Guest Posted October 5, 2009 Report Share Posted October 5, 2009 Prove that there does not exist any positive integer P which is a power of 2 with the proviso that the digits of P (in the base ten representation) can be permuted to form a different power of 2. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 5, 2009 Report Share Posted October 5, 2009 Thanks a bundle for this headache, KS! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 5, 2009 Report Share Posted October 5, 2009 can you simplify the wording on this one Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted October 5, 2009 Report Share Posted October 5, 2009 The difference of two base 10 numbers which are digit permutations of one another is always a multiple of 9. Note that, for a difference of two powers of 2 to be a multiple of 9, the powers must differ by a multiple of 6. This is easy to see if you think of the two numbers in base 2 -- 9 is 1001 and the difference of the powers of 2 is a sequence of 1s surrounded by 0s. But, powers of 2 differing by a multiple of 64 (26) makes them differ in the number of decimal digits they contain. Therefore, they cannot be permutations of each other. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 6, 2009 Report Share Posted October 6, 2009 Very succinct, SP. I am still pondering one aspect of your proof. Well done. Quote Link to comment Share on other sites More sharing options...
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Prove that there does not exist any positive integer P which is a power of 2 with the proviso that the digits of P (in the base ten representation) can be permuted to form a different power of 2.
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