A Complicated Number

Let the number given be $N = 19^{17^{15^a}}$ . We need to find $N \bmod 100$ .

$\large \begin{aligned} N & \equiv 19^{\color{#3D99F6} 17^{15^a} \bmod \lambda (100)} \text{ (mod 100)} & \small \color{#3D99F6} \text{Since }\gcd(19, 100) = 1 \text{, Euler's theorem applies.} \\ & \equiv 19^{\color{#3D99F6} 17^{15^a} \bmod 20} \text{ (mod 100)} & \small \color{#3D99F6} \text{Carmichael's lambda }\lambda (100) = 20 \\ & \equiv 19^{17^{\color{#3D99F6} 15^a \bmod 4} \bmod 20} \text{ (mod 100)} & \small \color{#3D99F6} \text{Again }\gcd(17, 20) = 1 \text{ and }\lambda (20) = 4 \\ & \equiv 19^{17^{\color{#3D99F6}3} \bmod 20} \text{ (mod 100)} & \small \color{#3D99F6} \text{See note 1.} \\ & \equiv 19^{\color{#3D99F6}13} \text{ (mod 100)} & \small \color{#3D99F6} \text{See note 2.} \\ & \equiv (20-1)^{13} \text{ (mod 100)} \\ & \equiv {\color{#3D99F6}(260-1)} \text{ (mod 100)} & \small \color{#3D99F6} \text{Last two terms of binomial expansion,} \\ & \equiv \boxed{59} \text{ (mod 100)} & \small \color{#3D99F6} \text{earlier terms are divisible by 100.} \end{aligned}$

Notes:

1. $15^a \equiv (16-1)^a \equiv {\color{#3D99F6} - 1} \equiv 3 \text{ (mod 4)} \quad \small \color{#3D99F6} \text{Since }a \text{ is odd.}$

2. $17^3 \equiv (20-3)^3 \equiv -27 \equiv - 7 \equiv 13 \text{ (mod 20)}$

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References:

• Euler's theorem
• Carmichael's lambda

It can be easily found that $19^{5}≡99≡-1$ mod $100$ . This means that $99^{10}≡1$ mod $100$ and that $19^{10n+k}≡19^{k}$ for positive integers $n,k$ .

Therefore, we need to find $17^{15^{...^{3^{1}}}}$ , or equivalently $7^{15^{...^{3^{1}}}}$ , mod $10$ .

As $7^{4}≡1$ mod $10$ , we need to find $15^{13^{...^{3^{1}}}}$ mod $4$ .

The above is equivalent to $(-1)^{13^{11^{...^{3^{1}}}}}$ mod $4$ . As $13^{11^{...^{3^{1}}}}$ is odd, $15^{13^{11^{...^{3^{1}}}}}≡-1≡3$ mod $4$ .

We can then deduce that $17^{15^{...^{3^{1}}}}≡7^{3}≡3$ mod $10$ .

Therefore, $19^{17^{15^{...^{3^{1}}}}}≡19^{3}≡\boxed{59}$ mod $100$ .