Waste less time on Facebook — follow Brilliant.
×

Strings of Zeros

81 is a square number. Find the minimum number of zeros needed to be inserted between 8 and 1 such that it makes another perfect square, or prove that it is impossible.

Note by Bryan Lee Shi Yang
2 years, 9 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

Not possible.

Suppose there does exist some \( n \) where \( n^2 = 800...001 \). Then since \( 800..001 \) is odd, \( n \) must. Therefore, \( n = 2k + 1 \). Then,

\( (2k + 1)^2 = 800...001 \)

\( 4k^2 + 4k + 1 = 800..001 \)

\( 4k(k+1) = 800...001 \)

\( k(k+1) = 200...000 \)

\( k(k+1) = 2 * 2^t*5^t \) for some \( t > 1 \)

\( k(k+1) = 2^{t+1}*5^t \)

Now, \( \gcd(k,k+1) = 1 \).Also, since \( 5^t > 2^{t+1} \), we must have \( k+1 = 5^t \) and \( k = 2^{t+1} \). But, we can easily prove that \( 5^t > 2^{t+1} + 1 \) for \( t > 1 \). Therefore, we have a contradiction, and there exist no integer \( n \) for which \( n^2 = 800...001 \)

Siddhartha Srivastava - 2 years, 9 months ago

Log in to reply

Nice solution :)

Calvin Lin Staff - 2 years, 8 months ago

Log in to reply

Nice solution

Azhaghu Roopesh M - 2 years, 9 months ago

Log in to reply

Comment deleted Mar 13, 2015

Log in to reply

@Bryan Lee Shi Yang Sorry , but did you mean that you salute me ? It's actually Siddhartha who posted the solution , so I guess you should be saluting him :)

Azhaghu Roopesh M - 2 years, 9 months ago

Log in to reply

Comment deleted Mar 13, 2015

Log in to reply

Comment deleted Mar 13, 2015

Log in to reply

@Bryan Lee Shi Yang I'm sorry but I'm not getting what you are trying to say

Azhaghu Roopesh M - 2 years, 9 months ago

Log in to reply

Comment deleted Mar 13, 2015

Log in to reply

@Bryan Lee Shi Yang Ok .

Azhaghu Roopesh M - 2 years, 9 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...