BMAD 63 Report post Posted February 16, 2015 Among all the numbers representable as 36^{k} - 5^{m} (k and m are natural numbers) find the smallest. Prove that it is really the smallest. Share this post Link to post Share on other sites

0 Barcallica 2 Report post Posted February 16, 2015 k=1 m=2 smallest 11. Share this post Link to post Share on other sites

0 Barcallica 2 Report post Posted February 16, 2015 Among all the numbers representable as 36^{k} - 5^{m} (k and m are natural numbers) find the smallest. Prove that it is really the smallest. You mean smallest natural? Share this post Link to post Share on other sites

0 BMAD 63 Report post Posted February 16, 2015 it does not need to be natural, only k and M have to. Share this post Link to post Share on other sites

0 gavinksong 11 Report post Posted February 16, 2015 But there can't be a smallest if you're including integers. Right? Share this post Link to post Share on other sites

0 Barcallica 2 Report post 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) Share this post Link to post Share on other sites

0 bonanova 81 Report post Posted February 17, 2015 Natural numbers are positive or non-negative Share this post Link to post Share on other sites

0 BMAD 63 Report post Posted February 17, 2015 I see the error in my clarification. Yes the result should be natural. That is my mistake. Share this post Link to post Share on other sites

0 BMAD 63 Report post Posted February 17, 2015 k=1 m=2 smallest 11. Now to prove your answer is the smallest. Share this post Link to post Share on other sites

0 Barcallica 2 Report post 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. 36^{k }= 5^{m} +1, left side is dividable by 4. Right side is not. Share this post Link to post Share on other sites

0 BMAD 63 Report post Posted February 18, 2015 I am missing how that proves k=1 and m=2 is the smallest Share this post Link to post Share on other sites

0 bonanova 81 Report post Posted February 18, 2015 k=m=0. Result is 0. Share this post Link to post Share on other sites

0 bonanova 81 Report post 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. Share this post Link to post Share on other sites

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