2015 But there can't be a smallest if you're including integers. Right? Quote Link to comment Share on other sites More sharing options...
0 Barcallica Posted February 17, 2015 Report Share Posted February 17, 2015 it does not need to be natural, only k and M have to. k=1 m=-infinite (though -infinite is not real number) Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 17, 2015 Report Share Posted February 17, 2015 Natural numbers are positive or non-negative Quote Link to comment Share on other sites More sharing options...
0 BMAD Posted February 17, 2015 Author Report Share Posted February 17, 2015 I see the error in my clarification. Yes the result should be natural. That is my mistake. Quote Link to comment Share on other sites More sharing options...
0 BMAD Posted February 17, 2015 Author Report Share Posted February 17, 2015 k=1 m=2 smallest 11. Now to prove your answer is the smallest. Quote Link to comment Share on other sites More sharing options...
0 Barcallica Posted February 18, 2015 Report Share Posted February 18, 2015 k=1 m=2 smallest 11. Now to prove your answer is the smallest. Last digit of result always is 1. We should prove it cannot be 1. 36k = 5m +1, left side is dividable by 4. Right side is not. Quote Link to comment Share on other sites More sharing options...
0 BMAD Posted February 18, 2015 Author Report Share Posted February 18, 2015 I am missing how that proves k=1 and m=2 is the smallest Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 18, 2015 Report Share Posted February 18, 2015 k=m=0. Result is 0. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 18, 2015 Report Share Posted February 18, 2015 I am missing how that proves k=1 and m=2 is the smallest Powers of 36 have last digit of 6; powers of 5 have last digit of 5. Their (positive) differences have a last digit of 1, so numbers other than 1 11 21 31 41 51 ... are excluded. k=1 and m=2 gives 11 - at worst the second smallest result. Since k-man showed that 1 must be exclude, 11 is smallest result. Quote Link to comment Share on other sites More sharing options...
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