Last two digits

Number Theory Level 2

Find the last two digits of $17^{28}$ .

The answer is 41.

1 solution

Chew-Seong Cheong
Oct 5, 2016

$\begin{aligned} 17^{28} & \equiv \color{#3D99F6}{(10+7)^{28}} \text{ (mod 100)} & \small \color{#3D99F6}{\text{Only last two terms are indivisible by 100}} \\ & \equiv 28 \cdot 10 \cdot 7^{27} + 7^{28} \text{ (mod 100)} \\ & \equiv 41 \cdot 7^{28} \text{ (mod 100)} \\ & \equiv 41 \cdot 49^{14} \text{ (mod 100)} \\ & \equiv 41 \cdot \color{#3D99F6}{(50-1)^{14}} \text{ (mod 100)} & \small \color{#3D99F6}{\text{Only last term is indivisible by 100}} \\ & \equiv 41 \cdot \color{#3D99F6}{(-1)^{14}} \text{ (mod 100)} \\ & \equiv \boxed{41} \text{ (mod 100)} \end{aligned}$

In the second step and third last step, have you expanded the binomial theorem and then removed the first 27 of the 29 terms because they are a factor of $10^2$ ? Or is it a technique in modulus questions?

Other than that, brilliant solution! Upvoted

Mahdi Raza - 8 months, 2 weeks ago

The first $27$ terms are a multiple of $10^2$ , hence leave a remainder of $0$ when divided by $100$ .

Chew-Seong Cheong - 8 months, 2 weeks ago

Last two digits

The answer is 41.

1 solution

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