# Find the simplest way to show that there exist positive integers $x,y,z$ that satisfy the equation $29x +30y+31z=366366$.

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Note by Pi Han Goh
4 years, 7 months ago

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Number of months with $29 , 30$ and $31$ days in $1001$ leap years.

- 4 years, 7 months ago

Wow that's quick!

- 4 years, 7 months ago

Thanks.

- 4 years, 7 months ago

awesome...

- 4 years, 7 months ago

Genius!

- 4 years, 7 months ago

Cool observation Sir!

- 4 years, 7 months ago

amazing !

- 4 years, 7 months ago

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.

- 4 years, 7 months ago

But Chicken McNugget didn't explicitly say that $y,z$ are positive.

- 4 years, 7 months ago

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

- 4 years, 7 months ago

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

- 4 years, 7 months ago

x=3955;y=4078;z=4171

- 4 years, 5 months ago

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

- 4 years, 5 months ago

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

- 4 years, 7 months ago

Won't HCF of $29$, $30$ and $31$ be $1$?

- 4 years, 5 months ago