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Note by Pi Han Goh 1 year, 6 months ago

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Number of months with \(29 , 30\) and \(31\) days in \(1001\) leap years. – Sudeep Salgia · 1 year, 6 months ago

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@Sudeep Salgia – Wow that's quick! – Pi Han Goh · 1 year, 6 months ago

@Pi Han Goh – Thanks. – Sudeep Salgia · 1 year, 6 months ago

@Sudeep Salgia – amazing ! – Karan Siwach · 1 year, 6 months ago

@Sudeep Salgia – Cool observation Sir! – Nihar Mahajan · 1 year, 6 months ago

@Sudeep Salgia – Genius! – Archit Boobna · 1 year, 6 months ago

@Sudeep Salgia – awesome... – Karan Shekhawat · 1 year, 6 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. – Daniel Liu · 1 year, 6 months ago

@Daniel Liu – But Chicken McNugget didn't explicitly say that \(y,z\) are positive. – Pi Han Goh · 1 year, 6 months ago

@Pi Han Goh – It does; or else it would just degenerate to Bezout's Identity. – Daniel Liu · 1 year, 6 months ago

@Daniel Liu – OH wait it does! Silly me! Thanks! I've found the second simplest solution. Yay! – Pi Han Goh · 1 year, 6 months ago

@Karthik Venkata – Note that I'm looking for positive integers \(x,y,z\) not integers \(x,y,z\). – Pi Han Goh · 1 year, 6 months ago

@Swapnil Das – How is this related to the question? – Pi Han Goh · 1 year, 6 months ago

x=3955;y=4078;z=4171; – Ovi Khan · 1 year, 5 months ago

x=3955;y=4078;z=4171 – Ovi Khan · 1 year, 5 months ago

see if HCF of 29,30,31 divides 366366 completely – Saket Sharan · 1 year, 6 months ago

@Saket Sharan – Won't HCF of \(29\), \(30\) and \(31\) be \(1\)? – Arulx Z · 1 year, 5 months ago

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## Comments

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TopNewestNumber of months with \(29 , 30\) and \(31\) days in \(1001\) leap years. – Sudeep Salgia · 1 year, 6 months ago

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– Pi Han Goh · 1 year, 6 months ago

Wow that's quick!Log in to reply

– Sudeep Salgia · 1 year, 6 months ago

Thanks.Log in to reply

– Karan Siwach · 1 year, 6 months ago

amazing !Log in to reply

– Nihar Mahajan · 1 year, 6 months ago

Cool observation Sir!Log in to reply

– Archit Boobna · 1 year, 6 months ago

Genius!Log in to reply

– Karan Shekhawat · 1 year, 6 months ago

awesome...Log in to reply

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 · 1 year, 6 months ago

Log in to reply

– Pi Han Goh · 1 year, 6 months ago

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

– Daniel Liu · 1 year, 6 months ago

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

– Pi Han Goh · 1 year, 6 months ago

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

Log in to reply

– Pi Han Goh · 1 year, 6 months ago

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

Log in to reply

– Pi Han Goh · 1 year, 6 months ago

How is this related to the question?Log in to reply

x=3955;y=4078;z=4171; – Ovi Khan · 1 year, 5 months ago

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x=3955;y=4078;z=4171 – Ovi Khan · 1 year, 5 months ago

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see if HCF of 29,30,31 divides 366366 completely – Saket Sharan · 1 year, 6 months ago

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– Arulx Z · 1 year, 5 months ago

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