# tens digit

• Nov 18th 2008, 05:56 AM
bearej50
tens digit
Determine the tens digit of 3^1999. Could you find a second solution?
• Nov 18th 2008, 10:16 AM
o_O
Finding the remainder once divided by 100 will give you the tens and ones digits. For example, $539 \equiv 39 \ (\text{mod 100})$.

So the problem becomes: $3^{1999} \equiv a \ (\text{mod 100})$

By Fermat's theorem: $3^{\varphi (100)} \equiv 1 \ (\text{mod } 100)$

But: $\varphi (100) = \varphi (2^2 \cdot 5^2) = \varphi (2^2) \varphi (5^2) = 2^1(2-1) \cdot 5^1(5-1)=40$

So: $3^{40} \equiv 1 \ (\text{mod } 100)$

Raise both sides by 50 (the closest we'll get the power to 1999):
\begin{aligned} \left(3^{40}\right)^{50} & \equiv 1^8 \ (\text{mod } 100) \\ 3^{2000} & \equiv 1 \ (\text{mod } 100) \\ & \ \ \vdots \end{aligned}
• Nov 18th 2008, 12:04 PM
bearej50
thank you so much