Solving stuff!

Relevant wiki: Euler's Theorem

Let the given number be $N$ . We need to find $N \bmod 1000$ .

$\begin{aligned} N & \equiv 7^{9729 \color{#3D99F6} \bmod \lambda (1000)} \text{ (mod 1000)} & \small \color{#3D99F6} \text{Since }\gcd(7,1000) = 1 \text{, Euler's theorem applies.} \\ & \equiv 7^{9729 \color{#3D99F6} \bmod 100} \text{ (mod 1000)} & \small \color{#3D99F6} \text{Carmichael lambda function }\lambda(1000) = 100 \\ & \equiv 7^{29} \text{ (mod 1000)} \\ & \equiv 7\cdot 49^{14} \text{ (mod 1000)} \\ & \equiv 7(50-1)^{14} \text{ (mod 1000)} \\ & \equiv 7 \left(\cdots + \frac {14\times 13}2 \cdot 50^2 - 14\cdot 50 + 1\right) \text{ (mod 1000)} \\ & \equiv 7 \left(\cdots + 500 - 700 + 1\right) \text{ (mod 1000)} \\ & \equiv 7 \left(-199 \right) \text{ (mod 1000)} \\ & \equiv -393 \text{ (mod 1000)} \\ & \equiv \boxed{607} \text{ (mod 1000)} \end{aligned}$

Since all of the answers start with "60," let's focus on the last digit.

The last digit cannot be even, since $\text{odd}\times \text{odd}\not= \text{even}$ .

Also, the last digit cannot be 5 because $(\text{non-multiple of 5})\times (\text{non-multiple of 5})\not=\text{multiple of 5}$ .

Therefore, the last digit of $7^n$ follows a 4-step repeating cycle: 7, 9, 3, 1. Since 9 will always come on an even $n$ , and 9,729 isn't even, that means the answer is 607. $~\beta_{\lceil \mid \rceil}$

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This kind of strategy is called "Process of elimination." This is very useful for those agonizing standardized tests.

The cyclicity for 7 is 4, i.e. every multiple of 4th power of 7 will have the same unit digit as $7^4$ , 1.

Thus, $7^{9728}$ will have it as 1, and $7^{9729}$ will have it as 7.