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Note by Pi Han Goh 3 years, 3 months ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

> 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"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

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Number of months with \(29 , 30\) and \(31\) days in \(1001\) leap years.

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Wow that's quick!

Thanks.

amazing !

Cool observation Sir!

Genius!

awesome...

Take \(x=1\) and by Chicken McNugget there exists \(y,z\) such that \(30y+31z=366337\).

That's the most straightforward way that immediately solves the problem as far as I know.

But Chicken McNugget didn't explicitly say that \(y,z\) are positive.

It does; or else it would just degenerate to Bezout's Identity.

@Daniel Liu – OH wait it does! Silly me! Thanks! I've found the second simplest solution. Yay!

Comment deleted Apr 15, 2015

Note that I'm looking for positive integers \(x,y,z\) not integers \(x,y,z\).

Comment deleted Aug 14, 2015

@Swapnil Das – How is this related to the question?

x=3955;y=4078;z=4171;

x=3955;y=4078;z=4171

see if HCF of 29,30,31 divides 366366 completely

Won't HCF of \(29\), \(30\) and \(31\) be \(1\)?

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestNumber of months with \(29 , 30\) and \(31\) days in \(1001\) leap years.

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Wow that's quick!

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Thanks.

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amazing !

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Cool observation Sir!

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Genius!

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awesome...

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Take \(x=1\) and by Chicken McNugget there exists \(y,z\) such that \(30y+31z=366337\).

That's the most straightforward way that immediately solves the problem as far as I know.

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But Chicken McNugget didn't explicitly say that \(y,z\) are positive.

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It does; or else it would just degenerate to Bezout's Identity.

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Comment deleted Apr 15, 2015

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Note that I'm looking for positive integers \(x,y,z\) not integers \(x,y,z\).

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Comment deleted Aug 14, 2015

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x=3955;y=4078;z=4171;

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x=3955;y=4078;z=4171

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see if HCF of 29,30,31 divides 366366 completely

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Won't HCF of \(29\), \(30\) and \(31\) be \(1\)?

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