×

# very interesting problem!

Prove that for every number consisting of $$3^{n}$$ equal digits is divisible by $$3^{n}$$.

Note by Kiran Patel
4 years, 7 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:

I'll prove this by induction. For $$n=1$$ it's obvious, as the digit sum is divisible by $$3$$. Now, let's asume the statement is true for some number with $$3^k$$ digits. The corresponding $$3^{k+1}$$ number is formed by multiplying the $$3^k$$ digit number by $$100...00100...001$$: a 1 followed by $$3^k-1$$ zeros, followed by a 1, followed by $$3^k-1$$ zeros, followed by a 1. This number has a digit sum of 3, so it is divisible by 3. Therefore, the new $$3^{k+1}$$ digit number is divisible by $$3^{k+1}$$. Thus, the statement is correct for all natural numbers $$n$$.

- 4 years, 7 months ago