Waste less time on Facebook — follow Brilliant.
×

Divisibility Practice

Prove that \((2014!)!\) is divisible by \( ((2014 \cdot 2013)!)^{2012!}\)

Note by Pratik Shastri
3 years, 1 month 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

Let us have \(2014!\) balls. Let us divide them in \(2012!\) groups, each group having \(2013.2014\) balls.

Let's assume that balls of each group are of different colour, but all balls of the same group are the of the same colour.

Number of ways of doing so is, \[\dfrac{(2014!)!}{((2014.2013)!)^{2012!}}\] Number of ways of doing something is always a positive integer. Hence Proved. \(\boxed{.}\)

Satvik Golechha - 3 years, 1 month ago

Log in to reply

Using Combinatorics for a Number Theory Problem is simply amazing!

Marc Vince Casimiro - 3 years, 1 month ago

Log in to reply

Excellent! +1

Pratik Shastri - 3 years, 1 month ago

Log in to reply

@Satvik Golechha this is great method see solution of @John Ashley Capellan here

Shivamani Patil - 3 years, 1 month ago

Log in to reply

superb way to do this......

Rajat Bhagat - 3 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...