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How do I convince my grandmother (who really hates mathematics) that there exist three positive integers $x,y,z$ that satisfy the equation $28x+30y+31z=3650$?

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    $\begingroup$ While it is an interesting question (and I like Gerhard's answer), I wouldn't say this is a puzzle. $\endgroup$ – BmyGuest Feb 1 '15 at 14:49
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    $\begingroup$ @BmyGuest, having seen the answer I'd say it is a puzzle. If it were just a plain vanilla maths question then it wouldn't be. $\endgroup$ – A E Feb 1 '15 at 19:26
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    $\begingroup$ You could just tell her the values of x, y, and z. If that doesn't convince her, nothing else will. $\endgroup$ – Set Big O Feb 1 '15 at 21:33
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    $\begingroup$ If she hates maths that much, she probably doesn't care enough to need convincing. $\endgroup$ – Pharap Feb 2 '15 at 0:09
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    $\begingroup$ I'm voting to close this question as off-topic because it really is straightforward math. $\endgroup$ – Set Big O Feb 2 '15 at 15:34
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You tell your Grandmother:

Take ten ordinary years. They contain exactly 3650 days, and these days are split among several months with 28, 30, and 31 days.

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    $\begingroup$ Ten ordinary years do not contain exactly 3650 days. They contain either 3652 or 3653, depending on which ten consecutive years you picked, due to leap years. The actual length of a solar year is not exactly 365 days, it is approximately 365.24 days. $\endgroup$ – Matthew Najmon Feb 2 '15 at 22:08
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    $\begingroup$ @MatthewNajmon I think the ordinary part means non-leap year... $\endgroup$ – QuyNguyen2013 Feb 2 '15 at 22:10
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    $\begingroup$ @MatthewNajmon They said nothing of consecutive years. $\endgroup$ – Vality Feb 3 '15 at 6:03
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    $\begingroup$ @QuyNguyen2013 Well, then picking out 10 non-consecutive years, your chances of at least one leap-year are high unless you specifically avoid them. If you are intentionally sorting against specific types of year, then the result doesn't really constitute an "ordinary" selection of years. Leap years are not some rare occurrence that can be discounted as outliers. They make up a very significant part (roughly a quarter) of the total population of years, more than enough for them to be relevant in defining what characteristics are "normal" or "ordinary" for a year to have. $\endgroup$ – Matthew Najmon Feb 3 '15 at 15:57
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    $\begingroup$ The actual exact time taken for a solar year is utterly irrelevant. This is a Maths problem, not a physical one. In mathematics, we deal with abstractions. The only property of the year we are exploiting is that it is well known even by grandmothers that 365 is a liners combination of 28, 30 and 31. The only reason we are giving the example of the year is that it's easier to understand this particular number theoretic property of 365. The actual amount of time in a solar year doesn't interest us. $\endgroup$ – user230452 Mar 20 '16 at 11:40
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Your grandmother can probably

answer the question for 365 (days),

and she can probably

multiply by 10.

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It's obviously the number of days in 10 non-leap years. and x, y, and z are 10, 40, and 70.

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