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Number $n$ is a natural number if there is $b_0,b_1, ..., b_k$ (the number 0-9) So that $n = \sum_0^m b_j2^{j^2}$. Is there for all the natural number this kind of representation? Is the representation of the number in this method is a single?

Nir Asaf
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  • This question is very unclear. I assume that a0,a1 .. ak is supposed to be $a_0,a_1,\dots,a_k$. But what do you mean when you call these numbers "decimals"? Also, the sum is unclear. Is this supposed to be $$\sum_{i=0}^ka_ki^2$$ or maybe $$\sum_{i=0}^ka_{k_i}^2$$ If you click on help, you may find links which show how to format mathematics on this website. – Gerry Myerson Oct 26 '13 at 22:49
  • I'm not sure the edit is right, as you can pull $a_k$ out from the sum, leaving $$n=a_k\sum_{i=0}^ki^2$$ and then the $a_j$, $0\le j\le k-1$, play no role. – Gerry Myerson Oct 26 '13 at 23:04
  • yes this is what i do I edit the question to Bit more complicated question – Nir Asaf Oct 27 '13 at 09:10
  • $b_0+2b_1+16b_3+512b_4+\dots$. How about $n=172$? – Gerry Myerson Oct 27 '13 at 09:26
  • So how do i prove that there is no representation? – Nir Asaf Oct 27 '13 at 09:51
  • I'm sure you can figure that out. First show that you can't use anything with $j\gt2$. Then ask yourself what numbers you can get with $m=2$. – Gerry Myerson Oct 27 '13 at 09:53
  • I sorry but i don't get it can you tell me $b_0,b_1, ..., b_k$ is number Between 1-9 but this is not like $b_0 = 0? – Nir Asaf Oct 27 '13 at 13:01
  • I don't understand your latest comment. Yes, $0\le b_j\le9$ for all $j$, and $b_0$ is allowed to be zero, but not required to be zero. Note that if $m=2$ and $b_0=b_1=b_2=9$ then $n=9(1)+9(2)+9(16)$, which is what number? – Gerry Myerson Oct 27 '13 at 22:18

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I expect the homework due date is past, so it may be OK to give a more complete answer than usual.

We're trying to represent $n$ as $$b_0+2b_1+16b_2+512b_3+\dots,\qquad0\le b_j\le9$$ If we try to represent $n=172$ that way, then we certainly need $b_3=b_4=\cdots=0$, so we are trying to get $$172=b_0+2b_1+16b_2$$ But we have $$b_0+2b_1+16b_2\le9+2\times9+16\times9=9+18+144=171$$ and we've reached a contradiction.

As to whether the representation is unique, for $n=9$ we can use $b_0=9$, $b_1=0$, or we can use $b_0=7$, $b_1=1$, or $b_0=5$, $b_1=2$, or....

Gerry Myerson
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