We denote by $C$ the set of integers of Cantor, that is to say the natural integers whose writing in base 3 does not admit any number 1.

Calculate $\sum \limits_{a\in C,a<3^{2017}} a^{103} \mod (2^{89}-1)$.

Source : les dattes à Dattier

  • 2
    $\begingroup$ For anyone who wonders: $2^{89}-1$ is prime. $\endgroup$ – ThomasR Jan 25 '17 at 22:22

So it's 98191387587393867840363579. You should have pretty good intuition.

Edit: explanation.

  1. Well, lets imagine the following situation: given an integer $A$ and integers $B[1], B[2], ..., B[k]$, we want to calculate following sum: $\sum_{c=1}^{k} (A+B[k])^p$ for some p.

  2. Lets look at the polynomial formula: $(a + b)^n$ = $\sum_{i, j, i + j = n} (\frac{n!}{i!j!}a^ib^j)$. Applying this to $A$ and $B[1], .., B[k]$ we get $$ S = \sum_{c=1}^{k} (A+B[k])^p = \sum_{i, j, i + j = p}\frac{p!}{i!j!}(A^i \sum_{c = 1}^{k}B[c]^j).$$ Thus knowing only the sum of the $j^{th}$ powers of each $B$ for all $0 \le j \le p$ we can calculate $S$.

  3. Ok, now lets calculate following: $dp[i][j]$ = sum of $j^{th}$ powers of all Cantor numbers $B[k] < 3^{i}$, or $dp[2017][103]$ is what we are looking for.

  4. To do this, one first can notice that following from point 2, $$dp[m][n] = dp[m - 1][n] + \sum_{i, j, i + j = n}\frac{n!}{i!j!}(2\cdot 3^{m-1})^i \cdot dp[m-1][j].$$ The proof is left as an exercise to the reader.

  5. Everything should be calculated mod $2^{89}-1$, starting with $dp[][]$ filled with zeroes and going into the ranges $1 \le i \le 2017$ and $0 \le j \le 103$.

This approach took about 5 minutes in Wolfram Mathematica 7.0 to finish and works in $O(2017 \cdot 104 \cdot 104 \cdot O(\text{long arithmetics}))$.

I used this code. Sorry if it's not great, I'm more used to C++.

Sorry for my English, I'm not a native speaker. If you have any questions left feel free to ask.

  • 2
    $\begingroup$ A little explanation definitely wouldn't hurt. $\endgroup$ – Ivan Neretin Jan 26 '17 at 21:18
  • $\begingroup$ My dear GaBuZoMeu, can you explain or give your algorithm ? $\endgroup$ – Dattier Jan 26 '17 at 21:47
  • 1
    $\begingroup$ Explained, answer edited $\endgroup$ – i.trofimow Jan 27 '17 at 0:14

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