Nine teens

The last three digits of 19n19^{n}is 001001. Define the second\color{#3D99F6}{\huge{second}} smallest solution for nn, or else to prove that it is impossible.

#NumberTheory

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

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list
  • bulleted
  • list

1. numbered
2. 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 1

paragraph 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 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

If 19n19^n ends in 001 001 , then 19n−1 19^n - 1 ends in 000 000 ,i.e, 19n−1 19^n - 1 is divisible by 1000 1000 . Or simply, 19n≡1(mod1000) 19^n \equiv 1 \pmod{1000} . By Euler's theorem, 19ϕ(1000)≡1(mod1000) 19^{\phi(1000)} \equiv 1 \pmod{1000} gives us an upper bound of ϕ(1000)=400 \phi(1000) = 400 .

We can do better though. We see that 192≡32≡1(mod8) 19^2 \equiv 3^2 \equiv 1 \pmod{8} . Also, 19ϕ(125)≡19100≡1(mod125) 19^{\phi(125)} \equiv 19^{100} \equiv 1 \pmod{125} . So we get a smaller solution 100 100 .

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

Siddhartha Srivastava - 6 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.

sanyam sood - 6 years ago
×

Problem Loading...

Note Loading...

Set Loading...