Japan Mathematical Olympiad

$\boxed{2042\cdot 19^{14}}$

First we show the following lemma.

Lemma. Let $a,b,c$ be integers satisfying $a\ne 0(\mathrm{mod}17)$, $b\ge 1$, $c\ge 4$. Then $a^{b^c}\equiv 1(\mathrm{mod}17)$ holds if and only if $a\equiv 1(\mathrm{mod}17)$ or $b$ is even.

Proof. If $a\equiv 1(\mathrm{mod}17)$, obviously $a^{b^c}\equiv 1(\mathrm{mod}17)$ holds. If $b$ is even, $c\ge 4$ shows $b^c$ is divisible by $16$. By Fermat's little theorem $a^{16}\equiv 1(\mathrm{mod}17)$ holds thus $a^{b^c}\equiv 1(\mathrm{mod}17)$. We show the converse. If $a^{b^c}\equiv 1(\mathrm{mod}17)$ holds, we can take minimum positive integer $d$ satisfying $a^d\equiv 1(\mathrm{mod}17)$. If $a\ne 1(\mathrm{mod}17)$ then $d\ne 1$. Assume $b^c$ is not divisible by $d$, then $b^c$ can be written in the form $b^c=sd+t$ with a non-negative integer $s$ and an integer $t$ satisfying $1\le t<d$ hence $$a^t\equiv (a^d{)}^s\cdot a^t=a^{b^c}\equiv 1(\mathrm{mod}17),$$ which contradicts to the minimality of $d$. Therefore $b^c$ is divisible by $d$. Similarly $16$ is divisible by $d$ and $d\ne 1$ shows $d$ is even. Hence $b^c$ is even and thus $b$ is even. ■

If $a_1\equiv 0(\mathrm{mod}17)$ or $a_2\equiv 0(\mathrm{mod}17)$ then the condition is not satisfied. From here we assume $a_1\ne 0(\mathrm{mod}17)$ and $a_2\ne 0(\mathrm{mod}17)$. Let $$c_1=a_3^{a_4^{a_5^{a_6^{a_7}}}},c_2=a_4^{a_5^{a_6^{a_7}}},$$ then $c_1\ge 2^2=4$, $c_2\ge 2^2=4$ holds. We have two cases; when $a_2$ is odd and when it is even.

When $a_2$ is odd.

Lemma shows $a_1^{a_2^{c_1}}\equiv 1(\mathrm{mod}17)$ if and only if $a_1\equiv 1(\mathrm{mod}17)$. Since $a_2\ne 18$, we have $a_2\ne 1(\mathrm{mod}17)$ and then lemma shows $a_2^{a_3^{c_2}}\equiv 1(\mathrm{mod}17)$ if and only if $a_3$ is even. Therefore the number of tuples satisfying the condition is $1\cdot 8\cdot 10\cdot 19^{14}=80\cdot 19^{14}$.

When $a_2$ is even.

We have $a_1\ne 0(\mathrm{mod}17)$ then lemma shows $a_1^{a_2^{c_1}}\equiv 1(\mathrm{mod}17)$. By lemma, $a_2^{a_3^{c_2}}\equiv 1(\mathrm{mod}17)$ holds if and only if $a_2\equiv 1(\mathrm{mod}17)$ or $a_3$ is even. Therefore the number of tuples satisfying the condition is $18\cdot 1\cdot 19^{15}+18\cdot 9\cdot 10\cdot 19^{14}=1962\cdot 19^{14}$.

Hence the total number of tuples satisfying the condition is $80\cdot 19^{14}+1962\cdot 19^{14}=2042\cdot 19^{14}$.