\begin{proof} We compute modulo $100$: \begin{align*} 9^4 &= 81^2 = 6561 \equiv 61,\\ 9^8 &\equiv 61^2 = 4021 \equiv 21,\\ 9^{10} &=9^8\cdot 9^2 \equiv 21 \cdot 81 = 1701 \equiv 1 \pmod {100}. \end{align*} Therefore $$9^{1500} = (9^{10})^{150} \equiv 1 \pmod 100.$$ The last two digits of $9^{1500}$ are $01$. \end{proof} With Sagemath: \begin{verbatim} sage: a = Mod(9, 100) sage: a^1500 1 sage: a.multiplicative_order() 10 \end{verbatim}

Homepage › Solution manuals › David S. Dummit › Abstract Algebra › Exercise 0.3.5 ($9^{1500} \mod 100$)

Exercise 0.3.5 ($9^{1500} \mod 100$)

Proof. We compute modulo $100$ :

\begin{array}{l} 9^{4} & = 8 1^{2} = 6561 \equiv 61, \\ 9^{8} & \equiv 6 1^{2} = 4021 \equiv 21, \\ 9^{10} & = 9^{8} \cdot 9^{2} \equiv 21 \cdot 81 = 1701 \equiv 1 (𝑚𝑜𝑑 100) . \end{array}

Therefore

9^{1500} = {(9^{10})}^{150} \equiv 1 (𝑚𝑜𝑑 1) 00 .

The last two digits of $9^{1500}$ are $01$ . □

With Sagemath:

sage: a = Mod(9, 100)
sage: a^1500
1
sage: a.multiplicative_order()
10