The sum of digits of a natural number is 3. What sum of digits can its cube have? I have found three cases: It either has only 3 and zeros or has 2+1 and zeros or 1+1+1+zeros. The problem is: How can I prove that the sum of digits is equal if there are zeros between the positive digits? (For example, I must prove that $201^3$ and $20001^3$ have the equal sum of digits.)
The sum of a cube's digits
3
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number-theory
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0You could note that $21 = 2\cdot 10 + 1$, while $2000001 = 2\cdot 10^6 + 1$, and use the binomial theorem. – 2017-01-07
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0I got $$(2*10^n+1)^3 = 8*10^{2*n} + 12*10^{2n} + 6*10^n+ 1$$. The sum is 1, but it must be 18. Where's the mistake? – 2017-01-07