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

Note by Pi Han Goh
2 years, 6 months ago

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

Sudeep Salgia - 2 years, 6 months ago

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Wow that's quick!

Pi Han Goh - 2 years, 6 months ago

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Thanks.

Sudeep Salgia - 2 years, 6 months ago

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amazing !

Karan Siwach - 2 years, 6 months ago

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Cool observation Sir!

Nihar Mahajan - 2 years, 6 months ago

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Genius!

Archit Boobna - 2 years, 6 months ago

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awesome...

Karan Shekhawat - 2 years, 6 months ago

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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.

Daniel Liu - 2 years, 6 months ago

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

Pi Han Goh - 2 years, 6 months ago

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

Daniel Liu - 2 years, 6 months ago

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

Pi Han Goh - 2 years, 6 months ago

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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 \(x,y,z\).

Pi Han Goh - 2 years, 6 months ago

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Comment deleted Aug 14, 2015

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@Swapnil Das How is this related to the question?

Pi Han Goh - 2 years, 6 months ago

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

Ovi Khan - 2 years, 4 months ago

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x=3955;y=4078;z=4171

Ovi Khan - 2 years, 4 months ago

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

Saket Sharan - 2 years, 6 months ago

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

Arulx Z - 2 years, 4 months ago

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