A partition of tenth powers into ten parts with equal sums

Question

Is there a natural number n for which the numbers $1^{10},2^{10},3^{10}\ldots, n^{10}$ we can put into 10 groups, such that the sum of the numbers in each group is the same?

Answer 1

$$\large n=10^{11}-1=99~999~999~999.$$

I'll show only hint/scheme on small examples.

A. Using base $3$, it is easy to show that one can put numbers $\color{gray}{0^2,}$ $1^2, 2^2, ..., (3^3-1)^2$ into $3$ such groups:

\begin{array}{ll|l|l} &group~ 1. & group~ 2. & group~ 3. \\ \hline &(000_3)^2 = 0^2 & (001_3)^2 = 1^2 & (002_3)^2 = 2^2 \\ &(011_3)^2 = 4^2 & (012_3)^2 = 5^2 & (010_3)^2 = 3^2 \\ &(022_3)^2 = 8^2 & (020_3)^2 = 6^2 & (021_3)^2 = 7^2 \\ \\ &(101_3)^2 = 10^2 & (102_3)^2 = 11^2 & (100_3)^2 = 9^2 \\ &(112_3)^2 = 14^2 & (110_3)^2 = 12^2 & (111_3)^2 = 13^2 \\ &(120_3)^2 = 15^2 & (121_3)^2 = 16^2 & (122_3)^2 = 17^2 \\ \\ &(202_3)^2 = 20^2 & (200_3)^2 = 18^2 & (201_3)^2 = 19^2 \\ &(210_3)^2 = 21^2 & (211_3)^2 = 22^2 & (212_3)^2 = 23^2 \\ &(221_3)^2 = 25^2 & (222_3)^2 = 26^2 & (220_3)^2 = 24^2 \\ \hline \sum(\bullet) : & 117 & 117 & 117 \\ \sum(\bullet)^2 : & 2067 & 2067 & 2067 \\ \end{array}

B. Using base $3$, same way one can put numbers $\color{gray}{0^3,}$ $1^3, 2^3, ..., (3^4-1)^3$ into $3$ such groups;
and one can put numbers $\color{gray}{0^4,}$ $1^4, 2^4, ..., (3^5-1)^4$ into $3$ such groups;
...
and one can put numbers $\color{gray}{0^{10},}$ $1^{10}, 2^{10}, ..., (3^{11}-1)^{10}$ into $3$ such groups.

C. Using base $10$, one can put numbers $\color{gray}{0^2,}$ $1^2, 2^2, ..., (10^3-1)^2$ into $10$ such groups;
and one can put numbers $\color{gray}{0^3,}$ $1^3, 2^3, ..., (10^4-1)^3$ into $10$ such groups;
and one can put numbers $\color{gray}{0^4,}$ $1^4, 2^4, ..., (10^5-1)^4$ into $10$ such groups;
...
and one can put numbers $\color{gray}{0^{10},}$ $1^{10}, 2^{10}, ..., (10^{11}-1)^{10}$ into $10$ such groups.

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I cannot answer if this number $n=10^{11}-1$ is the smallest possible. I hope it is the smallest possible.

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Examples (base $2$):

Using base $2$, one can construct interesting power sums identities:

\begin{array}{rclr} 0+3 & = & 1+2 & =3; \end{array}

\begin{array}{rclr} 0+3+5+6 & = & 1+2+4+7 & = 14; \\ 0^2+3^2+5^2+6^2 & = & 1^2+2^2+4^2+7^2 & =70; \end{array}

\begin{array}{rclr} 0+3+5+6+9+10+12+15 & = & 1+2+4+7+8+11+13+14 & =60; \\ \small 0^2+3^2+5^2+6^2+9^2+10^2+12^2+15^2 & = & \small 1^2+2^2+4^2+7^2+8^2+11^2+13^2+14^2 & =620; \\ \small 0^3+3^3+5^3+6^3+9^3+10^3+12^3+15^3 & = & \small 1^3+2^3+4^3+7^3+8^3+11^3+13^3+14^3 & =7200; \end{array}

\begin{array}{cccr} \small 0+3+5+6+9+10+12+15 & & \small 1+2+4+7+8+11+13+14 & \\ \small +17+18+20+23+24+27+29+30 & = & \small +16+19+21+22+25+26+28+31 & =248; \\ \scriptsize 0^2+3^2+5^2+6^2+9^2+10^2+12^2+15^2 & & \scriptsize 1^2+2^2+4^2+7^2+8^2+11^2+13^2+14^2 & \\ \scriptsize +17^2+18^2+20^2+23^2+24^2+27^2+29^2+30^2 & = & \scriptsize +16^2+19^2+21^2+22^2+25^2+26^2+28^2+31^2 & =5208; \\ \scriptsize 0^3+3^3+5^3+6^3+9^3+10^3+12^3+15^3 & & \scriptsize 1^3+2^3+4^3+7^3+8^3+11^3+13^3+14^3 & \\ \scriptsize +17^3+18^3+20^3+23^3+24^3+27^3+29^3+30^3 & = & \scriptsize +16^3+19^3+21^3+22^3+25^3+26^3+28^3+31^3 & =123008; \\ \scriptsize 0^4+3^4+5^4+6^4+9^4+10^4+12^4+15^4 & & \scriptsize 1^4+2^4+4^4+7^4+8^4+11^4+13^4+14^4 & \\ \scriptsize +17^4+18^4+20^4+23^4+24^4+27^4+29^4+30^4 & = & \scriptsize +16^4+19^4+21^4+22^4+25^4+26^4+28^4+31^4 & =3098760; \end{array}

$$...$$

Answer 2

$$\large n=10^{11}-1.$$

Proof by math. induction:

1. Note that one can put numbers $\color{gray}{0,} 1, 2, ..., 99$ into $10$ groups with equal sums:

\begin{array}{c} ~~~~ \color{gray}{00} + 11 + 22 + ... + 88 + 99 \\ = 01 + 12 + 23 + ... + 89 + 90 \\ = 02 + 13 + 24 + ... + 80 + 91 \\ = ... \\ = 09 + 10 + 21 + ... + 87 + 98. \end{array}

2. Suppose we can put $10p$ numbers $\color{gray}{0,} 1, 2, ..., 10p-1$ into $10$ groups \begin{array}{c|c|c|c|c|c} a_1 & b_1 & c_1 & ... & j_1 & k_1 \\ a_2 & b_2 & c_2 & ... & j_2 & k_2 \\ ... & ... & ... & ... & ... & ... \\ a_p & b_p & c_p & ... & j_p & k_p \\ \end{array} such that

$$ a_1+a_2+...+a_p = b_1+b_2+...+b_p = ~~~ ...~~~ = k_1+k_2+...+k_p ~~ = S_1; $$

$$ a_1^2+a_2^2+...+a_p^2 = b_1^2+b_2^2+...+b_p^2 = ~~~ ...~~~ = k_1^2+k_2^2+...+k_p^2 ~~ = S_2; $$

$$ a_1^l+a_2^l+...+a_p^l = b_1^l+b_2^l+...+b_p^l = ~~~ ...~~~ = k_1^l+k_2^l+...+k_p^l ~~ = S_l. $$

3. Denote $M = 10^l$. Build numbers \begin{array}{c|c|c|c} A_1 = 0\cdot M + a_1 & B_1 = 0\cdot M + b_1 & ... & K_1 = 0\cdot M + k_1 \\ A_2 = 0\cdot M + a_2 & B_2 = 0\cdot M + b_2 & ... & K_2 = 0\cdot M + k_2 \\ ... & ... & ... & ...\\ A_p = 0\cdot M + a_p & B_p = 0\cdot M + b_p & ... & K_p = 0\cdot M + k_p \\ \\ A_{p+1} = 1\cdot M + b_1 & B_{p+1} = 1\cdot M + c_1 & ... & K_{p+1} = 1\cdot M + a_1 \\ A_{p+2} = 1\cdot M + b_2 & B_{p+2} = 1\cdot M + c_2 & ... & K_{p+2} = 1\cdot M + a_2 \\ ... & ... & ... & ...\\ A_{p+p} = 1\cdot M + b_p & B_{p+p} = 1\cdot M + c_p & ... & K_{p+p} = 1\cdot M + a_p \\ \\ ......... & ......... & ... & ......... \\ \\ A_{9p+1} = 9\cdot M + k_1 & B_{9p+1} = 9\cdot M + a_1 & ... & K_{9p+1} = 9\cdot M + j_1 \\ A_{9p+2} = 9\cdot M + k_2 & B_{9p+2} = 9\cdot M + a_2 & ... & K_{9p+2} = 9\cdot M + j_2 \\ ... & ... & ... & ...\\ A_{9p+p} = 9\cdot M + k_p & B_{9p+p} = 9\cdot M + a_p & ... & K_{9p+p} = 9\cdot M + j_p \\ \end{array}

4. $~$ Prove that $$ A_1+A_2+...+A_{10p} = B_1+B_2+...+B_{10p} = ~~~ ...~~~ = K_1+K_2+...+K_{10p}; $$

$$ A_1^2+A_2^2+...+A_{10p}^2 = B_1^2+B_2^2+...+B_{10p}^2 = ~~~ ...~~~ = K_1^2+K_2^2+...+K_{10p}^2; $$

$$ A_1^l+A_2^l+...+A_{10p}^l = B_1^l+B_2^l+...+B_{10p}^l = ~~~ ...~~~ = K_1^l+K_2^l+...+K_{10p}^l; $$

$$ A_1^{l+1}+A_2^{l+1}+...+A_{10p}^{l+1} = B_1^{l+1}+B_2^{l+1}+...+B_{10p}^{l+1} = ~~~ ...~~~ = K_1^{l+1}+K_2^{l+1}+...+K_{10p}^{l+1}. $$

Denote $D_q = 0^q+1^q+2^q+...+9^q$ ($D_0=10$, $D_1=45$, ...).

Yes, indeed, $$ A_1+A_2+...+A_{10p} = B_1+B_2+...+B_{10p} = ~~~ ...~~~ = K_1+K_2+...+K_{10p} $$ $$ = (0+1+2+...+9)\cdot M + 10 S_1 = D_1 M + D_0 S_1. $$

$$ A_1^2+A_2^2+...+A_{10p}^2 = B_1^2+B_2^2+...+B_{10p}^2 = ~~~ ...~~~ = K_1^2+K_2^2+...+K_{10p}^2 $$ $$ = (0^2+1^2+2^2+...+9^2)\cdot M^2 + 2 \cdot(0+1+2+...+9)\cdot M \cdot S_1 + 10 S_2 $$ $$ = D_2 M^2 + 2 D_1 M S_1 + D_0 S_2. $$

$$ ... ... ... $$

$$ A_1^l+A_2^l+...+A_{10p}^l = B_1^l+B_2^l+...+B_{10p}^l = ~~~ ...~~~ = K_1^l+K_2^l+...+K_{10p}^l $$ $$ = D_l M^l + \binom{l}{l-1}D_{l-1}M^{l-1}S_1 + ... + \binom{l}{1}D_1M^1S_{l-1}+D_0 S_l $$ $$ =\sum_{q=0}^l \binom{l}{q}D_q M^q S_{l-q}. $$

$$ A_1^{l+1}+A_2^{l+1}+...+A_{10p}^{l+1} = B_1^{l+1}+B_2^{l+1}+...+B_{10p}^{l+1} = ~~~ ...~~~ = K_1^{l+1}+K_2^{l+1}+...+K_{10p}^{l+1} $$ $$ = D_{l+1} M^{l+1} + \binom{l+1}{l}D_{l}M^lS_1 + \binom{l+1}{l-1}D_{l-1}M^{l-1}S_2 + ... + \binom{l+1}{1}D_1M^1S_l $$ $$ + (a_1^{l+1}+a_2^{l+1}+...+a_p^{l+1}) + (b_1^{l+1}+b_2^{l+1}+...+b_p^{l+1}) + ... + (k_1^{l+1}+k_2^{l+1}+...+k_p^{l+1}) $$ $$ =\sum_{q=1}^{l+1} \binom{l+1}{q}D_q M^q S_{l+1-q} + \sum_{t=0}^{10p-1} t^{l+1}. $$

Proved.

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I am not sure that this $n$ is the smallest possible, but perhaps it is so.