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 k=1 m=2 smallest 11. 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 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...
Question
BMAD
Link to comment
Share on other sites
12 answers to this question
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.