Guest Posted August 16, 2008 Report Share Posted August 16, 2008 There is a number which is the cube of the sum of its digits ... What is the number ? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 16, 2008 Report Share Posted August 16, 2008 i found the answer pretty easily but i'll resist the urge to post because i cheated with the internet Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 16, 2008 Report Share Posted August 16, 2008 8x8x8=512 5+1+2=8 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 16, 2008 Report Share Posted August 16, 2008 1? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted August 16, 2008 Report Share Posted August 16, 2008 Pretty quickly I thought of a number that is the square of the sum of its digits. There might be quite a lot of them since it seems an easier problem then finding a cube.81 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 16, 2008 Report Share Posted August 16, 2008 4913 We could try to make a sequence of such numbers Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 16, 2008 Report Share Posted August 16, 2008 We could try to make a sequence of such numbers4913 I am almost certain that this is an exhaustive list (it is at least exhaustive out to one million): 1, 8, 17, 18, 26, 27 Quote Link to comment Share on other sites More sharing options...
0 Prime Posted August 17, 2008 Report Share Posted August 17, 2008 There is a number which is the cube of the sum of its digits ... What is the number ? I found 5 such numbers. After which I gave up further search and went back to watching olympics. The numbers I found are: 0, 1, 512, 4913, 5832 Quote Link to comment Share on other sites More sharing options...
0 Prime Posted August 17, 2008 Report Share Posted August 17, 2008 (edited) While it's a commercial break, here are couple more: 17576 and 19683 Interesting question: what is the largest possible such number? Edited August 17, 2008 by Prime Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 17, 2008 Report Share Posted August 17, 2008 While it's a commercial break, here are couple more: 17576 and 19683 Interesting question: what is the largest possible such number? I am almost certain that this is an exhaustive list (it is at least exhaustive out to one million): 1, 8, 17, 18, 26, 27 Answered already Quote Link to comment Share on other sites More sharing options...
0 Prime Posted August 17, 2008 Report Share Posted August 17, 2008 I am almost certain that this is an exhaustive list (it is at least exhaustive out to one million): 1, 8, 17, 18, 26, 27 1. You've missed 0 again. 2. Can there be any numbers greater than 1 million with that property? Better yet, any numbers greater than 157,464? 3. I can find many more numbers with that property, not shown in your "exhaustive list", all under 100,000. And I can find those without aid of a computer.(That's a new puzzle). Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 17, 2008 Report Share Posted August 17, 2008 1. You've missed 0 again. 2. Can there be any numbers greater than 1 million with that property? Better yet, any numbers greater than 157,464? 3. I can find many more numbers with that property, not shown in your "exhaustive list", all under 100,000. And I can find those without aid of a computer.(That's a new puzzle). 0 is a valid solution, but it is just a tricky value. This list has been analyzed by others besides me (I did write a computer program to check 1-1000000 and verify that it works that far as stated earlier). I refer you to http://www.research.att.com/~njas/sequences/A046459 Quote Link to comment Share on other sites More sharing options...
0 Prime Posted August 17, 2008 Report Share Posted August 17, 2008 0 is a valid solution, but it is just a tricky value. This list has been analyzed by others besides me (I did write a computer program to check 1-1000000 and verify that it works that far as stated earlier). I refer you to http://www.research.att.com/~njas/sequences/A046459 OK, so the the question #2 in my previous post may be too mathematical, so we don't want to tackle it here. But, surely, you can solve my question #3. That's just a puzzle. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 17, 2008 Report Share Posted August 17, 2008 My claim is that such numbers under 100000 do not exist. All numbers there are cubes of the sequence I showed you. If you know otherwise, please prove me wrong. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 17, 2008 Report Share Posted August 17, 2008 I got 1 and 512. Quote Link to comment Share on other sites More sharing options...
0 Prime Posted August 17, 2008 Report Share Posted August 17, 2008 My claim is that such numbers under 100000 do not exist. All numbers there are cubes of the sequence I showed you. If you know otherwise, please prove me wrong. You could give me a benefit of a doubt. And treat my proposition as a puzzle. May be I'll make a separate post of it. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 26, 2009 Report Share Posted July 26, 2009 (edited) 2. Can there be any numbers greater than 1 million with that property? Better yet, any numbers greater than 157,464? 3. I can find many more numbers with that property, not shown in your "exhaustive list", all under 100,000. And I can find those without aid of a computer.(That's a new puzzle). Let the number be N, and its sum of the digits be denoted by D. Then, by the problem, N = D^3. Again, by way of elementary number theory, it can easily be proved that the cube of any positive integer has the form: 0, 1, or -1(mod 9), and therefore: D^3 = 0, 1, -1(mod 9) -> N = 0, 1, -1(mod 9) …(i) Since, we know that: N = D (mod 9), in terms of (i), we must have: D = 0, 1, -1(mod 9) Let the number of digits in N be x. Then, Max (D)=9x, so that; max (N) = (9x)^3 = 729*x^3 = g(x) (say) Also, min(N) = 10^(x-1) = f(x) (say) Since (2/1)^3 =8 < 10 and (x+1)/x is decreasing in positive integer values of x, it follows that: ((x+1)/x)^3 < 10 for x>=1 and accordingly, f(x+1) > g(x+1), whenever f(x) > g(x) with x>=1 .......(ii) Now, we note that: f(7) = 10^6 > 729*(7^3) Accordingly, in terms of (ii), we must have; f(x) > g(x), whenever x>=7, so that the given problem does not have a solution whenever x>=7. Therefore, 1<= x<= 6, so that: 0<=N <= 999999, or: 0<= D^3 <= 999999 , or: 0<= D <= 99 and accordingly from (i), the possible values of D are: 0, 1, 8, 9, 10, 17, 18, 19, 26, 27, 28, 35, 36, 37, 44, 45, 46, 53, 54, 55, 62, 63, 64, 71, 72, 73, 80, 81, 82, 89, 90, 91, 98, 99 A calculator check for each of the above values for the equation N = D^3 reveals that, the valid values in conformity with the foregoing relationship and in consonance with all the provisions corresponding to the given problem, occurs at: D = 0, 1, 8, 17, 18, 26, 27 giving: N = 0, 1, 512, 4913, 5832, 17576, 19683 Consequently, the required numbers in conformity with all the provisions of the puzzle under reference are: 0, 1, 512, 4913, 5832, 17576 and 19683. ******* This agrees with the njas sequence # A046459 as mentioned by itsclueless in a subsequent post. Edited July 26, 2009 by K Sengupta Quote Link to comment Share on other sites More sharing options...
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There is a number which is the cube of the sum of its digits ... What is the number ?
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