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# 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
2 years, 1 month ago

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Number of months with $$29 , 30$$ and $$31$$ days in $$1001$$ leap years. · 2 years, 1 month ago

Wow that's quick! · 2 years, 1 month ago

Thanks. · 2 years, 1 month ago

amazing ! · 2 years, 1 month ago

Cool observation Sir! · 2 years, 1 month ago

Genius! · 2 years, 1 month ago

awesome... · 2 years, 1 month 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. · 2 years, 1 month ago

But Chicken McNugget didn't explicitly say that $$y,z$$ are positive. · 2 years, 1 month ago

It does; or else it would just degenerate to Bezout's Identity. · 2 years, 1 month ago

OH wait it does! Silly me! Thanks! I've found the second simplest solution. Yay! · 2 years, 1 month ago

Comment deleted Apr 15, 2015

Note that I'm looking for positive integers $$x,y,z$$ not integers $$x,y,z$$. · 2 years, 1 month ago

Comment deleted Aug 14, 2015

How is this related to the question? · 2 years, 1 month ago

x=3955;y=4078;z=4171; · 2 years ago

x=3955;y=4078;z=4171 · 2 years ago

see if HCF of 29,30,31 divides 366366 completely · 2 years, 1 month ago

Won't HCF of $$29$$, $$30$$ and $$31$$ be $$1$$? · 2 years ago