Waste less time on Facebook — follow Brilliant.
×

An RMO problem

Prove that the ten's digit in \(3^x\) is always even, where \(x\) is a natural number.

Note by Dev Sharma
2 years, 4 months ago

No vote yet
1 vote

  Easy Math Editor

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 \( 2 \times 3 \)
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} \( \boxed{123} \)

Comments

Sort by:

Top Newest

We can use binomial theorem to prove this. 3^x = ( 1 + 2 ) ^ x

Hari Om Sharma - 2 years, 4 months ago

Log in to reply

See that 3^0 = 1, 3^1 = 3, 3^2 = 9, 3^3 = 27, and 3^4=81 conclude us with:

\(3^{4n+k}= 81^n 3^k\)

For any natural number a, b, and a digit number y, notice that:

\((20a+1)(20b+y) = 400ab+20(ay+b)+y)\)

400ab only determine hundreds and upper and y is a digit, so both has nothing to do with the tens. The term 20(ay+b) is the one that fix the tens as an even number. By using all of this, we are done.

Gian Sanjaya - 2 years, 4 months ago

Log in to reply

why did you take 20a +1 . 20b +y

Dev Sharma - 2 years, 4 months ago

Log in to reply

Applying that with a=4 and the fact from 3^0, 3^1, 3^2, 3^3, and 3^4 (all of them has even tens). Hasn't it prove the original statement?

Gian Sanjaya - 2 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...