Last two digits of this big number

Number Theory Level 2

$\large 27^{41}$

What are the last two digits of the number above?

The answer is 27.

1 solution

Chew-Seong Cheong
Aug 10, 2020

$\begin{aligned} 27^{41} & \equiv 27^{41 \bmod \blue{\phi(100)}} \text{ (mod 100)} & \small \blue{\text{Since }\gcd(27,100) = 1 \text{, Euler's theorem applies.}} \\ & \equiv 27^{41 \bmod \blue{40}} \text{ (mod 100)} & \small \blue{\text{Euler's totient function }\phi(100) = 40} \\ & \equiv 27^1 \text{ (mod 100)} \\ & \equiv \boxed{27} \text{ (mod 100)} \end{aligned}$

References:

Euler's theorem
Euler's totient function

I was doing the course and I actually don't quite get why we can just put (mod 40) next to the exponent. Can you help explain? Thank you.

Barry Leung - 10 months ago

Barry, please read up references I have provided.

Chew-Seong Cheong - 10 months ago

The way I solved it is to rewrite $27^{41}$ as $27^{40} \times 27^{1}$ . Since $27^{40} ≡ 1$ (mod 100). So 27 is the last two digits.

Barry Leung - 10 months ago

@Barry Leung – It is the same thing I do.

Chew-Seong Cheong - 10 months ago

@Barry Leung – Because $27^{40n} \equiv 27^0 \equiv 1 \text{ (mod 100)}$ only the remainder $41 \bmod 40 =1$ counts.

Chew-Seong Cheong - 10 months ago

a mod b= a^(mod phi(b)) mod b

assume gcd(a,b) = 1

In this case, a = 27^41

Pi Han Goh - 10 months ago

Last two digits of this big number

The answer is 27.

1 solution

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