Determine the last three digits of the number ## 7^{999} ##

[Math100]

Homework Statement

Determine the last three digits of the number $7^{999}$.
[Hint: $7^{4n}\equiv (1+400)^{n}\equiv 1+400n\pmod {1000}$.]

Relevant Equations

None.

Observe that $7^{4n}\equiv (7^{4})^{n}\equiv (401)^{n}\equiv (1+400)^{n}\equiv 1+400n\pmod {1000}$.
Thus
\begin{align*}
&7^{999}\equiv [(7^{4})^{249}\cdot 7^{3}]\pmod {1000}\\
&\equiv [(1+400\cdot 249)\cdot 7^{3}]\pmod {1000}\\
&\equiv [(1+99600)\cdot 7^{3}]\pmod {1000}\\
&\equiv [(1+600)\cdot 7^{3}]\pmod {1000}\\
&\equiv (601\cdot 343)\pmod {1000}\\
&\equiv 206143\pmod {1000}\\
&\equiv 143\pmod {1000}.\\
\end{align*}
Therefore, the last three digits of the number $7^{999}$ are $143$.

 

[fresh_42]

Yes, right. Just one question: Why is $(1+400)^n\equiv 1+400 n \pmod{1000}?$

 

[Math100]

Yes, right. Just one question: Why is $(1+400)^n\equiv 1+400 n \pmod{1000}?$

I was thinking about it but I don't know. Can you tell me why?

 

[fresh_42]

I was thinking about it but I don't know. Can you tell me why?

Yes, it is the binomial formula.
\begin{align*}
(x+y)^n &=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k\\&=x^n+n\cdot x^{n-1}y+ \dfrac{n(n-1)}{2!}x^{n-2}y^2+\ldots+\dfrac{n(n-1)}{2!}x^{2}y^{n-2}+n\cdot xy^{n-1} +y^n\\[10pt]
(1+400)^n&=1+n\cdot 1^{n-1}\cdot 400+\dfrac{n(n-1)}{2!}\cdot x^{n-2}\cdot 400^2+\ldots + n\cdot 1\cdot 400^{n-1}+400^n
\end{align*}
Now, look at the zeros. After $1+400n$ are always at least four of them in each term. So they do not contribute anything modulo $1000.$

The coefficients $1, n, \dfrac{n(n-1)}{2!},\dfrac{n(n-1)(n-2)}{3!},\ldots$ are the numbers in Pascal's triangle.