# Nine teens

The last three digits of $$19^{n}$$is $$001$$. Define the $\color{BLUE}{\huge{second}}$ smallest solution for $$n$$, or else to prove that it is impossible.

Note by Bryan Lee Shi Yang
3 years, 3 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

If $$19^n$$ ends in $$001$$, then $$19^n - 1$$ ends in $$000$$,i.e, $$19^n - 1$$ is divisible by $$1000$$. Or simply, $$19^n \equiv 1 \pmod{1000}$$. By Euler's theorem, $$19^{\phi(1000)} \equiv 1 \pmod{1000}$$ gives us an upper bound of $$\phi(1000) = 400$$.

We can do better though. We see that $$19^2 \equiv 3^2 \equiv 1 \pmod{8}$$. Also, $$19^{\phi(125)} \equiv 19^{100} \equiv 1 \pmod{125}$$. So we get a smaller solution $$100$$.

At this point, I can't find a better solution. Wolfram Alpha gives the smallest solution as $$n = 50k$$ where $$k$$ is an integer, so the answer to your question is $$100$$.

- 3 years, 3 months ago

19= 20 - 1 . when we do any even power of (20-1) it will end up with last digit 1 succeeded by 3 zeroes if n is a multiple of 50 (binomial expansion) non multiple may give 1 or 2 zeroes before 1 but multiple of 50 gives 3 zeroes . so ans is 100.

- 3 years, 1 month ago