[Guest]

There is a number which is the cube of the sum of its digits ... What is the number ?

[Guest]

i found the answer pretty easily but i'll resist the urge to post because i cheated with the internet

[Guest]

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8x8x8=512

5+1+2=8

[Guest]

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1?

[bonanova]

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.

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81

[Guest]

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4913

We could try to make a sequence of such numbers

[Guest]

  OstapBender said:

We could try to make a sequence of such numbers

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4913

I am almost certain that this is an exhaustive list (it is at least exhaustive out to one million):

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1, 8, 17, 18, 26, 27

[Prime]

  Ben Law said:

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.

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The numbers I found are:

0, 1, 512, 4913, 5832

[Prime]

While it's a commercial break, here are couple more:

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17576 and 19683

Interesting question: what is the largest possible such number?

Edited August 17, 2008 by Prime

[Guest]

  Prime said:

While it's a commercial break, here are couple more:

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17576 and 19683

Interesting question: what is the largest possible such number?

  itsclueless said:

I am almost certain that this is an exhaustive list (it is at least exhaustive out to one million):

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1, 8, 17, 18, 26, 27

Answered already

[Prime]

  itsclueless said:

I am almost certain that this is an exhaustive list (it is at least exhaustive out to one million):

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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).

[Guest]

  Prime said:

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

[Prime]

  itsclueless said:

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.

[Guest]

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.

[Guest]

I got 1 and 512.

[Prime]

  itsclueless said:

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.

[Guest]

  On 8/17/2008 at 4:52 AM, Prime said:

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).

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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