Straight-forward...

1. Simplifying the expression:

$\large 9^{327} \implies 3^{654}$

1. Find the value of x in:

$3^{654} \equiv x \pmod{100}$

1. Find the value of n in this; to make it easier for further calculations:

$\large 3^n \equiv 1 \pmod{100}$

Using the Euler's totient function, we can conclude that:

$\large n = \phi (100) = 40 \text{ This works becuase gcd (3, 100) = 1 }$ .

$\large \therefore \text{ We know that } 3^{40} \equiv 1 \pmod{100} \\ \large \implies 3^{640} \equiv 1 \pmod{100} \\ \large \implies 3^{640} \times 3^{14} \equiv 1 \times 69 \pmod{100} \\ \large \color{#20A900} \implies 3^{654} \equiv \boxed{ 69 } \pmod{100}$

$\begin{aligned} 9^{327} & \equiv 9^{\color{#3D99F6} 327 \bmod \lambda (100)} \text{ (mod 100)} & \small \color{#3D99F6} \text{Since }\gcd(9, 100) = 1\text{, Euler's theorem applies.} \\ & \equiv 9^{\color{#3D99F6} 327 \bmod 20} \text{ (mod 100)} & \small \color{#3D99F6} \text{Carmichael's lambda function }\lambda(100) = 20 \\ & \equiv 9^7 \text{ (mod 100)} \\ & \equiv (10-1)^7 \text{ (mod 100)} \\ & \equiv (70-1) \text{ (mod 100)} \\ & \equiv \boxed{69} \text{ (mod 100)} \end{aligned}$

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References :

• Euler's theorem
• Carmichael's lambda function

Note that :

$9^0 \equiv 01 \pmod{100}$

$9^1 \equiv 09 \pmod{100}$

$9^2 \equiv 81 \pmod{100}$

$9^3 \equiv 29 \pmod{100}$

$9^4 \equiv 61 \pmod{100}$

$9^5 \equiv 49 \pmod{100}$

$9^6 \equiv 41 \pmod{100}$

$9^7 \equiv 69 \pmod{100}$

$9^8 \equiv 21 \pmod{100}$

$9^9 \equiv 89 \pmod{100}$

$9^{10} \equiv 01 \pmod{100} \equiv 9^0 \pmod{100}$

$\space$

In addition, $327 = 32*10 + 7$ ,

so $9^{327} \equiv 9^7 \equiv 69 \pmod{100}$ .

Consider the binomial expansion of $9^{327} = (10 - 1)^{327} = \displaystyle \sum_{k=0}^{327} \dbinom{327}{k} 10^{327 - k} \times (-1)^{k}$ . All the terms in the expansion will have a factor of $100$ except the last two, so to find the last two digits we only need to consider the last two terms of the expansion, namely

$\dbinom{327}{326} 10^{1} \times (-1)^{326} + \dbinom{327}{327} 10^{0} \times (-1)^{327} = 327 \times 10 - 1 = 3269 \Longrightarrow 9^{327} \pmod{100} = \boxed{69}$ .

9^1=09(%100)

9^2=-19(%100)

9^3=29(%100)

9^4=-39(%100)

So, we can see that the series is following a pattern which is:

9^n=((-1)^((n+1)%10) (10 ((n-1)%10)+9)(%100)

Hence, we can conclude that 9^327=69