67th Romanian Mathematical Olympiad

Suppose that $S_1,S_2,\dots ,S_{18}$ leave the same remainder when divided by 5. Since $S_1=a_1+(a_2+a_3+a_4+a_5)$ and $S_2=(a_2+a_3+a_4+a_5)+a_6$, if $S_1$ and $S_2$ leave the same remainder when divided by 5, then $a_1$ and $a_6$ leave the same remainder when divided by 5. In the same way:

1. $a_1,a_6,a_{11},a_{16}$ and $a_{21}$ leave the same remainder, $x$, when divided by 5. 2. $a_2,a_7,a_{12},a_{17}$ leave the same remainder, $a$, when divided by 5. 3. $a_3,a_8,a_{13},a_{18}$ leave the same remainder, $b$, when divided by 5. 4. $a_4,a_9,a_{14},a_{19}$ leave the same remainder, $c$, when divided by 5. 5. $a_5,a_{10},a_{15},a_{20}$ leave the same remainder, $d$, when divided by 5.

Since $\{a_1,a_2,\dots ,a_{21}\}=\{1,2,\dots ,21\}$, there are 5 remainders equal to 1 and 4 remainders equal to each of 2, 3, 4 or 0. It follows $x=1$ and $\{a,b,c,d\}=\{0,2,3,4\}$.

Now $S_1=\mathcal{M}5+1+0+2+3+4=\mathcal{M}5$ and $S_{18}=\mathcal{M}5+b+\mathcal{M}5+c+\mathcal{M}5+d+\mathcal{M}5+1+\mathcal{M}5+1=\mathcal{M}5+2+(a+b+c+d)-a=\mathcal{M}5+11-a$.

Since $S_1$ and $S_{18}$ leave the same remainder when divided by 5, it follows that $11-a=\mathcal{M}5$, whence $a=1$ – contradiction.