Waste less time on Facebook — follow Brilliant.
×

GCD property

Note by Shivamani Patil
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

\(c | (a+b) \rightarrow c|a \wedge c|b\) is wrong. Easily counterexample as \(a = 4, b = 5, c = 3)\).

Samuraiwarm Tsunayoshi - 3 years, 1 month ago

Log in to reply

Hint: use Bezout's identity

Samuraiwarm Tsunayoshi - 3 years, 1 month ago

Log in to reply

Another hint: \((a,b) = 1\) if and only if \(ax+by = 1\).

Samuraiwarm Tsunayoshi - 3 years, 1 month ago

Log in to reply

@Samuraiwarm Tsunayoshi Hey,@Samuraiwarm Tsunayoshi check my sol in Kartik Sharma's solution's replay box.

Shivamani Patil - 3 years, 1 month ago

Log in to reply

Let \(a = {{p}_{1}}^{{e}_{1}}{{p}_{2}}^{{e}_{2}}{{p}_{3}}^{{e}_{3}}..... {{p}_{n}}^{{e}_{n}}\)

Now, \(a = 0 mod {p}_{i}\)

We know that gcd(a,b) = 1, so none of the \({p}_{i}\) divide b

\(cn = a + b\)

\(cn mod {p}_{1} = 0 + x\)

\(cn = x mod {p}_{1}\)

And so on, till \({p}_{n}\) which tells us that \({p}_{i}\) doesn't divide c. Therefore there is no common divisor of c and a, so, \(gcd(a,c) = 1\)

Similarly we can show for b.

@shivamani patil Check out and now I think it is fine.

Kartik Sharma - 3 years, 1 month ago

Log in to reply

If gcd(a,b) then \(ax+by=1\) also \(cn=a+b\) for some \(n\) because \(c|a+b\).Therefore \(cn-b=a\),substituting \(cn-b=a\) in first equation we get \(x(cn-b)+by=1=cnx-bx+by=c(nx)+b(y-x)\) which implies gcd(c,b)=1.Similarly gcd(c,a)=1.@Kartik Sharma how is it??Your's is good.

Shivamani Patil - 3 years, 1 month ago

Log in to reply

I knew about this method too but that is already discussed by Samuraiwarm Tsunayoshi.

Kartik Sharma - 3 years, 1 month ago

Log in to reply

@Kartik Sharma But I solved it independently @Kartik Sharma

Shivamani Patil - 3 years, 1 month ago

Log in to reply

Yep, looks good.

Samuraiwarm Tsunayoshi - 3 years, 1 month ago

Log in to reply

Note that \( a|b \) implies that there exists an integer \( k \) suck that \( ka = b \), not \( kb = a \). So your assumption that \( c = n(a+b) \) is wrong. It should be \( cn = a+b \).

Siddhartha Srivastava - 3 years, 1 month ago

Log in to reply

Oh damn yes ! I knew it but just a silly mistake. Now, I am gonna edit it.

Kartik Sharma - 3 years, 1 month ago

Log in to reply

@Kartik Sharma Looking forward to your solution!

Samuraiwarm Tsunayoshi - 3 years, 1 month ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...